A Solution Manual For A First Course In Probability

A Solution Manual for:
A First Course In Probability
by Sheldon M. Ross.
John L. Weatherwax∗
February 7, 2012
Introduction
Here you’ll find some notes that I wrote up as I worked through this excellent book. I’ve
worked hard to make these notes as good as I can, but I have no illusions that they are perfect.
If you feel that that there is a better way to accomplish or explain an exercise or derivation
presented in these notes; or that one or more of the explanations is unclear, incomplete,
or misleading, please tell me. If you find an error of any kind – technical, grammatical,
typographical, whatever – please tell me that, too. I’ll gladly add to the acknowledgments
in later printings the name of the first person to bring each problem to my attention.
Acknowledgements
Special thanks to (most recent comments are listed first): Mark Chamness, Dale Peterson,
Doug Edmunds, Marlene Miller, John Williams (several contributions to chapter 4), Timothy
Alsobrooks, Konstantinos Stouras, William Howell, Robert Futyma, Waldo Arriagada, Atul
Narang, Andrew Jones, Vincent Frost, and Gerardo Robert for helping improve these notes
and solutions. It should be noted that Marlene Miller made several helpful suggestions
on most of the material in Chapter 3. Her algebraic use of event “set” notation to solve
probability problems has opened my eyes to this powerful technique. It is a tool that I wish
to become more proficient with.
All comments (no matter how small) are much appreciated. In fact, if you find these notes
useful I would appreciate a contribution in the form of a solution to a problem that is not yet
∗
wax@alum.mit.edu
1

worked in these notes. Sort of a “take a penny, leave a penny” type of approach. Remember:
pay it forward.
Miscellaneous Problems
The Crazy Passenger Problem
The following is known as the “crazy passenger problem” and is stated as follows. A line of
100 airline passengers is waiting to board the plane. They each hold a ticket to one of the 100
seats on that flight. (For convenience, let’s say that the k-th passenger in line has a ticket
for the seat number k.) Unfortunately, the first person in line is crazy, and will ignore the
seat number on their ticket, picking a random seat to occupy. All the other passengers are
quite normal, and will go to their proper seat unless it is already occupied. If it is occupied,
they will then find a free seat to sit in, at random. What is the probability that the last
(100th) person to board the plane will sit in their proper seat (#100)?
If one tries to solve this problem with conditional probability it becomes very difficult. We
begin by considering the following cases if the first passenger sits in seat number 1, then all
the remaining passengers will be in their correct seats and certainly the #100’th will also.
If he sits in the last seat #100, then certainly the last passenger cannot sit there (in fact he
will end up in seat #1). If he sits in any of the 98 seats between seats #1 and #100, say seat
k, then all the passengers with seat numbers 2, 3, . . . , k −1 will have empty seats and be able
to sit in their respective seats. When the passenger with seat number k enters he will have
as possible seating choices seat #1, one of the seats k + 1, k + 2, . . . , 99, or seat #100. Thus
the options available to this passenger are the same options available to the first passenger.
That is if he sits in seat #1 the remaining passengers with seat labels k+1, k+2, . . ., 100 can
sit in their assigned seats and passenger #100 can sit in his seat, or he can sit in seat #100
in which case the passenger #100 is blocked, or finally he can sit in one of the seats between
seat k and seat #99. The only difference is that this k-th passenger has fewer choices for
the “middle” seats. This k passenger effectively becomes a new “crazy” passenger.
From this argument we begin to see a recursive structure. To fully specify this recursive
structure lets generalize this problem a bit an assume that there are N total seats (rather
than just 100). Thus at each stage of placing a k-th crazy passenger we can choose from
• seat #1 and the last or N-th passenger will then be able to sit in their assigned seat,
since all intermediate passenger’s seats are unoccupied.
• seat # N and the last or N-th passenger will be unable to sit in their assigned seat.
• any seat before the N-th and after the k-th. Where the k-th passenger’s seat is taken
by a crazy passenger from the previous step. In this case there are N −1−(k+1)+1 =
N − k − 1 “middle” seat choices.

If we let p(n, 1) be the probability that given one crazy passenger and n total seats to select
from that the last passenger sits in his seat. From the argument above we have a recursive
structure give by
p(N, 1) =
1
N
(1) +
1
N
(0) +
1
N
N−1
X
k=2
p(N − k, 1)
=
1
N
+
1
N
N−1
X
k=2
p(N − k, 1) .
where the first term is where the first passenger picks the first seat (where the N will sit
correctly with probability one), the second term is when the first passenger sits in the N-th
seat (where the N will sit correctly with probability zero), and the remaining terms represent
the first passenger sitting at position k, which will then require repeating this problem with
the k-th passenger choosing among N − k + 1 seats.
To solve this recursion relation we consider some special cases and then apply the principle
of mathematical induction to prove it. Lets take N = 2. Then there are only two possible
arrangements of passengers (1, 2) and (2, 1) of which one (the first) corresponds to the second
passenger sitting in his assigned seat. This gives
p(2, 1) =
1
2
.
If N = 3, then from the 3! = 6 possible choices for seating arrangements
(1, 2, 3) (1, 3, 2) (2, 3, 1) (2, 1, 3) (3, 1, 2) (3, 2, 1)
Only
(1, 2, 3) (2, 1, 3) (3, 2, 1)
correspond to admissible seating arrangements for this problem so we see that
p(3, 1) =
3
6
=
1
2
.
If we hypothesis that p(N, 1) = 1
2
for all N, placing this assumption into the recursive
formulation above gives
p(N, 1) =
1
N
+
1
N
N−1
X
k=2
1
2
=
1
2
.
Verifying that indeed this constant value satisfies our recursion relationship.

Chapter 1 (Combinatorial Analysis)
Chapter 1: Problems
Problem 1 (counting license plates)
Part (a): In each of the first two places we can put any of the 26 letters giving 262
possible
letter combinations for the first two characters. Since the five other characters in the license
plate must be numbers, we have 105
possible five digit letters their specification giving a
total of
262
· 105
= 67600000 ,
total license plates.
Part (b): If we can’t repeat a letter or a number in the specification of a license plate then
the number of license plates becomes
26 · 25 · 10 · 9 · 8 · 7 · 6 = 19656000 ,
total license plates.
Problem 2 (counting die rolls)
We have six possible outcomes for each of the die rolls giving 64
= 1296 possible total
outcomes for all four rolls.
Problem 3 (assigning workers to jobs)
Since each job is different and each worker is unique we have 20! different pairings.
Problem 4 (creating a band)
If each boy can play each instrument we can have 4! = 24 ordering. If Jay and Jack can
play only two instruments then we will assign the instruments they play first with 2! possible
orderings. The other two boys can be assigned the remaining instruments in 2! ways and
thus we have
2! · 2! = 4 ,
possible unique band assignments.

Problem 5 (counting telephone area codes)
In the first specification of this problem we can have 9 − 2 + 1 = 8 possible choices for the
first digit in an area code. For the second digit there are two possible choices. For the third
digit there are 9 possible choices. So in total we have
8 · 2 · 9 = 144 ,
possible area codes. In the second specification of this problem, if we must start our area
codes with the digit “four” we will only have 2 · 9 = 18 area codes.
Problem 6 (counting kittens)
The traveler would meet 74
= 2401 kittens.
Problem 7 (arranging boys and girls)
Part (a): Since we assume that each person is unique, the total number of ordering is given
by 6! = 720.
Part (b): We have 3! orderings of each group of the three boys and girls. Since we can put
these groups of boys and girls in 2! different ways (either the boys first or the girls first) we
have
(2!) · (3!) · (3!) = 2 · 6 · 6 = 72 ,
possible orderings.
Part (c): If the boys must sit together we have 3! = 6 ways to arrange the block of boys.
This block of boys can be placed either at the ends or in between any of the individual 3!
orderings of the girls. This gives four locations where our block of boys can be placed we
have
4 · (3!) · (3!) = 144 ,
possible orderings.
Part (d): The only way that no two people of the same sex can sit together is to have the
two groups interleaved. Now there are 3! ways to arrange each group of girls and boys, and
to interleave we have two different choices for interleaving. For example with three boys and
girls we could have
g1b1g2b2g3b3 vs. b1g1b2g2b3g3 ,
thus we have
2 · 3! · 3! = 2 · 62
= 72 ,
possible arrangements.

Problem 8 (counting arrangements of letters)
Part (a): Since “Fluke” has five unique letters we have 5! = 120 possible arrangements.
Part (b): Since “Propose” has seven letters of which four (the “o”’s and the “p”’s) repeat
we have
7!
2! · 2!
= 1260 ,
arrangements.
Part (c): Now “Mississippi” has eleven characters with the “i” repeated four times, the “s”
repeated four times and the “p” repeated two times, so we have
11!
4! · 4! · 2!
= 34650 ,
possible rearranges.
Part (d): “Arrange” has seven characters with a double “a” and a double “r” so it has
7!
2! · 2!
= 1260 ,
different arrangements.
Problem 9 (counting colored blocks)
Assuming each block is unique we have 12! arrangements, but since the six black and the
four red blocks are not distinguishable we have
12!
6! · 4!
= 27720 ,
Problem 10 (seating people in a row)
Part (a): We have 8! = 40320 possible seating arrangements.
Part (b): We have 6! ways to place the people (not including A and B). We have 2! ways
to order A and B. Once the pair of A and B is determined, they can be placed in between
any ordering of the other six. For example, any of the “x”’s in the expression below could
be replaced with the A B pair
x P1 x P2 x P3 x P4 x P5x P6 x .

Giving seven possible locations for the A,B pair. Thus the total number of orderings is given
by
2! · 6! · 7 = 10080 .
Part (c): To place the men and women according to the given rules, the men and women
must be interleaved. We have 4! ways to arrange the men and 4! ways to arrange the
women. We can start our sequence of eight people with a woman or a man (giving two
possible choices). We thus have
2 · 4! · 4! = 1152 ,
Part (d): Since the five men must sit next to each other their ordering can be specified in
5! = 120 ways. This block of men can be placed in between any of the three women, or at
the end of the block of women, who can be ordered in 3! ways. Since there are four positions
we can place the block of men we have
5! · 4 · 3! = 2880 ,
Part (e): The four couple have 2! orderings within each pair, and then 4! orderings of the
pairs giving a total of
(2!)4
· 4! = 384 ,
total orderings.
Problem 11 (counting arrangements of books)
Part (a): We have (3 + 2 + 1)! = 6! = 720 arrangements.
Part (b): The mathematics books can be arranged in 2! ways and the novels in 3! ways.
Then the block ordering of mathematics, novels, and chemistry books can be arranged in 3!
ways resulting in
(3!) · (2!) · (3!) = 72 ,
Part (c): The number of ways to arrange the novels is given by 3! = 6 and the other three
books can be arranged in 3! ways with the blocks of novels in any of the four positions in
between giving
4 · (3!) · (3!) = 144 ,

Problem 12 (counting awards)
Part (a): We have 30 students to choose from for the first award, and 30 students to choose
from for the second award, etc. So the total number of different outcomes is given by
305
= 24300000
Part (b): We have 30 students to choose from for the first award, 29 students to choose
from for the second award, etc. So the total number of different outcomes is given by
30 · 29 · 28 · 27 · 26 = 17100720
Problem 13 (counting handshakes)
With 20 people the number of pairs is given by

20
2

= 190 .
Problem 14 (counting poker hands)
A deck of cards has four suits with thirteen cards each giving in total 52 cards. From these
52 cards we need to select five to form a poker hand thus we have

52
5

= 2598960 ,
unique poker hands.
Problem 15 (pairings in dancing)
We must first choose five women from ten in

10
5

possible ways, and five men from 12
in

12
5

ways. Once these groups are chosen then we have 5! pairings of the men and
women. Thus in total we will have

10
5

12
5

5! = 252 · 792 · 120 = 23950080 ,
possible pairings.

Problem 16 (forced selling of books)
Part (a): We have to select a subject from three choices. If we choose math we have

6
2

= 15 choices of books to sell. If we choose science we have

7
2

= 21 choices of
books to sell. If we choose economics we have

4
2

= 6 choices of books to sell. Since each
choice is mutually exclusive in total we have 15 + 21 + 6 = 42, possible choices.
Part (b): We must pick two subjects from

3
2

= 3 choices. If we denote the letter “M”
for the choice math the letter “S” for the choice science, and the letter “E” for the choice
economics then the three choices are
(M, S) (M, E) (S, E) .
For each of the choices above we have 6 · 7 + 6 · 4 + 7 · 4 = 94 total choices.
Problem 17 (distributing gifts)
We can choose seven children to give gifts to in

10
7

ways. Once we have chosen the
seven children, the gifts can be distributed in 7! ways. This gives a total of

10
7

· 7! = 604800 ,
possible gift distributions.
Problem 18 (selecting political parties)
We can choose two Republicans from the five total in

5
2

ways, we can choose two
Democrats from the six in

6
2

ways, and finally we can choose three Independents from
the four in

4
3

ways. In total, we will have

5
2

·

6
2

·

4
3

= 600 ,
different committees.

Problem 19 (counting committee’s with constraints)
Part (a): We select three men from six in

6
3

, but since two men won’t serve together
we need to compute the number of these pairings of three men that have the two that won’t
serve together. The number of committees we can form (with these two together) is given
by
2
2

·

4
1

= 4 .
So we have
6
3

− 4 = 16 ,
possible groups of three men. Since we can choose

8
3

= 56 different groups of women,
we have in total 16 · 56 = 896 possible committees.
Part (b): If two women refuse to serve together, then we will have

2
2

·

6
1

groups
with these two women in them from the

8
3

ways to draw three women from eight. Thus
we have
8
3

−

2
2

·

6
1

= 56 − 6 = 50 ,
possible groupings of woman. We can select three men from six in

6
3

= 20 ways. In
total then we have 50 · 20 = 1000 committees.
Part (c): We have

8
3

·

6
3

total committees, and

1
1

·

7
2

·

1
1

·

5
2

= 210 ,
committees containing the man and women who refuse to serve together. So we have

8
3

·

6
3

−

1
1

·

7
2

·

1
1

·

5
2

= 1120 − 210 = 910 ,
total committees.

Problem 20 (counting the number of possible parties)
Part (a): There are a total of

8
5

possible groups of friends that could attend (assuming
no feuds). We have

2
2

·

6
3

sets with our two feuding friends in them, giving

8
5

−

2
2

·

6
3

= 36
possible groups of friends
Part (b): If two fiends must attend together we have that

2
2

6
3

if the do attend
the party together and

6
5

if they don’t attend at all, giving a total of

2
2

6
3

+

6
5

= 26 .
Problem 21 (number of paths on a grid)
From the hint given that we must take four steps to the right and three steps up, we can
think of any possible path as an arraignment of the letters ”U” for up and “R” for right.
For example the string
U U U R R R R ,
would first step up three times and then right four times. Thus our problem becomes one of
counting the number of unique arrangements of three “U”’s and four “R”’s, which is given
by
7!
4! · 3!
= 35 .
Problem 22 (paths on a grid through a specific point)
One can think of the problem of going through a specific point (say P) as counting the
number of paths from the start A to P and then counting the number of paths from P to
the end B. To go from A to P (where P occupies the (2, 2) position in our grid) we are
looking for the number of possible unique arrangements of two “U”’s and two “R”’s, which
is given by
4!
2! · 2!
= 6 ,
possible paths. The number of paths from the point P to the point B is equivalent to the
number of different arrangements of two “R”’s and one “U” which is given by
3!
2! · 1!
= 3 .

From the basic principle of counting then we have 6 · 3 = 18 total paths.
Problem 23 (assignments to beds)
Assuming that twins sleeping in different bed in the same room counts as a different arraign-
ment, we have (2!) · (2!) · (2!) = 8 possible assignments of each set of twins to a room. Since
there are 3! ways to assign the pair of twins to individual rooms we have 6 · 8 = 48 possible
assignments.
Problem 24 (practice with the binomial expansion)
This is given by
(3x2
+ y)5
=
5
X
k=0

5
k

(3x2
)k
y5−k
.
Problem 25 (bridge hands)
We have 52! unique permutations, but since the different arrangements of cards within a
given hand do not matter we have
52!
(13!)4
,
possible bridge hands.
Problem 26 (practice with the multinomial expansion)
This is given by the multinomial expansion
(x1 + 2x2 + 3x3)4
=
X
n1+n2+n3=4

4
n1 , n2 , n3

xn1
1 (2x2)n2
(3x3)n3
The number of terms in the above summation is given by

4 + 3 − 1
3 − 1

=

6
2

=
6 · 5
2
= 15 .

Problem 27 (counting committees)
This is given by the multinomial coefficient

12
3 , 4 , 5

= 27720
Problem 28 (divisions of teachers)
If we decide to send n1 teachers to school one and n2 teachers to school two, etc. then the
total number of unique assignments of (n1, n2, n3, n4) number of teachers to the four schools
is given by
8
n1 , n2 , n3 , n4

.
Since we want the total number of divisions, we must sum this result for all possible combi-
nations of ni, or
X
n1+n2+n3+n4=8

8
n1 , n2 , n3 , n4

= (1 + 1 + 1 + 1)8
= 65536 ,
possible divisions.
If each school must receive two in each school, then we are looking for

8
2 , 2 , 2 , 2

=
8!
(2!)4
= 2520 ,
orderings.
Problem 29 (dividing weight lifters)
We have 10! possible permutations of all weight lifters but the permutations of individual
countries (contained within this number) are irrelevant. Thus we can have
10!
3! · 4! · 2! · 1!
=

10
3 , 4 , 2 , 1

= 12600 ,
possible divisions. If the united states has one competitor in the top three and two in
the bottom three. We have

3
1

possible positions for the US member in the first three
positions and

3
2

possible positions for the two US members in the bottom three positions,
giving a total of
3
1

3
2

= 3 · 3 = 9 ,

combinations of US members in the positions specified. We also have to place the other coun-
tries participants in the remaining 10 − 3 = 7 positions. This can be done in

7
4 , 2 , 1

=
7!
4!·2!·1!
= 105 ways. So in total then we have 9 · 105 = 945 ways to position the participants.
Problem 30 (seating delegates in a row)
If the French and English delegates are to be seated next to each other, they can be can be
placed in 2! ways. Then this pair constitutes a new “object” which we can place anywhere
among the remaining eight people, i.e. there are 9! arrangements of the eight remaining
people and the French and English pair. Thus we have 2·9! = 725760 possible combinations.
Since in some of these the Russian and US delegates are next to each other, this number
over counts the true number we are looking for by 2 · 28! = 161280 (the first two is for the
number of arrangements of the French and English pair). Combining these two criterion we
have
2 · (9!) − 4 · (8!) = 564480 .
Problem 31 (distributing blackboards)
Let xi be the number of black boards given to school i, where i = 1, 2, 3, 4. Then we must
have
P
i xi = 8, with xi ≥ 0. The number of solutions to an equation like this is given by

8 + 4 − 1
4 − 1

=

11
3

= 165 .
If each school must have at least one blackboard then the constraints change to xi ≥ 1 and
the number of such equations is give by

8 − 1
4 − 1

=

7
3

= 35 .
Problem 32 (distributing people)
Assuming that the elevator operator can only tell the number of people getting off at each
floor, we let xi equal the number of people getting off at floor i, where i = 1, 2, 3, 4, 5, 6.
Then the constraint that all people are off at the sixth floor means that
P
i xi = 8, with
xi ≥ 0. This has

n + r − 1
r − 1

=

8 + 6 − 1
6 − 1

=

13
5

= 1287 ,
possible distribution people. If we have five men and three women, let mi and wi be the
number of men and women that get off at floor i. We can solve this problem as the combi-
nation of two problems. That of tracking the men that get off on floor i and that of tracking

the women that get off on floor i. Thus we must have
6
X
i=1
mi = 5 mi ≥ 0
6
X
i=1
wi = 3 wi ≥ 0 .
The number of solutions to the first equation is given by

5 + 6 − 1
6 − 1

=

10
5

= 252 ,
while the number of solutions to the second equation is given by

3 + 6 − 1
6 − 1

=

8
5

= 56 .
So in total then (since each number is exclusive) we have 252 · 56 = 14114 possible elevator
situations.
Problem 33 (possible investment strategies)
Let xi be the number of investments made in opportunity i. Then we must have
4
X
i=1
xi = 20
with constraints that x1 ≥ 2, x2 ≥ 2, x3 ≥ 3, x4 ≥ 4. Writing this equation as
x1 + x2 + x3 + x4 = 20
we can subtract the lower bound of each variable to get
(x1 − 2) + (x2 − 2) + (x3 − 3) + (x4 − 4) = 20 − 2 − 2 − 3 − 4 = 9 .
Then defining v1 = x1 − 2, v2 = x2 − 2, v3 = x3 − 3, and v4 = x4 − 4, then our equation
becomes v1 + v2 + v3 + v4 = 9, with the constraint that vi ≥ 0. The number of solutions to
equations such as these is given by

9 + 4 − 1
4 − 1

=

12
3

= 220 .
Part (b): First we pick the three investments from the four possible in

4
3

= 4 possible
ways. The four choices are denoted in table 1, where a one denotes that we invest in that
option. Then investment choice number one requires the equation v2+v3+v4 = 20−2−3−4 =

choice v1 = x1 − 2 ≥ 0 v2 = x2 − 2 ≥ 0 v3 = x3 − 3 ≥ 0 v4 = x4 − 4 ≥ 0
1 0 1 1 1
2 1 0 1 1
3 1 1 0 1
4 1 1 1 0
Table 1: All possible choices of three investments.
11, and has

11 + 3 − 1
3 − 1

=

13
2

= 78 possible solutions. Investment choice number
two requires the equation v1 +v3 +v4 = 20−2−3−4 = 11, and again has

11 + 3 − 1
3 − 1

=

13
2

= 78 possible solutions. Investment choice number three requires the equation
v1 +v2 +v4 = 20−2−2−4 = 12, and has

12 + 3 − 1
3 − 1

=

14
2

= 91 possible solutions.
Finally, investment choice number four requires the equation v1 +v2+v3 = 20−2−2−3 = 13,
and has

13 + 3 − 1
3 − 1

=

15
2

= 105 possible solutions. Of course we could also invest
in all four opportunities which has the same number of possibilities as in part (a) or 220.
Then in total since we can do any of these choices we have 220 + 105 + 91 + 78 + 78 = 572
choices.
Chapter 1: Theoretical Exercises
Problem 1 (the generalized counting principle)
This can be proved by recursively applying the basic principle of counting.
Problem 2 (counting dependent experimental outcomes)
We have m choices for the outcome of the first experiment. If the first experiment returns
i as an outcome, then there are ni possible outcomes for the second experiment. Thus if
the experiment returns “one” we have n1 possible outcomes, if it returns “two” we have
n2 possible outcomes, etc. To count the number of possible experimental outcomes we can
envision a tree like structure representing the totality of possible outcomes, where we have m
branches leaving the root node indicating the m possible outcomes from the first experiment.
From the first of these branches we have n1 additional branches representing the outcome
of the second experiment when the first experimental outcome was a one. From the second
branch we have n2 additional branches representing the outcome of the second experiment
when the first experimental outcome was a two. We can continue this process, with the
m-th branch from the root node having nm leaves representing the outcome of the second
experiment when the first experimental outcome was a m. Counting all of these outcomes

we have
n1 + n2 + n3 + · · · + nm ,
total experimental outcomes.
Problem 3 (selecting r objects from n)
To select r objects from n, we will have n choices for the first object, n − 1 choices for the
second object, n − 2 choices for the third object, etc. Continuing we will have n − r + 1
choices for the selection of the r-th object. Giving a total of n(n − 1)(n − 2) · · ·(n − r + 1)
total choices if the order of selection matters. If it does not then we must divide by the
number of ways to rearrange the r selected objects i.e. r! giving
n(n − 1)(n − 2) · · ·(n − r + 1)
r!
,
possible ways to select r objects from n when the order of selection of the r object does not
matter.
Problem 4 (combinatorial explanation of

n
k

)
If all balls are distinguishable then there are n! ways to arrange all the balls. With in
this arrangement there are r! ways to uniquely arrange the black balls and (n − r)! ways
to uniquely arranging the white balls. These arrangements don’t represent new patterns
since the balls with the same color are in fact indistinguishable. Dividing by these repeated
patterns gives
n!
r!(n − r)!
,
gives the unique number of permutations.
Problem 5 (the number of binary vectors who’s sum is greater than k )
To have the sum evaluate to exactly k, we must select at k components from the vector x
to have the value one. Since there are n components in the vector x, this can be done in

n
k

ways. To have the sum exactly equal k + 1 we must select k + 1 components from x
to have a value one. This can be done in

n
k + 1

ways. Continuing this pattern we see
that the number of binary vectors x that satisfy
n
X
i=1
xi ≥ k

is given by
n
X
l=k

n
l

=

n
n

+

n
n − 1

+

n
n − 2

+ . . . +

n
k + 1

+

n
k

.
Problem 6 (counting the number of increasing vectors)
If the first component x1 were to equal n, then there is no possible vector that satisfies the
inequality x1 x2 x3 . . . xk constraint. If the first component x1 equals n − 1
then again there are no vectors that satisfy the constraint. The first largest value that
the component x1 can take on and still result in a complete vector satisfying the inequality
constraints is when x1 = n−k+1 For that value of x1, the other components are determined
and are given by x2 = n − k + 2, x3 = n − k + 3, up to the value for xk where xk = n.
This assignment provides one vector that satisfies the constraints. If x1 = n − k, then we
can construct an inequality satisfying vector x by assigning the k − 1 other components
x2, x3, up to xk by assigning the integers n − k + 1 , n − k + 2 , . . . n − 1 , n to the k − 1
components. This can be done in

k
1

ways. Continuing if x1 = n − k − 1, then we can
obtain a valid vector x by assign the integers n − k , n − k + 1 , . . . n − 1 , n to the k − 1
other components of x. This can be seen as an equivalent problem to that of specifying two
blanks from n − (n − k) + 1 = k + 1 spots and can be done in

k + 1
2

ways. Continuing
to decrease the value of the x1 component, we finally come to the case where we have n
locations open for assignment with k assignments to be made (or equivalently n − k blanks
to be assigned) since this can be done in

n
n − k

ways. Thus the total number of vectors
is given by
1 +

k
1

+

k + 1
2

+

k + 2
3

+ . . . +

n − 1
n − k − 1

+

n
n − k

.
Problem 7 (choosing r from n by drawing subsets of size r − 1)
Equation 4.1 from the book is given by

n
r

=

n − 1
r − 1

+

n − 1
r

.

Considering the right hand side of this expression, we have

n − 1
r − 1

+

n − 1
r

=
(n − 1)!
(n − 1 − r + 1)!(r − 1)!
+
(n − 1)!
(n − 1 − r)!r!
=
(n − 1)!
(n − r)!(r − 1)!
+
(n − 1)!
(n − 1 − r)!r!
=
n!
(n − r)!r!

r
n
+
n − r
n

=

n
r

,
and the result is proven.
Problem 8 (selecting r people from from n men and m women)
We desire to prove

n + m
r

=

n
0

m
r

+

n
1

m
r − 1

+ . . . +

n
r

m
0

.
We can do this in a combinatorial way by considering subgroups of size r from a group of
n men and m women. The left hand side of the above represents one way of obtaining this
identity. Another way to count the number of subsets of size r is to consider the number
of possible groups can be found by considering a subproblem of how many men chosen to
be included in the subset of size r. This number can range from zero men to r men. When
we have a subset of size r with zero men we must have all women. This can be done in

n
0

m
r

ways. If we select one man and r −1 women the number of subsets that meet
this criterion is given by

n
1

m
r − 1

. Continuing this logic for all possible subset of
the men we have the right hand side of the above expression.
Problem 9 (selecting n from 2n)
From problem 8 we have that when m = n and r = n that

2n
n

=

n
0

n
n

+

n
1

n
n − 1

+ . . . +

n
n

n
0

.
Using the fact that

n
k

=

n
n − k

the above is becomes

2n
n

=

n
0
2
+

n
1
2
+ . . . +

n
n
2
,
which is the desired result.

Problem 10 (committee’s with a chair)
Part (a): We can select a committee with k members in

n
k

ways. Selecting a chairper-
son from the k committee members gives
k

n
k

possible choices.
Part (b): If we choose the non chairperson members first this can be done in

n
k − 1

ways. We then choose the chairperson based on the remaining n − k + 1 people. Combining
these two we have
(n − k + 1)

n
k − 1

possible choices.
Part (c): We can first pick the chair of our committee in n ways and then pick k − 1
committee members in

n − 1
k − 1

. Combining the two we have
n

n − 1
k − 1

,
possible choices.
Part (d): Since all expressions count the same thing they must be equal and we have
k

n
k

= (n − k + 1)

n
k − 1

= n

n − 1
k − 1

.
Part (e): We have
k

n
k

= k
n!
(n − k)!k!
=
n!
(n − k)!(k − 1)!
=
n!(n − k + 1)
(n − k + 1)!(k − 1)!
= (n − k + 1)

n
k − 1

Factoring out n instead we have
k

n
k

= k
n!
(n − k)!k!
= n
(n − 1)!
(n − 1 − (k − 1))!(k − 1)!
= n

n − 1
k − 1

Problem 11 (Fermat’s combinatorial identity)
We desire to prove the so called Fermat’s combinatorial identity

n
k

=
n
X
i=k

i − 1
k − 1

=

k − 1
k − 1

+

k
k − 1

+ · · · +

n − 2
k − 1

+

n − 1
k − 1

.
Following the hint, consider the integers 1, 2, · · · , n. Then consider subsets of size k from n
elements as a sum over i where we consider i to be the largest entry in all the given subsets
of size k. The smallest i can be is k of which there are

k − 1
k − 1

subsets where when we
add the element k we get a complete subset of size k. The next subset would have k + 1
as the largest element of which there are

k
k − 1

of these. There are

k + 1
k − 1

subsets
with k +2 as the largest element etc. Finally, we will have

n − 1
k − 1

sets with n the largest
element. Summing all of these subsets up gives

n
k

.
Problem 12 (moments of the binomial coefficients)
Part (a): Consider n people from which we want to count the total number of committees
of any size with a chairman. For a committee of size k = 1 we have 1 ·

n
1

= n possible
choices. For a committee of size k = 2 we have

n
2

subsets of two people and two choices
for the person who is the chair. This gives 2

n
2

possible choices. For a committee of size
k = 3 we have 3

n
3

, etc. Summing all of these possible choices we find that the total
number of committees with a chair is
n
X
k=1
k

n
k

.

Another way to count the total number of all committees with a chair, is to consider first
selecting the chairperson from which we have n choices and then considering all possible
subsets of size n − 1 (which is 2n−1
) from which to construct the remaining committee
members. The product then gives n2n−1
.
Part (b): Consider again n people where now we want to count the total number of com-
mittees of size k with a chairperson and a secretary. We can select all subsets of size k in

n
k

ways. Given a subset of size k, there are k choices for the chairperson and k choices
for the secretary giving k2

n
k

committees of size k with a chair and a secretary. The
total number of these is then given by summing this result or
n
X
k=1
k2

n
k

.
Now consider first selecting the chair which can be done in n ways. Then selecting the
secretary which can either be the chair or one of the n−1 other people. If we select the chair
and the secretary to be the same person we have n−1 people to choose from to represent the
committee. All possible subsets from as set of n−1 elements is given by 2n−1
, giving in total
n2n−1
possible committees with the chair and the secretary the same person. If we select a
different person for the secretary this chair/secretary selection can be done in n(n − 1) ways
and then we look for all subsets of a set with n − 2 elements (i.e. 2n−2
) so in total we have
n(n − 1)2n−2
. Combining these we obtain
n2n−1
+ n(n − 1)2n−2
= n2n−2
(2 + n − 1) = n(n + 1)2n−2
.
Equating the two we have
n
X
k=1

n
k

k2
= 2n−2
n(n + 1) .
Part (c): Consider now selecting all committees with a chair a secretary and a stenographer,
where each can be the same person. Then following the results of Part (b) this total number
is given by
Pn
k=1

n
k

k3
. Now consider the following situations and a count of how many
cases they provide.
• If the same person is the chair, the secretary, and the stenographer, then this combi-
nation gives n2n−1
total committees.
• If the same person is the chair and the secretary, but not the stenographer, then this
combination gives n(n − 1)2n−2
total committees.
• If the same person is the chair and the stenographer, but not the secretary, then this
total committees.
• If the same person is the secretary and the stenographer, but not the chair, then this
total committees.

• Finally, if no person has more than one job, then this combination gives n(n − 1)(n −
2)2n−3
total committees.
Adding all of these possible combinations up we find that
n(n − 1)(n − 2)2n−3
+ 3n(n − 1)2n−2
+ n2n−1
= n2
(n + 3)2n−3
.
Problem 13 (an alternating series of binomial coefficients)
From the binomial theorem we have
(x + y)n
=
n
X
k=0

n
k

xk
yn−k
.
If we select x = −1 and y = 1 then x + y = 0 and the sum above becomes
0 =
n
X
k=0

n
k

(−1)k
,
as we were asked to prove.
Problem 14 (committees and subcommittees)
Part (a): Pick the committee of size j in

n
j

ways. The subcommittee of size i from
these j can be selected in

j
i

ways, giving a total of

j
i

n
j

committees and
subcommittee. Now assume that we pick the subcommittee first. This can be done in

n
i

ways. We then pick the committee in

n − i
j − i

ways resulting in a total

n
i

n − i
j − i

.
Part (b): I think that the lower index on this sum should start at i (the smallest subcom-
mittee size). If so then we have
n
X
j=i

n
j

j
i

=
n
X
j=i

n
i

n − i
j − i

=

n
i
n
X
j=i

n − i
j − i

=

n
i
n−i
X
j=0

n − i
j

=

n
i

2n−i
.

Part (c): Consider the following manipulations of a binomial like sum
n
X
j=i

n
j

j
i

xj−i
yn−i−(j−i)
=
n
X
j=i

n
i

n − i
j − i

xj−i
yn−j
=

n
i
n
X
j=i

n − i
j − i

xj−i
yn−j
=

n
i
n−i
X
j=0

n − i
j

xj
yn−(j+i)
=

n
i
n−i
X
j=0

n − i
j

xj
yn−i−j
=

n
i

(x + y)n−i
.
In summary we have shown that
n
X
j=i

n
j

j
i

xj−i
yn−j
=

n
i

(x + y)n−i
for i ≤ n
Now let x = 1 and y = −1 so that x + y = 0 and using these values in the above we have
n
X
j=i

n
j

j
i

(−1)n−j
= 0 for i ≤ n .
Problem 15 (the number of ordered vectors)
As stated in the problem we will let Hk(n) be the number of vectors with components
x1, x2, · · · , xk for which each xi is a positive integer such that 1 ≤ xi ≤ n and the xi are
ordered i.e. x1 ≤ x2 ≤ x3 ≤ · · · ≤ xn
Part (a): Now H1(n) is the number of vectors with one component (with the restriction on
its value of 1 ≤ x1 ≤ n). Thus there are n choices for x1 so H1(n) = n.
We can compute Hk(n) by considering how many vectors there can be when the last compo-
nent i.e. xk has value of j. This would be the expression Hk−1(j), since we know the value
of the k-th component. Since j can range from 1 to n the total number of vectors with k
components (i.e. Hk(n)) is given by the sum of all the previous Hk−1(j). That is
Hk(n) =
n
X
j=1
Hk−1(j) .
Part (b): We desire to compute H3(5). To do so we first note that from the formula above
the points at level k (the subscript) depends on the values of H at level k − 1. To evaluate

this expression when n = 5, we need to evaluate Hk(n) for k = 1 and k = 2. We have that
H1(n) = n
H2(n) =
n
X
j=1
H1(j) =
n
X
j=1
j =
n(n + 1)
2
H3(n) =
n
X
j=1
H2(j) =
n
X
j=1
j(j + 1)
2
.
Thus we can compute the first few values of H2(·) as
H2(1) = 1
H2(2) = 3
H2(3) = 6
H2(4) = 10
H2(5) = 15 .
So that we find that
H3(5) = H2(1) + H2(2) + H2(3) + H2(4) + H2(5)
= 1 + 3 + 6 + 10 + 15 = 35 .
Problem 16 (the number of tied tournaments)
Part (a): See Table 2 for the enumerations used in computing N(3). We have denoted A,
B, and C by the people all in the first place.
Part (b): To argue the given sum, we consider how many outcomes there are when i-players
tie for last place. To determine this we have to choose the i players from n that will tie (which
can be done in

n
i

ways). We then have to distributed the remaining n − i players in
winning combinations (with ties allowed). This can be done recursively in N(n − i) ways.
Summing up all of these terms we find that
N(n) =
n
X
i=1

n
i

N(n − i) .
Part (c): In the above expression let j = n − i, then our limits on the sum above change
as follows
i = 1 → j = n − 1 and
i = n → j = 0 ,
so that the above sum for N(n) becomes
N(n) =
n−1
X
j=0

n
j

N(j) .

First Place Second Place Third Place
A,B,C
A,B C
A,C B
C,B A
A B,C
B C,A
C A,B
A B C
B C A
C A B
A C B
.
.
.
.
.
.
.
.
.
B A C
C B A
Table 2: Here we have enumerated many of the possible ties that can happen with three
people. The first row corresponds to all three in first place. The next three rows corresponds
to two people in first place and the other in second place. The third row corresponds to two
people in second place and one in first. The remaining rows correspond to one person in
each position. The ellipses (
.
.
.) denotes thirteen possible outcomes.
Part (d): For the specific case of N(3) we find that
N(3) =
2
X
j=0

3
j

N(j)
=

3
0

N(0) +

3
1

N(1) +

3
2

N(2)
= N(0) + 3N(1) + 3N(2) = 1 + 3(1) + 3(3) = 13 .
We also find for N(4) that
N(4) =
3
X
j=0

4
j

N(j)
=

4
0

N(0) +

4
1

N(1) +

4
2

N(2) +

4
3

N(3)
= N(0) + 4N(1) +
3 · 4
2
N(2) + 4N(3) = 1 + 4(1) + 6(3) + 4(13) = 75 .
Problem 17 (why the binomial equals the multinomial)
The expression

n
r

is the number of ways to choose r objects from n, leaving another
group of n − r objects. The expression

n
r, n − r

is the number of divisions of n distinct

objects into two groups of size r and of size n − r respectively. As these are the same thing
the numbers are equivalent.
Problem 18 (a decomposition of the multinomial coefficient)
To compute

n
n1, n2, n3, · · · , nr

we consider fixing one particular object from the n. Then
this object can end up in any of the r individual groups. If it appears in the first one then we
have

n − 1
n1 − 1, n2, n3, · · · , nr

, possible arrangements for the other objects. If it appears in
the second group then the remaining objects can be distributed in

n − 1
n1, n2 − 1, n3, · · · , nr

ways, etc. Repeating this argument for all of the r groups we see that the original multinomial
coefficient can be written as sums of these individual multinomial terms as

n
n1, n2, n3, · · · , nr

=

n − 1
n1 − 1, n2, n3, · · · , nr

+

n − 1
n1, n2 − 1, n3, · · · , nr

+ · · ·
+

n − 1
n1, n2, n3, · · · , nr − 1

.
Problem 19 (the multinomial theorem)
The multinomial therm is
(x1 + x2 + · · · + xr)n
=
X
n1+n2+···+nr=n

n
n1, n2, · · · , nr

xn1
1 xn2
2 · · ·xnr
r ,
which can be proved by recognizing that the product of (x1 + x2 + · · · + xr)n
will contain
products of the type xn1
1 xn2
2 · · · xnr
r , and recognizing that the number of such terms, i.e. the
coefficient in front of this term is a count of the number of times we can select n1 of the
variable x1’s, and n2 of the variable x2, etc from the n variable choices. Since this number
equals the multinomial coefficient we have proven the multinomial theorem.
Problem 20 (the number of ways to fill bounded urns)
Let xi be the number of balls in the ith urn. We must have xi ≥ mi and we are distributing
the n balls so that
Pr
i=1 xi = n. To solve this problem lets shift our variables so that each
must be greater than or equal to zero. Our constraint then becomes (by subtracting the

lower bound on xi)
r
X
i=1
(xi − mi) = n −
r
X
i=1
mi .
This expression motivates us to define vi = xi − mi. Then vi ≥ 0 so we are looking for the
number of solutions to the equation
r
X
i=1
vi = n −
r
X
i=1
mi ,
where vi must be greater than or equal to zero. This number is given by

n −
Pr
i=1 mi + r − 1
r − 1

.
Problem 21 (k zeros in an integer equation )
To find the number of solutions to
x1 + x2 + · · · + xr = n ,
where exactly k of the xr’s are zero, we can select k of the xi’s to be zero in

r
k

ways and
then count the number of solutions with positive (greater than or equal to one solutions)
for the remaining r − k variables. The number of solutions to the remaining equation is

n − 1
r − k − 1

ways so that the total number is the product of the two or

r
k

n − 1
r − k − 1

.
Problem 22 (the number of partial derivatives)
Let ni be the number of derivatives taken of the xith variable. Then a total order of n
derivatives requires that these componentwise derivatives satisfy
Pn
i=1 ni = n, with ni ≥ 0.
The number of such is given by

n + n − 1
n − 1

=

2n − 1
n − 1

.
Problem 23 (counting discrete wedges)
We require that xi ≥ 1 and that they sum to a value less than k, i.e.
n
X
i=1
xi ≤ k .

To count the number of solutions to this equation consider the number of equations with
xi ≥ 1 and
Pn
i=1 xi = k̂, which is
k̂ − 1
n − 1

so to calculate the number of equations to the requested problem we add these up for all
k̂ k. The number of solutions is given by
k
X
k̂=n

k̂ − 1
n − 1

with k n .
Chapter 1: Self-Test Problems and Exercises
Problem 1 (counting arrangements of letters)
Part (a): Consider the pair of A with B as one object. Now there are two orderings of this
“fused” object i.e. AB and BA. The remaining letters can be placed in 4! orderings and
once an ordering is specified the fused A/B block can be in any of the five locations around
the permutation of the letters CDEF. Thus we have 2 · 4! · 5 = 240 total orderings.
Part (b): We want to enforce that A must be before B. Lets begin to construct a valid
sequence of characters by first placing the other letters CDEF, which can be done in 4! = 24
possible ways. Now consider an arbitrary permutation of CDEF such as DFCE. Then if we
place A in the left most position (such as as in ADFCE), we see that there are five possible
locations for the letter B. For example we can have ABDFCE, ADBFCE, ADFBCE,
ADFCBE, or ADFCEB. If A is located in the second position from the left (as in DAFCE)
then there are four possible locations for B. Continuing this logic we see that we have a total
of 5 + 4 + 3 + 2 + 1 = 5(5+1)
2
= 15 possible ways to place A and B such that they are ordered
with A before B in each permutation. Thus in total we have 15 · 4! = 360 total orderings.
Part (c): Lets solve this problem by placing A, then placing B and then placing C. Now
we can place these characters at any of the six possible character locations. To explicitly
specify their locations lets let the integer variables n0, n1, n2, and n3 denote the number of
blanks (from our total of six) that are before the A, between the A and the B, between the
B and the C, and after the C. By construction we must have each ni satisfy
ni ≥ 0 for i = 0, 1, 2, 3 .
In addition the sum of the ni’s plus the three spaces occupied by A, B, and C must add to
six or
n0 + n1 + n2 + n3 + 3 = 6 ,
or equivalently
n0 + n1 + n2 + n3 = 3 .

The number of solutions to such integer equalities is discussed in the book. Specifically,
there are
3 + 4 − 1
4 − 1

=

6
3

= 20 ,
such solutions. For each of these solutions, we have 3! = 6 ways to place the three other
letters giving a total of 6 · 20 = 120 arrangements.
Part (d): For this problem A must be before B and C must be before D. Let begin to
construct a valid ordering by placing the letters E and F first. This can be done in two ways
EF or FE. Next lets place the letters A and B, which if A is located at the left most position
as in AEF, then B has three possible choices. As in Part (b) from this problem there are
a total of 3 + 2 + 1 = 6 ways to place A and B such that A comes before B. Following the
same logic as in Part (b) above when we place C and D there are 5 + 4 + 3 + 2 + 1 = 15
possible placements. In total then we have 15 · 6 · 2 = 180 possible orderings.
Part (e): There are 2! ways of arranging A and B, 2! ways of arranging C and D, and 2!
ways of arranging the remaining letters E and F. Lets us first place the blocks of letters
consisting of the pair A and B which can be placed in any of the positions around E and F.
There are three such positions. Next lets us place the block of letters consisting of C and
D which can be placed in any of the four positions (between the E, F individual letters, or
the A and B block). This gives a total number of arrangements of
2! · 2! · 2! · 3 · 4 = 96 .
Part (f): E can be placed in any of five choices, first, second, third, fourth or fifth. Then
the remaining blocks can be placed in 5! ways to get in total 5(5!) = 600 arrangement’s.
Problem 2 (counting seatings of people)
We have 4! arrangements of the Americans, 3! arrangements of the French, and 3! arrange-
ments of the Britch and then 3! arrangements of these groups giving
4! · 3! · 3! · 3! ,
Problem 3 (counting presidents)
Part (a): With no restrictions we must select three people from ten. This can be done in

10
3

ways. Then with these three people there are 3! ways to specify which person is the
president, the treasurer, etc. Thus in total we have

10
3

· 3! =
10!
7!
= 720 ,

possible choices.
Part (b): If A and B will not searve together we can construct the total number of choices
by considering clubs consisting of instances with A included but no B, B included by no A,
and finally neither A or B included. This can be represented as
1 ·

8
2

+ 1 ·

8
2

+ ·

8
3

= 112 .
This result needs to again be multipled by 3! as in Part (a) of this problem. When we do so
we find we obtain 672.
Part (c): In the same way as in Part (b) of this problem lets count first the number of clubs
with C and D in them and second the number of clubs without C and D in them. This
number is
8
1

+

8
3

= 64 .
Again multiplying by 3! we find a total number of 3! · 64 = 384 clubs.
Part (d): For E to be an officer means that E must be selected as a club member. The
number of other members that can be selected is given by

9
2

= 36. Again multiplying
this by 3! gives a total of 216 clubs.
Part (e): If for F to serve F must be a president we have two cases. The first is where F
serves and is the president and the second where F does not serve. When F is the president
we have two permutations for the jobs of the other two selected members. When F does not
serve, we have 3! = 6 possible permutations in assigning titles amoung the selected people.
In total then we have
2

9
2

+ 6

9
3

= 576 ,
possible clubs.
Problem 4 (anwsering questions)
She must select seven questions from ten, which can be done in

10
7

= 120 ways. If she
must select at least three from the first five then she can choose to anwser three, four or all
five of the questions. Counting each of these choices in tern, we find that she has

5
3

5
4

+

5
4

5
3

+

5
5

5
2

= 110 .
possible ways.

Problem 5 (dividing gifts)
We have

7
3

ways to select three gifts for the first child, then

4
2

ways to select two
gifts for the second, and finally

2
2

for the third child. Giving a total of

7
3

·

4
2

·

2
2

= 210 ,
arrangements.
Problem 6 (license plates)
We can pick the location of the three letters in

7
3

ways. Once these positions are selected
we have 263
different combinations of letters that can be placed in the three spots. From
the four remaining slots we can place 104
different digits giving in total

7
3

· 263
· 104
,
possible seven place license plates.
Problem 7 (a simple combinatorial argument)
Remember that the expression

n
r

counts the number of ways we can select r items from
n. Notice that once we have specified a particular selection of r items, by construction we
have also specified a particular selection of n − r items, i.e. the remaining ones that are
unselected. Since for each specification of r items we have an equivalent selection of n − r
items, the number of both i.e.

n
r

and

n
n − r

must be equal.
Problem 8 (counting n-digit numbers)
Part (a): To have no to consecutive digits equal, we can select the first digit in one of ten
possible ways. The next digit in one of nine possible ways (we can’t use the digit we selected
for the first position). For the third digit we have three possible choices, etc. Thus in total
we have
10 · 9 · 9 · · · 9 = 10 · 9n−1
,
possible digits.

Part (b): We now want to count the number of n-digit numbers where the digit 0 appears i
times. Lets pick the locations where we want to place the zeros. This can be done in

n
i

ways. We then have nine choices for the other digits to place in the other n − i locations.
This gives 9n−i
possible enoumerations for non-zero digits. In total then we have

n
i

9n−i
,
n digit numbers with i zeros in them.
Problem 9 (selecting three students from three classes)
Part (a): To choose three students from 3n total students can be done in

3n
3

ways.
Part (b): To pick three students from the same class we must first pick the class to draw
the student from. This can be done in

3
1

= 3 ways. Once the class has been picked we
have to pick the three students in from the n in that class. This can be done in

n
3

ways.
Thus in total we have
3

n
3

,
possible selections of three students all from one class.
Part (c): To get two students in the same class and another in a different class, we must
first pick the class from which to draw the two students from. This can be done in

3
1

= 3
ways. Next we pick the other class from which to draw the singleton student from. Since
there are two possible classes to select this student from this can be done in two ways. Once
both of these classes are selected we pick the individual two and one students from their
respective classes in

n
2

and

n
1

ways respectively. Thus in total we have
3 · 2 ·

n
2

n
1

= 6n
n(n − 1)
2
= 3n2
(n − 1) ,
ways.
Part (d): Three students (all from a different class) can be picked in

n
1
3
= n3
ways.
Part (e): As an identity we have then that

3n
3

= 3

n
3

+ 3n2
(n − 1) + n3
.

We can check that this expression is correct by expanding each side. Expanding the left
hand side we find that

3n
3

=
3n!
3!(3n − 3)!
=
3n(3n − 1)(3n − 2)
6
=
9n3
2
−
9n2
2
+ n .
While expanding the right hand side we find that
3

n
3

+ 3n2
(n − 1) + n3
= 3
n!
3!(n − 3)!
+ 3n3
− 3n2
+ n3
=
n(n − 1)(n − 2)
2
+ 4n3
− 3n2
=
n(n2
− 3n + 2)
2
+ 4n3
− 3n2
=
n3
2
−
3n2
2
+ n + 4n3
− 3n2
=
9n3
2
−
9n2
2
+ n ,
which is the same, showing the equivalence.
Problem 10 (counting five digit numbers with no triple counts)
Lets first enumerate the number of five digit numbers that can be constructed with no
repeated digits. Since we have nine choices for the first digit, eight choices for the second
digit, seven choices for the third digit etc. We find the number of five digit numbers with no
repeated digits given by 9 · 8 · 7 · 6 · 5 = 9!
4!
= 15120.
Now lets count the number of five digit numbers where one of the digits 1, 2, 3, · · · , 9 repeats.
We can pick the digit that will repeat in nine ways and select its position in the five digits
in

5
2

ways. To fill the remaining three digit location can be done in 8 · 7 · 6 ways. This
gives in total
9 ·

5
2

· 8 · 7 · 6 = 30240 .
Lets now count the number five digit numbers with two repeated digits. To compute this
we might argue as follows. We can select the first digit and its location in 9 ·

5
2

ways.
We can select the second repeated digit and its location in 8 ·

3
2

ways. The final digit
can be selected in seven ways, giving in total
9

5
2

· 8

3
2

· 7 = 15120 .
We note, however, that this analysis (as it stands) double counts the true number of five
digits numbers with two repeated digits. This is because in first selecting the first digit from

nine classes and then selecting the second digit from eight choices the total two digits chosen
can actually be selected in the opposite order but placed in same spots from among our five
digits. Thus we have to divide the above number by two giving
15120
2
= 7560 .
So in total we have by summing up all these mutually exclusive events we find that the total
number of five digit numbers allowing repeated digits is given by
9 · 8 · 7 · 6 · 5 + 9

5
2

· 8 · 7 · 6 +
1
2
· 9 ·

5
2

8

3
2

· 7 = 52920 .
Problem 11 (counting first round winners)
Lets consider a simple case first and then generalize this result. Consider some symbolic
players denoted by A, B, C, D, E, F. Then we can construct a pairing of players by first
selecting three players and then ordering the remaining three players with respect to the
first chosen three. For example, lets first select the players B, E, and F. Then if we want A
to play E, C to play F, and D to play B we can represent this graphically by the following
B E F
D A C ,
where the players in a given fixed column play each other. From this we can select three
different winners by selecting who wins each match. This can be done in 23
total ways.
Since we have two possible choices for the winner of the first match, two possible choices for
the winner of the second match, and finally two possible choices for the winner of the third
match. Thus two generalize this procedure to 2n people we must first select n players from
the 2n to for the “template” first row. This can be done in

2n
n

ways. We then must
select one of the n! orderings of the remaining n players to form matches with. Finally, we
must select winners of each match in 2n
ways. In total we would then conclude that we have

2n
n

· n! · 2n
=
(2n)!
n!
· 2n
,
total first round results. The problem with this is that it will double count the total number
of pairings. It will count the pairs AB and BA as distinct. To remove this over counting we
need to divide by the total number of ordered n pairs. This number is 2n
. When we divide
by this we find that the total number of first round results is given by
(2n)!
n!
.
Problem 12 (selecting committees)
Since we must select a total of six people consisting of at least three women and two men,
we could select a committee with four women and two mean or a committee with three

woman and three men. The number of ways of selecting this first type of committee is given
by

8
4

7
2

. The number of ways to select the second type of committee is given by

8
3

7
3

. So the total number of ways to select a committee of six people is

8
4

7
2

+

8
3

7
3

Problem 13 (the number of different art sales)
Let Di be the number of Dalis collected/bought by the i-th collector, Gi be the number of
van Goghs collected by the i-th collector, and finally Pi the number of Picassos’ collected by
the i-th collector when i = 1, 2, 3, 4, 5. Then since all paintings are sold we have the following
constraints on Di, Gi, and Pi,
5
X
i=1
Di = 4 ,
5
X
i=1
Gi = 5 ,
5
X
i=1
Pi = 6 .
Along with the requirements that Di ≥ 0, Gi ≥ 0, and Pi ≥ 0. Remembering that the
number of solutions to an equation like
x1 + x2 + · + xr = n ,
when xi ≥ 0 is given by

n + r − 1
r − 1

. Thus the number of solutions to the first equation
above is given by

4 + 5 − 1
5 − 1

=

8
4

= 70, the number of solutions to the second
equation is given by

5 + 5 − 1
5 − 1

=

9
4

= 126, and finally the number of solutions to
the third equation is given by

6 + 5 − 1
5 − 1

=

10
4

= 210. Thus the total number of
solutions is given by the product of these three numbers. We find that

8
4

9
4

10
4

= 1852200 ,
See the Matlab file chap 1 st 13.m for these calculations.
Problem 14 (counting vectors that sum to less than k)
We want to evaluate the number of solutions to
Pn
i=1 xi ≤ k for k ≥ n, and xi a positive
integer. Now since the smallest value that
Pn
i=1 xi can be under these conditions is given
when xi = 1 for all i and gives a resulting sum of n. Now we note that for this problem the
sum
Pn
i=1 xi take on any value greater than n up to and including k. Consider the number

of solutions to
Pn
i=1 xi = j when j is fixed such that n ≤ j ≤ k. This number is given by

j − 1
n − 1

. So the total number of solutions is given by summing this expression over j for
j ranging from n to k. We then find the total number of vectors (x1, x2, · · · , xn) such that
each xi is a positive integer and
Pn
i=1 xi ≤ k is given by
k
X
j=n

j − 1
n − 1

.
Problem 15 (all possible passing students)
With n total students, lets assume that k people pass the test. These k students can be
selected in

n
k

ways. All possible orderings or rankings of these k people is given by k!
so that the we have
n
k

k! ,
different possible orderings when k people pass the test. Then the total number of possible
test postings is given by
n
X
k=0

n
k

k! .
Problem 16 (subsets that contain at least one number)
There are

20
4

subsets of size four. The number of subsets that contain at least one of
the elements 1, 2, 3, 4, 5 is the complement of the number of subsets that don’t contain any of
the elements 1, 2, 3, 4, 5. This number is

15
4

, so the total number of subsets that contain
at least one of 1, 2, 3, 4, 5 is given by

20
4

−

15
4

= 4845 − 1365 = 3480 .
Problem 17 (a simple combinatorial identity)
To show that

n
2

=

k
2

+ k(n − k) +

n − k
2

for 1 ≤ k ≤ n ,

is true, begin by expanding the right hand side (RHS) of this expression. Using the definition
of the binomial coefficients we obtain
RHS =
k!
2!(k − 2)!
+ k(n − k) +
(n − k)!
2!(n − k − 2)!
=
k(k − 1)
2
+ k(n − k) +
(n − k)(n − k − 1)
2
=
1
2
k2
− k + kn − k2
+ n2
− nk − n − kn + k2
+ k

=
1
2
n2
− n

.
Which we can recognize as equivalent to

n
2

since from its definition we have that

n
2

=
n!
2!(n − 2)!
=
n(n − 1)
2
.
proving the desired equivalence. A combinatorial argument for this expression can be given
in the following way. The left hand side

n
2

represents the number of ways to select
two items from n. Now for any k (with 1 ≤ k ≤ n) we can think about the entire set of n
items as being divided into two parts. The first part will have k items and the second part
will have the remaining n − k items. Then by considering all possible halves the two items
selected could come from will yield the decomposition shown on the right hand side of the
above. For example, we can draw our two items from the initial k in the first half in

k
2

ways, from the second half (which has n−k elements) in

n − k
2

ways, or by drawing one
element from the set with k elements and another element from the set with n − k elements,
in k(n − k) ways. Summing all of these terms together gives

k
2

+ k(n − k) +

n − k
2

for 1 ≤ k ≤ n ,
as an equivalent expression for

n
2

.

Chapter 2 (Axioms of Probability)
Chapter 2: Problems
Problem 1 (the sample space)
The sample space consists of the possible experimental outcomes, which in this case is given
by
{(R, R), (R, G), (R, B), (G, R), (G, G), (G, B), (B, R), (B, G), (B, B)} .
If the first marble is not replaced then our sample space loses all “paired” terms in the above
(i.e. terms like (R, R)) and it becomes
{(R, G), (R, B), (G, R), (G, B), (B, R), (B, G)} .
Problem 2 (the sample space of continually rolling a die)
The sample space consists of all possible die rolls to obtain a six. For example we have
{(6), (1, 6), (2, 6), (3, 6), (4, 6), (5, 6), (1, 1, 6), (1, 2, 6), · · · , (2, 1, 6), (2, 2, 6) · · ·}
The points in En are all sequences of rolls with n elements in them, so that ∪∞
1 En is all
possible sequences ending with a six. Since a six must happen eventually, we have (∪∞
1 En)c
=
φ.
Problem 8 (mutually exclusive events)
Since A and B are mutually exclusive then P(A ∪ B) = P(A) + P(B).
Part (a): To calculate the probability that either A or B occurs we evaluate P(A ∪ B) =
P(A) + P(B) = 0.3 + 0.5 = 0.8
Part (b): To calculate the probability that A occurs but B does not we want to evaluate
P(AB). This can be done by considering
P(A ∪ B) = P(B ∪ (AB)) = P(B) + P(AB) ,
where the last equality is due to the fact that B and AB are mutually independent. Using
what we found from part (a) P(A ∪ B) = P(A) + P(B), the above gives
P(AB) = P(A) + P(B) − P(B) = P(A) = 0.3 .

Part (c): To calculate the probability that both A and B occurs we want to evaluate
P(A ∩ B), which can be found by using
P(A ∪ B) = P(A) + P(B) − P(A ∩ B) .
Using what we know in the above we have that
P(A ∩ B) = P(A) + P(B) − P(A ∪ B) = 0.8 − 0.3 − 0.5 = 0 ,
Problem 9 (accepting credit cards)
Let A be the event that a person carries the American Express card and B be the event that
a person carries the VISA card. Then we want to evaluate P(∪B), the probability that a
person carries the American Express card or the person carries the VISA card. This can be
calculated as
P(A ∪ B) = P(A) + P(B) − P(A ∩ B) = 0.24 + 0.64 − 0.11 = 0.77 .
Problem 10 (wearing rings and necklaces)
Let P(A) be the probability that a student wears a ring. Let P(B) be the probability that
a student wears a necklace. Then from the information given we have that
P(A) = 0.2
P(B) = 0.3
P((A ∪ B)c
) = 0.3 .
Part (a): We desire to calculate for this subproblem P(A ∪ B), which is given by
P(A ∪ B) = 1 − P((A ∪ B)c
) = 1 − 0.6 = 0.4 ,
Part (b): We desire to calculate for this subproblem P(AB), which can be calculated by
using the inclusion/exclusion identity for two sets which is
P(A ∪ B) = P(A) + P(B) − P(AB) .
so solving for P(AB) in the above we find that
P(AB) = P(A) + P(B) − P(A ∪ B) = 0.2 + 0.3 − 0.4 = 0.1 .
Problem 11 (smoking cigarettes v.s cigars)
Let A be the event that a male smokes cigarettes and let B be the event that a male smokes
cigars. Then the data given is that P(A) = 0.28, P(B) = 0.07, and P(AB) = 0.05.

Part (a): We desire to calculate for this subproblem P((A ∪ B)c
), which is given by (using
the inclusion/exclusion identity for two sets)
P((A ∪ B)c
) = 1 − P(A ∪ B)
= 1 − (P(A) + P(B) − P(AB))
= 1 − 0.28 − 0.07 + 0.05 = 0.7 .
Part (b): We desire to calculate for this subproblem P(B ∩ Ac
) We will compute this from
the identity
P(B) = P((B ∩ Ac
) ∪ (B ∩ A)) = P(B ∩ Ac
) + P(B ∩ A) ,
since the events B ∩ Ac
and B ∩ A are mutually exclusive. With this identity we see that
the event that we desire the probability of (B ∩ Ac
) is given by
P(B ∩ Ac
) = P(B) − P(A ∩ B) = 0.07 − 0.05 = 0.02 .
Problem 12 (language probabilities)
Let S be the event that a student is in a Spanish class, let F be the event that a student is
in a French class and let G be the event that a student is in a German class. From the data
given we have that
P(S) = 0.28 , P(F) = 0.26 , P(G) = 0.16
P(S ∩ F) = 0.12 , P(S ∩ G) = 0.04 , P(F ∩ G) = 0.06
P(S ∩ F ∩ G) = 0.02 .
Part (a): We desire to compute
P(¬(S ∪ F ∪ G)) = 1 − P(S ∪ F ∪ G) .
Define the event A to be A = S ∪ F ∪ G, then we will use the inclusion/exclusion identity
for three sets which expresses P(A) = P(S ∪ F ∪ G) in terms of set intersections as
P(A) = P(S) + P(F) + P(G) − P(S ∩ F) − P(S ∩ G) − P(F ∩ G) + P(S ∩ F ∩ G)
= 0.28 + 0.26 + 0.16 − 0.12 − 0.04 − 0.06 + 0.02 = 0.5 .
So that we have that P(¬(S ∪ F ∪ G)) = 1 − 0.5 = 0.5.
Part (b): Using the definitions of the events above for this subproblem we want to compute
P(S ∩ (¬F) ∩ (¬G)) , P((¬S) ∩ F ∩ (¬G)) , P((¬S) ∩ (¬F) ∩ G) .
As these are all of the same form, lets first consider P(S ∩ (¬F) ∩ (¬G)), which equals
P(S ∩(¬(F ∪G))). Now decomposing S into two disjoint sets S ∩(¬(F ∪G)) and S ∩(F ∪G)
we see that P(S) can be written as
P(S) = P(S ∩ (¬(F ∪ G))) + P(S ∩ (F ∪ G)) .

Now since we know P(S) if we knew P(S ∩(F ∪G)) we can compute the desired probability.
Distributing the intersection in S ∩ (F ∪ G), we see that we can write this set as
S ∩ (F ∪ G) = (S ∩ F) ∪ (S ∩ G) .
So that P(S ∩ (F ∪ G)) can be computed (using the inclusion/exclusion identity) as
P(S ∩ (F ∪ G)) = P((S ∩ F) ∪ (S ∩ G))
= P(S ∩ F) + P(S ∩ G) − P((S ∩ F) ∩ (S ∩ G))
= P(S ∩ F) + P(S ∩ G) − P(S ∩ F ∩ G)
= 0.12 + 0.04 − 0.02 = 0.14 .
Thus
P(S ∩ (¬(F ∪ G))) = P(S) − P(S ∩ (F ∪ G))
= 0.28 − 0.14 = 0.14 .
In the same way we find that
P((¬S) ∩ F ∩ (¬G)) = P(F) − P(F ∩ (S ∪ G))
= P(F) − (P(F ∩ S) + P(F ∩ G) − P(F ∩ S ∩ G)
= 0.26 − 0.12 − 0.06 + 0.02 = 0.1 .
and that
P((¬S) ∩ (¬F) ∩ G) = P(G) − P(G ∩ (S ∪ F))
= P(G) − (P(G ∩ S) + P(G ∩ F) − P(S ∩ F ∩ G)
= 0.16 − 0.04 − 0.06 + 0.02 = 0.08 .
With all of these intermediate results we can compute that the probability that a student is
taking exactly one language class is given by the sum of the probabilities of the three events
introduced at the start of this subproblem. We find that this sum is given by
0.14 + 0.1 + 0.08 = 0.32 .
Part (c): If two students are chosen randomly the probability that at least one of them is
taking a language class is the complement of the probability that neither is taking a language
class. From part a of this problem we know that fifty students are not taking a language
class, from the one hundred students at the school. Therefore the probability that we select
two students both not in a language class is given by

50
2

100
2
=
1225
4950
=
49
198
,
thus the probability of drawing two students at least one of which is in a language class is
given by
1 −
49
198
=
149
198
.

Problem 13 (the number of paper readers)
Before we begin to solve this problem lets take the given probabilities of intersections of
events and convert them into probabilities of unions of events. Then if we need these values
later in the problem we will have them. This can be done with the inclusion-exclusion
identity. For two general sets A and B the inclusion-exclusion identity is
P(A ∪ B) = P(A) + P(B) − P(A ∩ B) .
Using this we can evaluate the probabilities of union of events.
P(II ∪ III) = P(II) + P(III) − P(II ∩ III) = 0.3 + 0.05 − 0.04 = 0.31
P(I ∪ II) = P(I) + P(II) − P(I ∩ II) = 0.1 + 0.3 − 0.08 = 0.32
P(I ∪ III) = P(I) + P(III) − P(I ∩ III) = 0.1 + 0.05 − 0.02 = 0.13
P(I ∪ II ∪ III) = P(I) + P(II) + P(III) − P(I ∩ II) − P(I ∩ III)
− P(II ∩ III) + P(I ∩ II ∩ III)
= 0.1 + 0.3 + 0.05 − 0.08 − 0.02 − 0.04 + 0.01 = 0.32 .
We will now use these results in the following wherever needed.
Part (a): The requested proportion of people who read only one paper can be represented
from three disjoint probabilities/proportions:
1. P(I ∩ ¬II ∩ ¬III) which represents the proportion of people who only read paper I.
2. P(¬I ∩ II ∩ ¬III) which represents the proportion of people who only read paper II.
3. P(¬I ∩ ¬II ∩ III) which represents the proportion of people who only read paper III.
The sum of these three probabilities will be the total number of people who read only one
newspaper. To compute the first probability (P(I ∩ ¬II ∩ ¬III)) we begin by noting that
P(I ∩ ¬II ∩ ¬III) + P(I ∩ ¬(¬II ∩ ¬III)) = P(I) ,
which is true since we can obtain the event I by intersecting it with two sets that union to
the entire sample space i.e. ¬II ∩ ¬III, and its negation ¬(¬II ∩ ¬III). With this expression
we can evaluate our desired probability P(I∩¬II∩¬III) using the above. Simple subtraction
gives
P(I ∩ ¬II ∩ ¬III) = P(I) − P(I ∩ ¬(¬II ∩ ¬III))
= P(I) − P(I ∩ (II ∪ III))
= P(I) − P((I ∩ II) ∪ (I ∩ III)) .
Where the last two equations follows from the first by some simple set theory. Since the
problem statement gives the probabilities of the events I ∩ II and I ∩ III, to be able to
further evaluate the right hand side of the expression above requires the ability to compute

probabilities of unions of such sets. This can be done with the inclusion-exclusion identity
which for two general sets A and B is given by P(A ∪ B) = P(A) + P(B) − P(A ∩ B). Thus
the above desired probability then becomes
P(I ∩ ¬II ∩ ¬III) = P(I) − P(I ∩ II) − P(I ∩ III) + P((I ∩ II) ∩ (I ∩ III))
= P(I) − P(I ∩ II) − P(I ∩ III) + P(I ∩ II ∩ III)
= 0.1 − 0.08 − 0.02 + 0.01 = 0.01 ,
using the numbers provided. For the probability P(¬I ∩ II ∩ ¬III) of we can use the work
earlier with the substitutions
I → II
II → I .
Since in the first probability we computed the event not negated is event I, while in the
second probability this is event II. This substitution gives
P(¬I ∩ II ∩ ¬III) = P(II) − P(II ∩ I) − P(II ∩ III) + P(II ∩ I ∩ III)
= 0.3 − 0.08 − 0.04 + 0.01 = 0.19 ,
For the probability P(¬I ∩ ¬II ∩ III) of we can use the work earlier with the substitutions
I → III
III → I .
To give
P(¬I ∩ ¬II ∩ III) = P(III) − P(III ∩ II) − P(III ∩ I) + P(I ∩ II ∩ III)
= 0.05 − 0.04 − 0.02 + 0.01 = 0.00 .
Finally the number of people who read only one newspaper is given by
0.01 + 0.19 + 0.00 = 0.2 ,
so the number of people who read only one newspaper is given by 0.2 × 105
= 20, 000.
Part (b): The requested proportion of people who read at least two newspapers can be
represented from three disjoint probabilities/proportions:
1. P(I ∩ II ∩ ¬III)
2. P(I ∩ ¬II ∩ III)
3. P(¬I ∩ II ∩ III)
4. P(I ∩ II ∩ III)

We can compute each in the following ways. For the first probability we note that
P(¬I ∩ II ∩ III) + P(I ∩ II ∩ III) = P(II ∩ III)
= P(II) + P(III) − P(II ∪ III)
= 0.3 + 0.5 − 0.31 = 0.04 .
So that P(¬I ∩ II ∩ III) = 0.04 − P(I ∩ II ∩ III) = 0.04 − 0.01 = 0.03. Using this we find that
P(I ∩ ¬II ∩ III) = P(I ∩ III) − P(I ∩ II ∩ III)
= P(I) + P(III) − P(I ∪ III) − P(I ∩ II ∩ III)
= 0.1 + 0.5 − 0.13 − 0.01 = 0.01 ,
and that
P(I ∩ II ∩ ¬III) = P(I ∩ II) − P(I ∩ II ∩ III)
= P(I) + P(II) − P(I ∪ II) − P(I ∩ II ∩ III)
= 0.1 + 0.3 − 0.32 − 0.01 = 0.07 .
We also have P(I ∩ II ∩ III) = 0.01, from the problem statement. Combining all of this
information the total percentage of people that read at least two newspapers is given by
0.03 + 0.01 + 0.07 + 0.01 = 0.12 ,
so the total number of people is given by 0.12 × 105
= 12000.
Part (c): For this part we to compute P((I ∩ II) ∪ (III ∩ II)), which gives
P((I ∩ II) ∪ (III ∩ II)) = P(I ∩ II) + P(III ∩ II) − P(I ∩ II ∩ III)
= 0.08 + 0.04 − 0.01 = 0.11 ,
so the number of people read at least one morning paper and one evening paper is 0.11×105
=
11000.
Part (d): To not read any newspaper we are looking for
1 − P(I ∪ II ∪ III) = 1 − 0.32 = 0.68 ,
so the number of people is 68000.
Part (e): To read only one morning paper and one evening paper is expressed as
P(I ∪ II ∪ ¬III) + P(¬I ∩ II ∩ III) .
The first expression has been calculated as 0.01, while the second expansion has been calcu-
lated as 0.03 giving a total 0.04 giving a total of 40000 people who read I as their morning
paper and II as their evening paper or who read III as their morning paper and II as their
evening paper. This number excludes the number who read all three papers.

Problem 14 (an inconsistent study)
Following the hint given in the book, we let M denote the set of people who are married,
W the set of people who are working professionals, and G the set of people who are college
graduates. If we choose a random person and ask what the probability that he/she is either
married or working or a graduate we are looking to compute P(M ∪ W ∪ G). By the
inclusion/exclusion theorem we have that the probability of this event is given by
P(M ∪ W ∪ G) = P(M) + P(W) + P(G)
− P(M ∩ W) − P(M ∩ G) − P(W ∩ G)
+ P(M ∩ W ∩ G) .
From the given data each individual event probability can be estimated as
P(M) =
470
1000
, P(G) =
525
1000
, P(W) =
312
1000
and each pairwise event probability can be estimated as
P(M ∩ G) =
147
1000
, P(M ∩ W) =
86
1000
, P(W ∩ G) =
42
1000
Finally the three-way event probability can be estimated as
P(M ∩ W ∩ G) =
25
1000
.
Using these numbers in the inclusion/exclusion formula above we find that
P(M ∪ W ∪ G) = 0.47 + 0.525 + 0.312 − 0.147 − 0.086 − 0.042 + 0.025
= 1.057 1 ,
in contradiction to the rules of probability.
Problem 15 (probabilities of various poker hands)
Part (a): We must count the number of ways to obtain five cards of the same suit. We can
first pick the suit in

4
1

= 4 ways after which we must pick five cards in

13
5

ways.
So in total we have
4

13
5

= 5148 ,
ways to pick cards in a flush giving a probability of
4

13
5

52
5
= 0.00198 .

Part (b): We can select the first denomination “a” in thirteen ways with

4
2

ways to
obtain the faces for these two cards. We can select the second denomination “b” in twelve
ways with

4
1

possible faces, the third denomination in eleven ways with four faces, the
fourth denomination in ten ways again with four possible faces. The selection of the cards
“b”, “c”, and “d” can be permuted in any of the 3! ways and the same hand results. Thus
we have in total for the number of paired hands the following count
13

4
2

· 12

4
1

· 11

4
1

· 10

4
1

3!
= 1098240 .
Giving a probability of 0.42256.
Part (c): To calculate the number of hands with two pairs we have

13
1

4
2

ways to
select the “a” pair. Then

12
1

4
2

ways to select the “b” pair. Since first selecting the
“a” pair and then the “b” pair results in the same hand as selecting the “b” pair and then
the “a” pair this direct product over counts the total number of “a” and “b” pairs by 2! = 2.
Finally, we have

11
1

4
1

ways to pick the last card in the hand. Thus we have

13
1

4
2

·

12
1

4
2

2!
·

11
1

4
1

= 123552 ,
total number of hands. Giving a probability of 0.04754.
Part (d): We have

13
1

4
3

ways to pick the “a” triplet. We can then pick “b” in

12
1

·4 and pick “c” in

11
1

·4. This combination over counts by two so that the total
number of three of a kind hands is given by

13
1

·

4
3

12
1

· 4

11
1

· 4
2!
= 54912 ,
giving a probability of 0.021128.
Part (e): We have 13 ·

4
4

ways to pick the “a” denomination and twelve ways to pick
the second card with a possible four faces, giving in total 13 · 12 · 4 = 624 possible hands.
This gives a probability of 0.00024.

Problem 16 (poker dice probabilities)
Part (a): We can select the five numbers that will show on the face of the 5 die in

6
5

= 6
ways. We then have 5! ways to order these five selected numbers. This gives for a probability
6 · 5!
65
=
6!
65
= 0.09259 .
Another way to compute this is using the results from parts (b)-(g) for this problem our
probability of interest is
1 − Pb − Pc − Pd − Pe − Pf − Pg ,
where Pi is the probability computed during part “i” of this problem. Using the values
provided in the problem we can evaluate the above to 0.0925.
Part (b): So solve this problem we will think of the die’s outcome as being a numerical
specifications (one through six) of five “slots”. In this specification there are 65
total out-
comes for a trial with the five dice. To determine the number of one pair “hands”, we note
that we can pick the number in the pair in six ways and their locations from the five bins in

5
2

ways. Another number in the hand can be chosen from the five remaining numbers
and placed in any of the remaining bins in

3
1

ways. Continuing this line of reasoning
for the values and placements of the remaining two dice, we have
6 ·

5
2

· 5

3
1

· 4

2
1

· 3

1
1

,
as the number of ordered placements of our four distinct numbers. Since the ordered place-
ment of the three different singleton numbers does not matter we must divide this result by
3!, which results in a value of 3600. Then the probability of one pair is given by
3600
65
= 0.4629 .
Part (c): We specify the two numerical values to use in each of the two pairs in

6
2

ways. Then the location of the first pair in

5
2

, the location of the second pair in

3
2

ways, and finally the

4
1

to select the third number. When we multiply these we get

6
2

5
2

3
2

4
1

= 1800 .
Combined this gives a probability of obtaining two pair of
1800
65
= 0.2315 .

Part (d): We can pick the number for the digit that is repeated three times in six ways,
another digit in five ways and the final digit in four ways. The number of ways we can
place the three dice with the same numeric value is given by

5
3

ways. So the number of
permutations of these three numbers is given by
6 · 5 · 4 ·

5
3

= 1200 .
This gives a probability of 1200
65 = 0.154.
Part (e): Recall that a full house is five dice, three and two of which have the same numeric
value. We can choose the number shown on three die in 6 ways and their locations in the
five rolls in

5
3

ways. We then choose the number shown on the remaining two die in 5
ways. Thus the probability of a full houses is thus given by
6 · 5 ·

5
3

65
= 0.0386 .
Part (f): To get four dice with the same numeric value we must pick one special number
out of six in

6
1

ways representing the four common die. We then pick one more number
from the remaining five in

5
1

ways representing the number on the lone die. Thus we
have

6
1

·

5
1

ways to pick the two numbers to use in the selection of this hand. We
have

5
1

= 5 places in which we can place the lone die after which the location of the
common four is determined. Using this the count of the number of arrangements is given by

6
1

·

5
1

· 5 = 150 .
This gives a requested probability of 150
65 = 0.01929.
Part (g): If all five dice are the same then there are one of six possibilities (the six numbers
on a die). The total number of possible die throws is 65
= 7776 giving a probability to throw
this hand of
6
65
=
1
64
= 0.0007716 .
Problem 17 (randomly placing rooks)
A possible placement of a rook on the chess board can be obtained by specifying the row and
column at which we will locate our rook. Since there are eight rows and eight columns there

are 82
= 64 possible placements for a given rook. After we place each rook we obviously have
one less position where we can place the additional rooks. So the total number of possible
locations where we can place eight rooks is given by
64 · 63 · 62 · 61 · 60 · 59 · 58 · 57 ,
since the order of placement does not matter we must divide this number by 8! to get
64!
8!(64 − 8)!
=

64
8

= 4426165368 .
The number of locations where eight rooks can be placed who won’t be able to capture any
of the other is given by
82
· 72
· 62
· 52
· 42
· 32
· 22
· 12
,
Which can be reasoned as follows. The first rook can be placed in 64 different places. Once
this rook is located we cannot place the next rook in the same row or column that the first
rook holds. This leaves seven choices for a row and seven choices for a column giving a total
of 72
= 49 possible choices. Since the order of these choices does not matter we will need to
divide this product by 8! giving a total probability of
8!2
8!

64
8
= 9.109 10−6
,
in agreement with the book.
Problem 18 (randomly drawing blackjack)
The total number of possible two card hands is given by

52
2

. We can draw an ace in
one of four possible ways i.e. in

4
1

ways. For blackjack the other card must be a ten or
a jack or a queen or a king (of any suite) and can be drawn in

4 + 4 + 4 + 4
1

=

16
1

possible ways. Thus the number of possible ways to draw blackjack is given by

4
1

16
1

52
2
= 0.048265 .
Problem 19 (symmetric dice)
We can solve this problem by considering the disjoint events that both dice land on colors
given by red, black, yellow, or white. For the first die to land on red will happen with

probability 2/6, the same for the second die. Thus the probability that both die land on red
is given by
2
6
2
.
Summing up all the probabilities for all the possible colors, we have a total probability of
obtaining the same color on both dice given by

2
6

2
6

+

2
6

2
6

+

1
6

1
6

+

1
6

1
6

=
5
18
.
Problem 20 (blackjack against a dealer)
We assume that blackjack means the player gets an ace and a king, queen, a jack or a ten
on the initial draw, and ignore the cases where the ace is used with a value of one and the
player may draw another card. In that case, the probability that either the player or the
dealer gets blackjack (independent of the other player) is just

4
1

16
1

52
2
= 0.048265 .
Let A and B be the events that player A or B gets blackjack. In the above we calculated
P(A) and P(B). We want to calculate P((A ∪ B)c
) = 1 − P(A ∪ B). This last event is
P(A ∪ B) = P(A) + P(B) − P(AB) .
Thus we need to calculate P(AB). This can be done as
P(AB) = P(B|A)P(A) =

3
1

15
1

50
2
P(A) = 0.001773 .
We thus find that P((A ∪ B)c
) = 1 − (2(0.048265) − 0.00177) = 0.9052.
Problem 21 (the number of children)
Part (a): Let Pi be the probability that the family chosen has i children. Then we see from
the numbers provided that P1 = 4
20
= 1
5
, P2 = 8
20
= 2
5
, P3 = 5
20
= 1
4
, and P4 = 1
20
, assuming
a uniform probability of selecting any given family.
Part (b): We have
4(1) + 8(2) + 5(3) + 2(4) + 1(5) = 4 + 16 + 15 + 8 + 5 = 48 ,
total children. Then the probability a random child comes from a family with i children is
given by (and denoted by Pi) is P1 = 4
48
, P2 = 16
48
, P3 = 15
48
, P4 = 8
48
, and P5 = 5
48
.

1 2 3 4 5 6
1 0 1 1 1 1 1
2 0 0 1 1 1 1
3 0 0 0 1 1 1
4 0 0 0 0 1 1
5 0 0 0 0 0 1
6 0 0 0 0 0 0
Table 3: The elements of the sample space where the second die is strictly larger in value
than the first.
Problem 22 (shuffling a deck of cards)
To have the ordering exactly the same we must have k heads in a row (which leave the first k
cards unmoved) followed by n − k tails in a row (which will move the cards k + 1, k + 2, . . . n
to the end sequentially). We can do this for any k = 0 to k = n. The probability of getting
k heads followed by n − k tails is

1
2
k
1
2
n−k
=

1
2
n
Now since each of these outcomes is mutually exclusive to compute the total probability we
can sum this result for k = 0 to k = n to get
n
X
k=0

1
2
n
=
n + 1
2n
.
Problem 23 (a larger roll than the first)
We begin by constructing the sample space of possible outcomes. These numbers are com-
puted in table 3, where the row corresponds to the outcome of the first die through and
the column corresponds to the outcome of the second die through. In each square we have
placed a one if the number on the second die is strictly larger than the first. Since each
element of our sample space has a probability of 1/36, by enumeration we find that
15
36
=
5
12
,
is our desired probability.
Problem 24 (the probability the sum of the dice is i)
As in Problem 23 we can explicitly enumerate these probabilities by counting the number
of times each occurrence happens, in Table 4 we have placed the sum of the two dice in the

1 2 3 4 5 6
1 2 3 4 5 6 7
2 3 4 5 6 7 8
3 4 5 6 7 8 9
4 5 6 7 8 9 10
5 6 7 8 9 10 11
6 7 8 9 10 11 12
Table 4: The possible values for the sum of the values when two dice are rolled.
center of each square. Then by counting the number of squares where are sum equals each
number from two to twelve, we have
P2 =
1
36
, P7 =
6
36
=
1
6
P3 =
2
36
=
1
18
, P8 =
5
36
P4 =
3
36
=
1
12
, P9 =
4
36
=
1
9
P5 =
4
36
=
1
9
, P10 =
3
36
=
1
12
P6 =
5
36
, P11 =
2
36
=
1
18
, P12 =
1
36
. (1)
Problem 25 (rolling a five before a seven)
A sum of five has a probability of P5 = 2
18
= 1
9
of occurring. A sum of seven has a
probability of P7 = 1
6
of occurring, so the probability that neither a five or a seven is
given by 1 − 1
9
− 1
6
= 13
18
. Following the hint we let En be the event that a five occurs on the
n-th roll and no five or seven occurs on the n − 1-th rolls up to that point. Then
P(En) =

13
18
n−1
1
9
,
since we want the probability that a five comes first, this can happen at roll number one
(n = 1), at roll number two (n = 2) or any subsequent roll. Thus the probability that a five
comes first is given by
∞
X
n=1

13
18
n−1
1
9
=
1
9
∞
X
n=0

13
18
n
=
1
9
1
1 − 13
18
=
2
5
= 0.4 .

Problem 26 (winning at craps)
From Problem 24 we have computed the individual probabilities for various sum of two
random dice. Following the hint, let Ei be the event that the initial dice sum to i and that
the player wins. We can compute some of these probabilities immediately P(E2) = P(E3) =
P(E12) = 0, and P(E7) = P(E11) = 1. We now need to compute P(Ei) for i = 4, 5, 6, 8, 9, 10.
Again following the hint define Ei,n to be the event that the player initial sum is i and wins
on the n-th subsequent roll. Then
P(Ei) =
∞
X
n=1
P(Ei,n) ,
since if we win, it must be either on the first, or second, or third, etc roll after the initial
roll. We now need to calculate the P(Ei,n) probabilities for each n. As an example of this
calculation first lets compute P(E4,n) which means that we initially roll a sum of four and
the player wins on the n-th subsequent roll. We will win if we roll a sum of a four or loose
if we roll a sum of a seven, while if roll anything else we continue, so to win when n = 1 we
see that
P(E4,1) =
1 + 1 + 1
36
=
1
12
,
since to get a sum of four we can roll pairs consisting of (1, 3), (2, 2), and (3, 1).
To compute P(E4,2) the rules of craps state that we will win if a sum of four comes up (with
probability 1
12
) and loose if a sum of a seven comes up (with probability 6
36
= 1
6
) and continue
playing if anything else is rolled. This last event (continued play) happens with probability
1 −
1
12
−
1
6
=
3
4
.
Thus P(E4,2) = 3
4
1
12
= 1
16
. Here the first 3
4
is the probability we don’t roll a four or a
seven on the n = 1 roll and the second 1
12
comes from rolling a sum of a four on the second
roll (where n = 2). In the same way we have for P(E4,3) the following
P(E4,3) =

3
4
2
1
12
.
Here the first two factors of 3
4
are from the two rolls that “keep us in the game”, and the
factor of 1
12
, is the roll that allows us to win. Continuing in this in this manner we see that
P(E4,4) =

3
4
3
1
12
,
and in general we find that
P(E4,n) =

3
4
n−1
1
12
for n ≥ 1 .
To compute P(Ei,n) for other i, the derivations just performed, only change in the probabil-
ities required to roll the initial sum. We thus find that for other initial rolls (heavily using

the results of Problem 24) that
P(E5,n) =
1
9

1 −
1
9
−
1
6
n−1
=
1
9

13
18
n−1
P(E6,n) =
5
36

1 −
5
36
−
1
6
n−1
=
5
36

25
36
n−1
P(E8,n) =
5
36

1 −
5
36
−
1
6
n−1
=
5
36

25
36
n−1
P(E9,n) =
1
9

1 −
1
9
−
1
6
n−1
=
1
9

13
18
n−1
P(E10,n) =
1
12

1 −
1
12
−
1
6
n−1
=
1
12

3
4
n−1
.
To compute P(E4) we need to sum the results above. We have that
P(E4) =
1
12
X
n≥1

3
4
n−1
=
1
12
X
n≥0

3
4
n
=
1
12
1
1 − 3
4
=
1
3
.
Note that this also gives the probability for P(E10). For P(E5) we find P(E5) = 2
5
, which
also equals P(E9). For P(E6) we find that P(E6) = 5
11
, which also equals P(E8). Then our
probability of winning craps is given by summing all of the above probabilities weighted by
the associated priors of rolling the given initial roll. We find by defining Ii to be the event
that the initial roll is i and W the event that we win at craps that
P(W) = 0 P(I2) + 0 P(I3) +
1
3
P(I4) +
4
9
P(I5) +
5
9
P(I6)
+ 1 P(I7) +
5
9
P(I8) +
4
9
P(I9) +
1
3
P(I10) + 1 P(I11) + 0 P(I12) .
Using the results of Exercise 25 to evaluate P(Ii) for each i we find that the above summation
gives
P(W) =
244
495
= 0.49292 .
These calculations are performed in the Matlab file chap 2 prob 26.m.
Problem 27 (drawing the first red ball)
We want the probability that A selects the first red ball. Since A draws first he will select
a red ball on the first draw with probability 3
10
. If he does not select a red ball B will draw
next and he must not draw a red ball (or the game will stop). The probability that A draws
a red ball on the third total draw is then
P3 =

1 −
3
10

1 −
3
9

3
8

.

Continuing this pattern we see that for A to draw a ball on the fifth total draw will happen
with probability
P5 =

1 −
3
10

1 −
3
9

1 −
3
8

1 −
3
7

3
6

,
and finally on the seventh total draw with probability
P7 =

1 −
3
10

1 −
3
9

1 −
3
8

1 −
3
7

1 −
3
6

1 −
3
5

3
4

.
If player A does not get a red ball after seven draws he will not draw a red ball before
player B. The total probability that player A draws a red ball first is given by the sum
of all these individual probabilities of these mutually exclusive events. In the Matlab code
chap 2 prob 27.m we evaluate this sum and find the probability that A wins given by
P(A) =
7
12
.
So the corresponding probability that B wins is 1 − 7
12
= 5
12
showing the benefit to being the
first “player” in a game like this.
Problem 28 (sampling colored balls from an urn)
Part (a): We want the probability that each ball will be of the same color. This is given by

5
3

+

6
3

+

8
3

5 + 6 + 8
3
= 0.08875 .
Part (b): The probability that all three balls are of different colors is given by

5
1

6
1

8
1

19
3
= 0.247 .
If we replace the ball after drawing it, then the probabilities that each ball is the same color
is now given by
5
19
3
+

6
19
3
+

8
19
3
= 0.124 .
while if we want three balls of different colors, then this happens with probability given by
3!

5
19

6
19

8
19

= 0.2099 .

Problem 29
Warning: Here are some notes I had on this problem. I’ve not had the time to check these
in as much detail as I would have liked. Caveat emptor.
Part (a): The probability we obtain two while balls is given by
n
n + m

n − 1
m + n − 1

.
The probability that we obtain two black balls is given by
m
m + n

m − 1
m + n − 1

,
so the probability of two balls of the same color then is
n(n − 1)
(m + n)(m + n − 1)
+
m(m − 1)
(m + n)(m + n − 1)
=
n(n − 1) + m(m − 1)
(m + n)(m + n − 1)
,
Part (b): Now we replace the balls after we draw them so the probability we draw two
white balls is then
n
m + n

n
m + n

,
and for black balls we have
m
m + n

m
m + n

,
So in total then we have
n2
+ m2
(m + n)2
=
n2
+ m2
m2 + 2mn + n2
.
Part (c): We expect to have a better chance of getting two balls of the same color in
Part (b) of this problem since we have an additional white or black ball in the pot to draw
on the second draw. Thus we want to show that
n2
+ m2
(m + n)2
≥
n(n − 1) + m(m − 1)
(m + n)(m + n − 1)
.
We will perform reversible manipulations to derive an equivalent expression. If the reduced
expression is true, then the original expression is true. We begin by canceling the factor 1
m+n
to give
n2
+ m2
m + n
≥
n(n − 1) + m(m − 1)
m + n − 1
.
multiplying by the common denominator we obtain the following sequence of transformations
(m2
+ n2
)(m + n − 1) ≥ (m + n)(n(n − 1) + m(m − 1))
m3
+ m2
n − m2
+ n2
m + n3
− n2
≥ n(nm − m + n2
− n) + m(m2
− m + nm − n)
m3
+ m2
n − m2
+ n2
m + n3
− n2
≥ mn2
− mn + n3
− n2
+ m3
− m2
+ nm2
− nm
n2
m ≥ −mn + nm2
− nm ,
by dividing by mn we get n ≥ −1 + n − 1 or the inequality 0 ≥ −2 which is true showing
that the original inequality is true.

Problem 30 (the chess club)
Part (a): For Rebecca and Elise to be paired they must first be selected onto their respected
schools chess teams and then be paired in the tournament. Thus if S is the event that the
sisters play each other then
P(S) = P(R)P(E)P(Paired|R, E) ,
where R is the event that that Rebecca is selected for her schools chess team and E is the
event that Elise is selected for her schools team and Paired is the event that the two sisters
play each other. Computing these probabilities we have
P(R) =

1
1

7
3

8
4
=
1
2
,
and
P(E) =

1
1

8
3

9
4
=
4
9
,
and finally
P(Paired) =
1 · 3!
4!
=
1
4
.
so that P(S) = 1
2
· 4
9
· 1
4
= 1
18
.
Part (b): The event that Rebecca and Elise are chosen and then do not play each other
will occur with a probability of
P(R)P(E)P(Pairedc
|R, E) =
1
2
·
4
9

1 −
1
4

=
1
6
.
Part (c): For this part we can have either (and these events are mutually exclusive) Rebecca
picked to represent her school or Elise picked to represent her school but not both and not
neither. Since

1
1

7
3

is the number of ways to choose the team A with Rebecca as
a member and

8
4

are the number of ways to choose team B without having Elise as a
member, their product is the number of ways of choosing the first option above. This given
a probability of
1
1

7
3

8
4
·

8
4

9
4
=
5
18
.

In the same way the other probability is given by

7
4

8
4
·

1
1

8
3

9
4
=
2
9
.
Thus the probability we are after is the sum of the two probabilities above and is given by
9
18
= 1
2
.
Problem 31 (selecting basketball teams)
Part (a): On the first draw we will certainly get one of the team members. Then on the
second draw we must get any team member but the one that we just drew. This happens
with probability 2
3
. Finally, we must get the team member we have not drawn in the first
two draws. This happens with probability 1
3
. In total then, the probability to draw an entire
team is given by
1 ·
2
3
·
1
3
=
2
9
.
Part (b): The probability the second player plays the same position as the first drawn
player is given by 1
3
, while the probability that the third player plays the same position as
the first two is given by 1
3
. Thus this event has a probability of
1
3
·
1
3
=
1
9
.
Problem 32 (a girl in the i-th position)
We can compute all permutations of the b + g people that have a girl in the i-th spot as
follows. We have g choices for the specific girl we place in the i-th spot. Once this girl is
selected we have b + g − 1 other people to place in the b + g − 1 slots around this i-th spot.
This can be done in (b+ g −1)! ways. So the total number of ways to place a girl at position
i is g(b + g − 1)!. Thus the probability of finding a girl in the i-th spot is given by
g(b + g − 1)!
(b + g)!
=
g
b + g
.
Problem 33 (a forest of elk)
After tagging the initial elk we have 5 tagged elk from 20. When we capture four more elk
the probability we get two tagged elk is the number of ways we can select two tagged elks

(from 5) and two untagged elks (from 20 − 5 = 15) divided by the number of ways to select
four elk from 20. This probability p is given by
p =

5
2

15
2

20
4
=
70
323
.
Problem 34 (the probability of a Yarborough)
We must not have a ten, a jack, a queen, a king, or an ace (a total of 5 face cars) in our
hand of thirteen cards. The number of ways to select a hand that does not have any of these
cards is equivalent to selecting thirteen cards from among a set that does not contain any of
the cards mentioned above. Specifically this number is

52 − 4 − 4 − 4 − 4 − 4
13

52
13
=

32
13

52
13
= 0.000547 ,
a relatively small probability.
Problem 35 (selecting psychiatrists for a conference)
The probability that at least one psychologist is choose is given by considering all selections
of sets of psychologists that contain at least one

30
2

24
1

+

30
1

24
2

+

30
0

24
3

54
3
= 0.8363 .
Where in the numerator we have enumerated all possible selections of three people such that
at least one psychologist is chosen.

Problem 36 (choosing two identical cards)
Part (a): We have

52
2

possible ways to draw two cards from the 52 total. For us to
draw two aces, this can be done in

4
2

ways. Thus our probability is given by

4
2

52
2
= 0.00452 .
Part (b): For the two cards to have the same value we can pick the value to represent in
thirteen ways and the two cards in

4
2

ways. Thus our probability is given by
13

4
2

52
2
= 0.0588 .
Problem 37 (solving enough problems on an exam)
Part (a): In this part of the problem imagine that we label the 10 questions as “known” or
“unknown”. Since the student knows how to solve 7 of the 10 problems, we have 7 known
problems and 3 unknown questions. If we imagine the teacher selecting the 5 exam questions
randomly then the probability that the student answers all 5 selected problems correctly is
the probability that we draw 5 known questions from a “set” of 7 known and 3 unknown
questions. This later probability is given by

7
5

3
0

10
5
=
1
12
= 0.083333 .
Part (b): To answer at least four of the questions correctly will happen if the student
answers 5 questions correctly (with probability given above) or 4 questions correctly. In the
same way as above this later probability is given by

7
4

3
1

10
5
=
5
12
.

Thus the probability that the student answers at least four of the problems correctly is the
sum of these two probabilities or
5
12
+
1
12
=
1
2
.
Problem 38 (two red socks)
We are told that three of the socks are red so that n − 3 are not red. When we select two
socks, the probability that they are both red is given by
3
n
·
2
n − 1
.
If we want this to be equal to 1
2
we must solve for n in the following expression
3
n
·
2
n − 1
=
1
2
⇒ n2
− n = 12 .
Using the quadratic formula this has a solution given by
n =
1 ±
p
1 + 4(1)(12)
2(1)
=
1 ± 7
2
.
Taking the positive solution we have that n = 4.
Problem 39 (five different hotels)
When the first person checks into the hotel, the next person will check into a different hotel
with probability 4
5
. The next person will check into a different hotel with probability 3
5
.
Thus the probability that we check into three different hotels is given by
4
5
·
3
5
=
12
25
= 0.48 .
Problem 41 (obtaining a six at least once)
This is the complement of the probability that a six never appears or
1 −

5
6
4
= 0.5177 .

Problem 42 (double sixes)
The probability that a double six appear at least once is the complement of the probability
that a double six never appears. The probability of not seeing a double six is given by
1 − 1
36
= 35
36
, so the probability that a double six appears at least once in n throws is given
by
1 −

35
36
n
.
To make this probability at least 1/2 we need to have
1 −

35
36
n
≥
1
2
.
which gives when we solve for n
n ≥
ln(1
2
)
ln(35
36
)
≈ 24.6 ,
so we should take n = 25.
Problem 43 (the probability you are next to me)
Part (a): The number of ways to arrange N people is N!. To count the number of permu-
tation of the other people and the “pair” A and B consider A and B as fused together as
one unit (say AB) to be taken with the other N − 2 people. So in total we have N − 2 + 1
things to order. This can be done in (N − 1)! ways. Note that for every permutation we
also have two orderings of A and B i.e. AB and BA so we have 2(N − 1)! orderings where
A and B are fused together. The the probability we have A and B fused together is given
by 2(N−1)!
N!
= 2
N
.
Part (b): If the people are arraigned in a circle there are (N−1)! unique arrangements of the
total people. The number of arrangement as in part (a) is given by 2(N−2+1−1)! = 2(N−2)!
so our probability is given by
2(N − 2)!
(N − 1)!
=
2
N − 1
.
Problem 44 (people between A and B)
Note that we have 5! orderings of the five individual people.
Part (a): The number of permutations that have one person between A and B can be
determined as follows. First pick the person to put between A and B from our three choices
C, D, and E. Then pick the ordering of A and B i.e AB or BA. Then considering this AB
object as one object we have to place it with two other people in 3! ways. Thus the number

of orderings with one person between A and B is given by 3 · 2 · 3!, giving a probability of
this event of
3 · 2 · 3!
5!
= 0.3 .
Part (b): Following Part (a) we can pick the two people from the three remaining in

3
2

= 3 (ignoring order) ways. Since the people can be ordered in two different ways and
A and B on the outside can be ordered in two different ways, we have 3 · 2 · 2 = 12 ways
to create the four person “object” with A and B on the outside. This can ordered with the
remaining single person in two ways. Thus our probability is given by
2 · 12
5!
=
1
5
.
Part (c): To have three people between A and B, A and B must be on the ends with 3! = 6
possible ordering of the remaining people. Thus with two orderings of A and B we have a
probability of
2 · 6
5!
=
1
10
.
Problem 45 (trying keys at random)
Part (a): If unsuccessful keys are removed as we try them, then the probability that the
k-th attempt opens the door can be computed by recognizing that all attempts up to (but
not including) the k-th have resulted in failures. Specifically, if we let N be the random
variable denoting the attempt that opens the door we see that
P{N = 1} =
1
n
P{N = 2} =

1 −
1
n

1
n − 1
P{N = 3} =

1 −
1
n

1 −
1
n − 1

1
n − 2
.
.
.
P{N = k} =

1 −
1
n

1 −
1
n − 1

· · ·

1 −
1
n − (k − 2)

1
n − (k − 1)
.
We can check that this result is a valid expression to represent a probability by selecting a
value for n and verifying that when we sum the above over k for 1 ≤ k ≤ n we sum to one.
A verification of this can be found in the Matlab file chap 2 prob 45.m, along with explicit
calculations of the mean and variance of N. A much simpler expression, making the above
Matlab script rather silly, is obtained if we simplify the above expressions by multiplying all
factors together. When we do that we see that we obtain
P{N = k} =
1
n
.

Part (b): If unsuccessful keys are not removed then the probability that the correct key is
selected at draw k is a geometric random with parameter p = 1/n. Thus our probabilities
are given by P{N = k} = (1 − p)k−1
p, and we have for a geometric random variable an
expectation and a variance given by
E[N] =
1
p
= n
Var(N) =
1 − p
p2
= n(n − 1) .
The above expression of (1−p)k−1
p represents the probability that we fail to find the correct
key (with probability 1 − p) in the trials 1, 2, · · · , k − 1 and then on the k trial find the
correct key (with probability p).
Problem 46 (the birthdays of people in a room)
The probability that at least two people share the same birthday is the complement of
the probability that no two people have the same birthday (or that all people have distinct
birthdays). Let En the the event that n people have at least one birthday in common. Based
on the above if n = 2 then the probability that at least two people share the same birthday
is
P(E2) = 1 −
11
12
.
That three people share the same birthday is
P(E3) = 1 −
11
12
10
11
= 1 −
10
12
.
That 4 people share the same birthday is
P(E4) = 1 −
11
12
10
11
9
10
= 1 −
9
12
.
It seems the pattern for general n is
P(En) = 1 −
12 − (n − 1)
12
= 1 −
13 − n
12
.
We want to pick n such that P(En) ≥ 1
2
. From the above when we solve for n this means
that n ≥ 5.
Problem 47 (strangers in a room)
There are 1212
possible ways to distribute birthdays to all people. We next want to count
the number of ways to distribute birthdays so that no two are the same. This can be done
in 12! ways. Thus we get a probability of
P =
12!
1212
=
11!
1211
.

Problem 48 (certain birthdays)
As each person can have his birthday assigned to one of the twelve calender months we have
T = 1220
possible ways to assign birthday months to people. This will be the denominator
in the probability we seek. We now need to compute the number of ways we can get the
desired distribution of months and people requested in the problem. We can select the four
months that are to have two birthdays in

12
4

ways and after this the four months are
to have three birthdays in

8
4

ways. Thus the number of selections of months M can be
done in
M =

12
4

8
4

=
12!
8!4!
·
8!
4!4!
=
12!
4!3 = 34650 ,
ways. Once the months are specified we need to select the people that will have their
birthdays in these selected months. Since we need to put two men in the first selected four
months and then three men in the second selected four months we can do that in N ways
where N is given by
N =

20
2

18
2

16
2

14
2

×

12
3

9
3

6
3

3
3

=
20!
18!2!
·
18!
16!2!
·
16!
14!2!
·
14!
12!2!
×
12!
9!3!
·
9!
6!3!
·
6!
3!3!
·
3!
3!0!
=
20!
2!4
3!4 = 1.173274 1014
.
Using these results we get a probability of
P =
NM
T
≈ 0.0010604 ,
the same as in the back of the book.
Problem 49 (men and women)
The only way to have equal numbers of men in each group is to have three men in each
group (and thus three women in each group). We have

6
3

ways to select the men (and
the same number of ways to select the women). The probability is then given by
P =

6
3

6
3

12
6
=
202
924
= 0.4329 .

Problem 50 (hands of bridge with with spades)
We have

52
13

ways to draw the first hand. If we want to have 5 spades we can select
these in

13
5

ways and then the additional cards for this hand in

52 − 13
13 − 5

=

39
8

ways. The hand with five spades is then drawn with a probability

13
5

39
8

52
13
= 0.12469 .
After this hand is drawn we need do draw the second hand. We want this hand to have the
remaining 8 spades and can be drawn with probability

8
8

31
5

39
13
= 2.0918 10−5
.
Then the probability that both of these events happen simultaneously is then given by their
product or 2.6084 10−6
.
Problem 51 (n balls in N compartments)
If we put m balls in the first component we have to place the remaining n−m balls in N −1
compartments. This can be done in (N − 1)n−m
ways. We can select our m balls to place in
the first compartment in

n
m

ways. Combining these two gives a probability of

n
m

(N − 1)n−m
Nn
.
Another way to view this problem is to consider the event that a given one of our n balls
land in the first (or any specific compartment) as a success that happens with probability
p = 1
N
. Then the probability we have m success from our n trials is a binomial random
variable giving
P(M = m) =

n
m

1
N
m
1 −
1
N
n−m
,
the same as earlier.

Problem 52 (a closet with shoes)
Part (a): We have

20
8

ways of selecting our eight shoes. Since we don’t want any
matching pairs for this part we can select 8 pairs from the 10 pairs available in

10
8

ways. In each pair we can select either the right or the left shoe. This gives
P =

10
8

28

20
8
= 0.09 .
Part (b): We select one pair in include in

10
1

ways. Then the pairs the other shoes
will come from in

9
6

ways and the left-right shoe in 26
ways giving
P =

10
1

9
6

26

20
8
= 0.4267 .
Problem 53 (four married couples in a row)
This problem is very much like the Example 5n from the book. We let Ei be the event that
couple i sits next to each other. The event that at least one couple sits next to each other is
∪iEi. The probability that no couple sits next to each other is then 1−P(∪iEi). To evaluate
P(∪iEi) we will use the inclusion-exclusion lemma which for our four couples is given by
P(∪4
i=1Ei) =
4
X
i=1
P(Ei) −
X
ij
P(EiEj) +
X
ijk
P(EiEjEk) −
X
ijkl
P(EiEjEkEl) . (2)
We now need to compute each of these joint probabilities. To do that first consider P(Ei).
First given the 8 total people there are 8! ways of arranging all the people in a row. We want
to count the number of these that have couple i sitting next to each other. If we consider
this couple “fused” together there are then 7 objects that can be placed in a line such that
the couple is sitting together (the 6 other people and the one couple object). This gives 7!
ways of arranging this 7 objects. We have then two ways to permute the husband and wife
in the couple giving
P(Ai) =
2 · 7!
8!
.
Next consider the evaluation of P(AiAj). We again have 8! for the denominator of this
probability. To compute the numerator we again imagine fuse two couples together giving

8 − 4 + 2 = 6 objects to place. This can be done in 6! way. We can permute the husband
and wife in each pair in 2 · 2 = 22
ways. Thus we find
P(AiAj) =
22
· 6!
8!
.
In general, following the same logic we have for r couples
P(AiAj · · · Ak) =
2r
(8 − r)!
8!
.
Now by symmetry all of the probabilities in the individual sums are the same and that there
are

4
r

for r = 1, 2, 3, 4 terms respectively in each of the sums in Equation 2 above. Thus
using what we have so far the probability that at least one couple sits together is given by
P(∪iEi) =

4
1

2 · 7!
8!
−

4
2

22
· 6!
8!
+

4
3

23
· 5!
8!
−

4
4

24
4!
8!
= 1 −
12
35
,
Thus the probability we seek is given by 1 − (1 − 12
35
) = 12
35
.
Problem 54 (a bridge hand that is void in at least one suit)
We want the probability that a given hand of bridge is void in at least one suit which
means the hand could be void in more than one suit. The error in the suggested calculation
probability given is that it gives the probability that the hand is void in one (and only one
suit), thus it underestimates the probability of interest. Let Ei be the event that the hand
is void in the suit i for i = 1, 2, 3, 4. Then the probability we want is P(∪4
i=1Ei) which we
can calculate by using the inclusion-exclusion identity given in this case by
P(∪4
i=1Ei) =
4
X
i=1
P(Ei) −
X
ij
P(EiEj) +
X
ijk
P(EiEjEk) . (3)
To do this we need to be able to evaluate the joint probabilities P(Ei), P(EiEj), and
P(EiEjEk) for i = 1, 2, 3, 4. Note there is no terms P(EiEjEkEl) since we must be dealt
some cards. We start with P(Ei) where we fix the value of i and
P(Ei) =

39
13

52
13
= 0.01279 .
Next we have
P(EiEj) =

26
13

52
13
= 1.63785 10−5
.

and finally
P(EiEjEk) =

13
13

52
13
= 1.57476 10−12
.
are

4
r

for r = 1, 2, 3 terms respectively in each of the sums in Equation 3 above. Thus
we get
P(∪4
i=1Ei) =

4
1

(0.01279)−

4
2

(1.63785 10−5
)+

4
3

(1.57476 10−12
) = 0.0510655208 .
Problem 55 (hands of cards)
Part (a): We want the probability that a given hand of bridge has the ace and king in at
least one suit. Let Ei be the event that the hand has an ace and a king in the suit i for
i = 1, 2, 3, 4. Then the probability we want is P(∪4
i=1Ei) which we can calculate by using
the inclusion-exclusion identity given in this case by
P(∪4
i=1Ei) =
4
X
i=1
P(Ei) −
X
ij
P(EiEj) +
X
ijk
P(EiEjEk) −
X
ijkl
P(EiEjEkEl) . (4)
To do this we need to be able to evaluate the joint probabilities P(Ei), P(EiEj), P(EiEjEk),
and P(EiEjEkEl) for i, j, k, and l for 1, 2, 3, 4. We start with P(Ei) where we fix the value
of i and
P(Ei) =

50
11

52
13
= 0.0588235 .
Next we have
P(EiEj) =

48
9

52
13
= 0.002641 .
and
P(EiEjEk) =

46
7

52
13
= 8.4289 10−5
,
and finally
P(EiEjEkEl) =

44
5

52
13
= 1.7102 10−6
,

are

4
r

for r = 1, 2, 3 terms respectively in each of the sums in Equation 4 above. Thus
we get
P(∪4
i=1Ei) =

4
1

(0.0588235) −

4
2

(0.002641) +

4
3

(8.4289 10−5
) −

4
4

(1.7102 10−6
)
= 0.0808910 .
Part (b): In the same way as before we let Ei be the event that the hand is missing all four
suits from the ith denominator 1 ≤ i ≤ 13. We then when we fix the indices i, j, k, and l
P(Ei) =

48
9

52
13
= 0.00264
P(EiEj) =

52 − 2(4)
13 − 2(4)

52
13
= 1.71021 10−6
P(EiEjEk) =

52 − 3(4)
13 − 3(4)

52
13
= 6.29907 10−11
,
are

13
r

for r = 1, 2, 3 . . . 13 terms respectively in each of the sums in the inclusion-
exclusion identity. Thus we get
P(∪13
i=1Ei) =

13
1

(0.00264)−

13
2

(1.71021 10−6
)+

13
3

(6.29907 10−11
) = 0.034200 .
Problem 1 (set identities)
To prove this let x ∈ E ∩ F then by definition x ∈ E and therefore x ∈ E ∪ F. Thus
E ∩ F ⊂ E ∪ F.
Problem 2 (more set identities)
If E ⊂ F then x ∈ E implies that x ∈ F. If y ∈ Fc
, then this implies that y /
∈ F which
implies that y ∈ E, for if y was in E then it would have to be in F which we know it is not.

Problem 3 (more set identities)
We want to prove that F = (F ∩ E) ∪ (F ∩ Ec
). We will do this using the standard proof
where we show that each set in the above is a subset of the other. We begin with x ∈ F.
Then if x ∈ E, x will certainly be in F ∩ E, while if x /
∈ E then x will be in F ∩ Ec
. Thus
in either case (x ∈ E or x /
∈ E) x will be in the set (F ∩ E) ∪ (F ∩ Ec
).
If x ∈ (F ∩ E) ∪ (F ∩ Ec
) then x is in either F ∩ E, F ∩ Ec
, or both by the definition of
the union operation. Now x cannot be in both sets or else it would simultaneously be in E
and Ec
, so x must be in one of the two sets only. Being in either set means that x ∈ F and
we have that the set (F ∩ E) ∪ (F ∩ Ec
) is a subset of F. Since each side is a subset of the
other we have shown set equality.
To prove that E ∪F = E ∪(Ec
∩F), we will begin by letting x ∈ E ∪F, thus x is an element
of E or an element of F or of both. If x is in E at all then it is in the set E ∪ (Ec
∩ F). If
x /
∈ E then it must be in F to be in E ∪ F and it will therefore be in Ec
∩ F. Again both
sides are subsets of the other and we have shown set equality.
Problem 6 (set expressions for various events)
Part (a): This would be given by the set E ∩ Fc
∩ Gc
.
Part (b): This would be given by the set E ∩ G ∩ Fc
.
Part (c): This would be given by the set E ∪ F ∪ G.
Part (d): This would be given by the set
((E ∩ F) ∩ Gc
) ∪ ((E ∩ G) ∩ Fc
) ∪ ((F ∩ G) ∩ Ec
) ∪ (E ∩ F ∩ G) .
This expresses the fact that satisfy this criterion by being inside two other events or by being
inside three events.
Part (e): This would be given by the set E ∩ F ∩ G.
Part (f): This would be given by the set (E ∪ F ∪ G)c
.
Part (g): This would be given by the set
(E ∩ Fc
∩ Gc
) ∪ (Ec
∩ F ∩ Gc
) ∪ (Ec
∩ Fc
∩ G)
Part (h): At most two occur is the complement of all three taking place, so this would be
given by the set (E ∩F ∩G)c
. Note that this includes the possibility that none of the events
happen.

Part (i): This is a subset of the sets in Part (d) (i.e. without the set E ∩ F ∩ G) and is
given by the set
((E ∩ F) ∩ Gc
) ∪ ((E ∩ G) ∩ Fc
) ∪ ((F ∩ G) ∩ Ec
) .
Part (j): At most three of them occur must be the entire samples space since we only have
three events total.
Problem 7 (set simplifications)
Part (a): We have that (E ∪ F) ∩ (E ∪ Fc
) = E.
Part (b): For the set
(E ∩ F) ∩ (Ec
∪ F) ∩ (E ∪ Fc
)
We begin with the set
(E ∩ F) ∩ (Ec
∪ F) = ((E ∩ F) ∩ Ec
) ∪ (E ∩ F ∩ F)
= ∅ ∪ (E ∩ F)
= E ∩ F .
So the above becomes
(E ∩ F) ∩ (E ∪ Fc
) = ((E ∩ F) ∩ E) ∪ ((E ∩ F) ∩ Fc
)
= (E ∩ F) ∪ ∅
= E ∩ F .
Part (c): We find that
(E ∪ F) ∩ (F ∪ G) = ((E ∪ F) ∩ F) ∪ ((E ∪ F) ∩ G)
= F ∪ ((E ∩ G) ∪ (F ∩ G))
= (F ∪ (E ∩ G)) ∪ (F ∪ (F ∩ G))
= (F ∪ (E ∩ G)) ∪ F
= F ∪ (E ∩ G) .
Problem 8 (counting partitions)
Part (a): As a simple example, we begin by considering all partition of the elements {1, 2, 3}.
We have
{{1}, {2}, {3}}, {{1, 2, 3}}, {{1}, {2, 3}}, {{2}, {1, 3}}, {{3}, {1, 2}} ,
giving a count of five different partitions.

Part (b): Following the hint this result can be derived as follows. We select one of the
n + 1 items in our set of n + 1 items to be denoted as special. With this item held out we
partition the remaining n items into two sets a set of size k and its complement a set of size
n − k (we can take k values from {0, 1, 2, . . ., n}). Each of these partitions has n or fewer
elements. Specifically, the set of size k has Tk partitions. Lumping our special item with the
set of size n − k we obtain a set of size n − k + 1. Grouped with the set of size k we have
a partition of our original set of size n + 1. Since the number of k subset elements can be
chosen in

n
k

ways we have
1 +
n
X
k=1

n
k

Tk ,
possible partitions of the set {1, 2, . . . , n, n + 1}. Note that the one in the above formulation
represents the k = 0 set and corresponds to the relatively trivial partition consisting of the
entire set itself.
Problem 10
From the inclusion/exclusion principle we have
P(E ∪ F ∪ G) = P(E) + P(F) + P(G) − P(E ∩ F) − P(E ∩ G) − P(F ∩ G)
+ P(E ∩ F ∩ G)
Now consider the following decompositions of sets into mutually exclusive components
E ∩ F = (E ∩ F ∩ Gc
) ∪ (E ∩ F ∩ G)
E ∩ G = (E ∩ G ∩ Fc
) ∪ (E ∩ G ∩ F)
F ∩ G = (F ∩ G ∩ Ec
) ∪ (F ∩ G ∩ E) .
Since each set above is mutually exclusive we have that
P(E ∩ F) = P(E ∩ F ∩ Gc
) + P(E ∩ F ∩ G)
P(E ∩ G) = P(E ∩ G ∩ Fc
) + P(E ∩ G ∩ F)
P(F ∩ G) = P(F ∩ G ∩ Ec
) + P(F ∩ G ∩ E) .
Adding these three sets we have that
P(E∩F)+P(E∩G)+P(F∩G) = P(E∩F∩Gc
)+P(E∩F∩Fc
)+P(F∩G∩Ec
)+3P(E∩F∩G) ,
which when put into the inclusion/exclusion identity above gives the desired result.
Problem 11 (Bonferroni’s inequality)
From the inclusion/exclusion identity for two sets we have
P(E ∪ F) = P(E) + P(F) − P(EF) .

Since P(E ∪ F) ≤ 1, the above becomes
P(E) + P(F) − P(EF) ≤ 1 .
or
P(EF) ≥ P(E) + P(F) − 1 ,
which is known as Bonferroni’s inequality. From the numbers given we find that
P(EF) ≥ 0.9 + 0.8 − 1 = 0.7 .
Problem 12 (exactly one of E or F occurs)
Exactly one of the events E or F occurs is given by the probability of the set
(EFc
) ∪ (Ec
F) .
Since the two sets above are mutually exclusive the probability of this set is given by
P(EFc
) + P(Ec
F) .
Since E = (EFc
) ∪ (EF), we then have that P(E) can be expressed as
P(E) = P(EFc
) + P(EF) .
In the same way we have for P(F) the following
P(F) = P(Ec
F) + P(EF) .
so the above expression for our desired event (exactly one of E or F occurring) using these
two expressions for P(E) and P(F) is given by
P(EFc
) + P(Ec
F) = P(E) − P(EF) + P(F) − P(EF)
= P(E) + P(F) − 2P(EF) ,
as requested.
Problem 13 (E and not F)
Since E = EF ∪ EFc
, and both sets on the right hand side of this equation are mutually
exclusive we find that
P(E) = P(EF) + P(EFc
) ,
or solving for P(EFc
) we find
P(EFc
) = P(E) − P(EF) ,
as expected.

Problem 15 (drawing k white balls from r total)
This is given by
Pk =

M
k

N
r − k

M + N
r
for k ≤ r .
Problem 16 (more Bonferroni)
From Bonferroni’s inequality for two sets P(EF) ≥ P(E) + P(F) − 1, when we apply this
identity recursively we see that
P(E1E2E3 · · ·En) ≥ P(E1) + P(E2E3 · · · En) − 1
≥ P(E1) + P(E2) + P(E3E4 · · · En) − 2
≥ P(E1) + P(E2) + P(E3) + P(E4 · · · En) − 3
≥ · · ·
≥ P(E1) + P(E2) + · · · + P(En) − (n − 1) .
That the final term is n − 1 can be verified to be correct by evaluating this expression for
n = 2 which yields the original Bonferroni inequality.
Problem 18 (the number of sequences with no consecutive heads)
If the first flip lands tails then we have fn−1 sequences that have n total flips and no con-
secutive heads (and that all start with a tail). If instead we get a head on the first flip then
we cannot get a head on the second flip or we will have had two consecutive heads. In other
words we must flip a tail for the second flip in order to count these sequences. Thus we have
fn−2 additional sequences that we must count in this case. In total then we find
fn = fn−1 + fn−2 .
Note that f1 = 2 since we can toss either a head or a tail to not get two consecutive heads.
We note that f2 = 3 since we can through a HT, TT, or a TH and not get two consecutive
heads. When we take n = 2 in the above we get
f2 = f1 + f0 ⇒ 3 = 2 + f0 ,
so f0 = 1. The probability is given by Pn = fn
2n . Thus we need to compute f10 using the
above recursion relationship.

Problem 19
k-balls will be with drawn if there are r − 1 red balls in the first k − 1 draws and the kth
draw is the rth red ball. This happens with probability
P =

n
r − 1

m
k − 1 − (r − 1)

n + m
k − 1
·

n − (r − 1)
1

n + m − (k − 1)
1

=

n
r − 1

m
k − 1 − (r − 1)

n + m
k − 1
·

n − (r − 1)
n + m − (k − 1)

.
Here the first probability is that required to obtain r−1 red balls from n and k−1−(r−1) =
k −r blue balls from m. The next probability is the one requested to obtain the last kth red
ball.
Problem 21 (counting total runs)
Following the example from 5o if we assume that we have an even number of total runs i.e.
say 2k, then we have two cases for the distribution of the win and loss runs. The wins and
losses runs must be interleaved since we have the same number of each i.e. k, so we can start
with a loosing block and end with a winning block or start with a winning block and end
with a loosing block as in the following diagram
L L . . . L , W W . . . W , L . . . L , W W . . . W
W W . . . W , L L . . . L , W . . . W , L L . . . L .
In either case, the number of wins including all winning streaks i must sum to the total
number of wins n and the number of losses in all loosing streaks i must sum to the total
number of losses. In equations, using xi to denote the number of wins in the i-th winning
streak and yi to denote the number of losses in the i-th loosing streak we have that
x1 + x2 + . . . + xk = n
y1 + y2 + . . . + yk = m .
Under the constraint that xi ≥ 1 and yi ≥ 1 since we are told that we have exactly k wins
and losses (and therefore can’t remove any of the unknowns. The number of solutions to the
first and second equation above are given by

n − 1
k − 1

and

m − 1
k − 1

.
Giving a total count on the number of possible situations where we have k winning streaks
and k loosing streaks of
2 ·

n − 1
k − 1

·

m − 1
k − 1

Note that the “two” in the above formulation accounts for the two possibilities, i.e. we begin
with a winning or loosing streak. Combined this give a probability of
2 ·

n − 1
k − 1

·

m − 1
k − 1

n + m
n
.
If instead we are told that we have a total of 2k+1 runs as an outcome we could have one more
winning streak than loosing streak or corresponding one more loosing streak than winning
streak. Assuming that we have one more winning streak than loosing our distribution of
wins and looses looks schematically like the following
W W . . . W , L L . . . L , W W . . . W , L . . . L , W W . . . W
Then counting the total number of wins and losses with our xi and yi variables we must have
in this case
x1 + x2 + . . . + xk + xk+1 = n
y1 + y2 + . . . + yk = m .
The first equation has

n − 1
k + 1 − 1

=

n − 1
k

solutions and the second has

m − 1
k − 1

.
If instead we have one more loosing streak than winning our distribution of wins and looses
looks schematically like the following
L L . . . L , W W . . . W , L L . . . L , W . . . W , L L . . . L
Then counting the total number of wins and losses with our xi and yi variables we must have
in this case
x1 + x2 + . . . + xk = n
y1 + y2 + . . . + yk + yk+1 = m .
The first equation has

n − 1
k − 1

solutions and the second has

m − 1
k + 1 − 1

=

m − 1
k

.
Since either of these two mutually exclusive cases can occur the total number is given by

n − 1
k

·

m − 1
k − 1

+

n − 1
k − 1

·

m − 1
k

.
Giving a probability of

n − 1
k

·

m − 1
k − 1

+

n − 1
k − 1

·

m − 1
k

n + m
n
.
as expected.

Problem 1 (a cafeteria sample space)
Part (a): We have two choices for the entree, three choices for the starch, and four choices
for the dessert giving 2 · 3 · 4 = 24 total outcomes in the sample space.
Part (b): Now we have two choices for the entrees, and three choices for the starch giving
six total outcomes.
Part (c): Now we have three choices for the starch and four choices for the desert giving
12 total choices.
Part (d): The event A ∩ B means that we pick chicken for the entree and ice cream for the
desert, so the three possible outcomes correspond to the three possible starches.
Part (e): We have two choices for an entree and four for a desert giving eight possible
choices.
Part (f): This event is a dinner of chicken, rice, and ice cream.
Problem 2 (purchasing suits and ties)
Let Su, Sh , and T be the events that a person purchases a suit, a shirt, and a tie respectively.
Then the problem gives the information that
P(Su) = 0.22 P(Sh) = 0.3 P(T) = 0.28
P(Su ∩ Sh) = 0.11 P(Su ∩ T) = 0.14 P(Sh ∩ T) = 0.1
and P(Su ∩ Sh ∩ T) = 0.06.
Part (a): This is the event P((Su ∪ Sh ∪ T)c
), which we see is given by
P((Su ∪ Sh ∪ T)c
) = 1 − P(Su ∪ Sh ∪ T)
= 1 − P(Su) − P(Sh) − P(T) + P(Su ∩ Sh) + P(Su ∩ T)
+ P(Sh ∩ T) − P(Su ∩ Sh ∩ T)
= 1 − 0.22 − 0.3 − 0.28 + 0.11 + 0.14 + 0.1 − 0.06 = 0.49 .
Part (b): Exactly one item means that we want to evaluate each of the following three
mutually exclusive events
P(Su ∩ Sc
h ∩ Tc
) and P(Sc
u ∩ Sh ∩ Tc
) and P(Sc
u ∩ Sc
h ∩ T)

and add the resulting probabilities up. We note that problem thirteen from this chapter
was solved in this same way. To compute this probability we will begin by computing the
probability that two or more items were purchased. This is the event
(Su ∩ Sh) ∪ (Su ∩ T) ∪ (Sh ∩ T) ,
which we denote by E2 for shorthand. Using the inclusion/exclusion identity we have that
the probability of the event E2 is given by
P(E2) = P(Su ∩ Sh) + P(Su ∩ T) + P(Sh ∩ T)
− P(Su ∩ Sh ∩ Su ∩ T) − P(Su ∩ Sh ∩ Sh ∩ T) − P(Su ∩ T ∩ Sh ∩ T)
+ P(Su ∩ Sh ∩ Su ∩ T ∩ Sh ∩ T)
= P(Su ∩ Sh) + P(Su ∩ T) + P(Sh ∩ T)
− P(Su ∩ Sh ∩ T) − P(Su ∩ Sh ∩ T) − P(Su ∩ Sh ∩ T) + P(Su ∩ Sh ∩ T)
= P(Su ∩ Sh) + P(Su ∩ T) + P(Sh ∩ T) − 2P(Su ∩ Sh ∩ T)
= 0.11 + 0.14 + 0.1 − 2(0.06) = 0.23 .
If we let E0 and E1 be the events that we purchase no items or one item, then the probability
that we purchase exactly one item must satisfy
1 = P(E0) + P(E1) + P(E2) ,
which we can solve for P(E1). We find that
P(E1) = 1 − P(E0) − P(E2) = 1 − 0.49 − 0.23 = 0.28 .
Problem 3 (the fourteenth card is an ace)
Since the probability that any one specific card is the fourteenth is 1/52 and we have four
ways of getting an ace in the fourteenth spot we have a probability given by
4
52
=
1
13
.
Another way to solve this problem is to recognized that we have 52! ways of ordering the
52 cards in the deck. Then the number of ways that the fourteenth card can be an ace is
given by the fact that we have four choices for the ace in the fourteenth position and then
the requirement that we need to place 52 − 1 = 51 other cards in 51! ways so we have a
probability of
4(51!)
52!
=
4
52
=
1
13
.
To have the first ace occurs in the fourteenth spot we have to pick thirteen cards to place in
the thirteen slots in front of this ace (from the 52 − 4 = 48 “non” ace cards). This can be
done in
48 · 47 · 46 · · ·(48 − 13 + 1) = 48 · 47 · 46 · · ·36 ,
ways. Then we have four choices for the ace to pick in the fourteenth spot, then finally we
have to place the remaining 52−14 = 38 cards in 38! ways. Thus our probability is given by
(48 · 47 · 46 · · ·36) · 4 · (38!)
52!
= 0.03116 .

Problem 4 (temperatures)
Let A = {tLA = 70} be the event that the temperature in LA is 70. Let B = {tNY = 70}
be the event that the temperature in NY is 70. Let C = {max(tLA, tNY ) = 70} be the event
that the max of the two temperatures is 70. Let D = {min(tLA, tNY ) = 70} be the event
that the min of the two temperatures is 70. We note that C ∩D = A∩B and C ∪D = A∪B.
Then we want to compute P(D). Since
P(C ∪ D) = P(C) + P(D) − P(C ∩ D) ,
by the inclusion/exclusion identity for two sets. We also have
P(C ∪ D) = P(A ∪ B) = P(A) + P(B) − P(A ∩ B)
= P(A) + P(B) − P(C ∪ D)
By the relationship C ∪ D = A ∪ B and the inclusion/exclusion identity for A and B. We
can equate these two expressions to obtain
P(A) + P(B) − P(C ∩ D) = P(C) + P(D) − P(C ∩ D) ,
or
P(D) = P(A) + P(B) − P(C) = 0.3 + 0.4 − 0.2 = 0.5 .
Problem 5 (the top four cards)
Part (a): There are 52! arrangements of the cards. Then we have 52 choices for the first
card, 52 − 4 = 48 choices for the second card, 52 − 4 − 4 = 42 choices for the third card etc.
This gives a probability of
52 · 48 · 42 · 38(52 − 4)!
52!
= 0.613 .
Part (b): For different suits we have 52! total arrangements and to impose that constraint
that the top four all have different suits we have 52 choices for the first and then 52−13 = 39
choices for the second card, 39−13 = 26 choices for the third card etc. This gives a probability
of
52 · 39 · 26 · (52 − 4)!
52!
= 0.1055 .
Problem 6 (balls of the same color)
We have this probability given by

3
1

4
1

6
1

10
1
+

3
1

6
1

6
1

10
1
=
1
2
.

Where the first term is the probability that the first ball drawn is red and the second term
is the probability that the second ball is drawn is black.
Problem 7 (the state lottery)
Part (a): We have
1

40
8
= 1.3 10−8
,
since there is only one way to get all eight numbers.
Part (b): We have

8
7

40 − 8
1

40
8
=

8
7

32
1

40
8
= 3.3 10−6
.
Part (c): To solve this part we now need the probability of selecting six numbers which is
given by
8
6

40 − 8
2

40
8
,
which must be added to the probabilities in Part (a) and Part (b).
Problem 8 (committees)
Part (a): We have
3
1

4
1

4
1

3
1

3 + 4 + 4 + 3
4
=
3 · 4 · 4 · 3

14
4
.
Part (b): We have
4
2

4
2

14
4
.

Part (c): We can have no sophomores and four junior or one sophomore and three juniors
or two sophomores and two juniors or three sophomores and one juniors or four sophomores
and zero juniors. So our probability is given by

4
0

4
4

+

4
1

4
3

+

4
2

4
2

+

4
3

4
1

+

4
4

4
0

14
4

From Problem 9 on Page 19 with

n
k

=

n
n − k

, the sum in the numerator is given
by
2(4)
4

=

8
4

.
Problem 9 (number of elements in various sets)
Both of these claims follow directly from the inclusion-exclusion identity if we assume that
every element in our finite universal set S (with n elements) is equally likely and has prob-
ability 1/n.
Problem 10 (horse experiments)
We have N(A) = 3 · 5! = 3 · 120 = 360. We have N(B) = 5! = 120, and
N(A ∩ B) = 2 · 4! = 2 · 4 · 3 · 2 · 1 = 2(12 · 2) = 48 .
The union gives
N(A ∪ B) = N(A) + N(B) − N(A ∩ B) = 720 + 120 − 48 = 432 .
Problem 11
We have

52
5

possible five card hands from our fifty-two cards. To have one card from
each of the four suits we need to count the number of ways to select one club from the
thirteen available, (this can be done in

13
1

ways) one spade from the thirteen available,
(this can be done in

13
1

ways), one spade from the thirteen available in

13
1

ways
etc. The last card can be selected in

52 − 4
1

,

ways. Thus we have

13
1
4
48
1

possible hands containing one card from each suit,
where the order of the choice made in the

48
1

selections and the corresponding selection
from the

13
1

that has a suit that matches the

48
1

selection mater. To better explain
this say when picking clubs we get the three card. When we pick from the 48 remaining
cards (after having selected a card of each suit) assume we select a four of clubs. This hand
is equivalent to having picked the four of clubs first and then the three of clubs. So we must
divide the above by a 2! giving a probability of
1
2!

13
1
4
48
1

52
5
= 0.2637 .
Problem 12 (basketball choices)
We have 10! ways of permutations of all the player (frontcourt and backcourt considered the
same). Grouping the players list into pairs, we have five pairs and since the order within
each pair does not matter we have 10!
25 divisions of the ten players into a first roommate pair,
a second roommate pair etc. Since the ordering of the roommate pairs does not matter we
have 10!
255!
pairs of roommates to choose from. Now there are

6
2

4
2

,
ways of selecting the two frontcourt and backcourt player and 2! ways of assigning them.
We then have to create roommate pairs from only frontcourt and backcourt players. For the
frontcourt we use the following logic to derive the number of pairs of total players
4!
22(2!)
= 3 .
For the backcourt players we have
2!
21(1!)
= 1 ,
so we have a probability of

6
2

4
2

· 2 · 3 · 1
10!
255!
= 0.5714 .

Problem 13 (random letter)
The same letter could be chosen if and only if it comes from one of R, E, or V . The
probability of R is chosen from both words is

2
7

1
8

=
2
56
.
The probability of E is chosen from both words is

3
7

1
8

=
3
56
.
Finally the probability of V is chosen from both words is
1
7

1
8

=
1
56
.
So the total probability is the sum of all the above probabilities or
6
56
=
3
28
.
Problem 14 (Boole’s inequality)
We begin by decomposing the countable union of sets Ai
A1 ∪ A2 ∪ A3 . . .
into a countable union of disjoint sets Cj. Define these disjoint sets as
C1 = A1
C2 = A2A1
C3 = A3(A1 ∪ A2)
C4 = A4(A1 ∪ A2 ∪ A3)
.
.
.
Cj = Aj(A1 ∪ A2 ∪ A3 ∪ · · · ∪ Aj−1)
Then by construction
A1 ∪ A2 ∪ A3 · · · = C1 ∪ C2 ∪ C3 · · · ,
and the Cj’s are disjoint, so that we have
Pr(A1 ∪ A2 ∪ A3 ∪ · · ·) = Pr(C1 ∪ C2 ∪ C3 ∪ · · · ) =
X
j
Pr(Cj) .
Since Pr(Cj) ≤ Pr(Aj), for each j, this sum is bounded above by
X
j
Pr(Aj) ,

Problem 15
From the fact that ∩iAi is a set, its probability must be less than or equal to 1, that is
1 ≥ P(∩iAi) = P((∪iAc
i )c
) = 1 − P(∪iAc
i ) .
By Boole’s inequality we also have that
P(∪iAc
i ) ≤
∞
X
i=1
P(Ac
i ) =
∞
X
i=1
(1 − P(Ai)) .
But since P(Ai) = 1 each term in this sum is zeros and P(∪iAc
i ) ≤ 0. Thus
1 ≥ P(∩iAi) ≥ 1 − 0 = 1 ,
showing that P(∩iAi) = 1.
Problem 16 (the number of non-empty partitions of size k)
Let Tk(n) be the number of partitions of the set {1, 2, 3, · · · , n} into k nonempty subsets.
Computing this number can be viewed as a counting the number of partitions with the
singleton set {1} in them, and counting the number of partitions without the singleton set
{1} in them. If {1} is in a singleton set then we have used up one subset and are now looking
at the number of partitions of a set of size n − 1. Thus the number of partitions where {1}
is a singleton set must be Tk−1(n − 1). The number of partition where the element one is in
a partition is given by kTk(n−1), since Tk(n−1) gives the number of k partitions of a set of
size n − 1 and we can insert the element 1 into any of these k sets to derive a k partition of
a set of n. Adding these two mutually exclusion results we obtain the following expression
for Tk(n)
Tk(n) = Tk−1(n − 1) + kTk(n − 1) .
Problem 17 (drawing balls from an urn)
Consider the complementary probability that no balls of a given color are chosen. For
example let R be the event that no red balls are chosen, W the event that no white balls
are chosen and B the event that no blue ball is chosen. The the desired probability is given
by the complement of P(R ∪ W ∪ B). By the inclusion/exclusion identity we have
P(R∪W ∪B) = P(R)+P(W)+P(B)−P(R∩W)−P(R∩B)−P(W ∩B)+P(R∩W ∩B) .

Now the individual probabilities are given by
P(R) =

13
5

18
5
, P(W) =

5 + 7
5

18
5
=

12
5

18
5

P(B) =

5 + 6
5

18
5
=

11
5

18
5

P(R ∩ W) =

7
5

18
5
, P(R ∩ B) =

6
5

18
5

P(W ∩ B) =

5
5

18
5
P(R ∩ W ∩ B) = 0 .
Then adding these results together gives P(R ∪ W ∪ B), and the desired result is 1 − P(R ∪
W ∪ B).

Chapter 3 (Conditional Probability and Independence)
Notes on the Text
The probability that event E happens before event F (page 93)
Let A be the event that E occurs before F (E and F are mutually exclusive). Here we are
envisioning independent trials where E or F or (E ∪ F)c
are the only possible occurrences
of each experiment. Then conditioning on each of these three events we have that
P(A) = P(A|E)P(E) + P(A|F)P(F) + P(A|(E ∪ F)c
)P((E ∪ F)c
)
= P(E) + (1 − P(E) − P(F))P(A) .
Since P(A|E) = 1, P(A|F) = 0 and P(A|(E ∪ F)c
) = P(A). Solving for P(A) gives
P(A) =
P(E)
P(E) + P(F)
. (5)
From the symmetry of that equation we have that the probability that F happens before
event E is then
P(Ac
) =
P(F)
P(E) + P(F)
.
The duration of play problem (page 98)
I found this section of the text difficult to understand at first and wrote this simple expla-
nation to help myself understand things better. In the ending arguments of example 4k,
Ross applies the gamblers ruin problem to the duration of play problem of Huygens. In the
duration of play problem if an eleven is thrown (with probability of 27
216
) the player B wins
a point, if a fourteen is thrown (with probability of 15
216
) the player A wins a point, while
anything else results in a continuation of the game. Since the outcome that A wins a point
will only happen if a fourteen is thrown before an eleven we need to compute that probability
to apply to the gamblers ruin problem. The probability that a fourteen is thrown before an
eleven is given by example 4h and equals
P(E)
P(E) + P(F)
=
15
216
27
216
+ 15
216
=
15
42
,
the number given for p in the text.

Problem Solutions
Problem 1 (fair dice)
Let E be the event that at least one die is a six and F the event that the two die lands on
different numbers. Then P(E|F) = P (EF )
P (F )
. The event F can be any of the following pairs
(1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 3), (2, 4), (2, 5), (2, 6), (3, 1), (3, 2), (3, 4),
(3, 5), (3, 6), (4, 1), (4, 2), (4, 3), (4, 5), (4, 6), (5, 1), (5, 2), (5, 3), (5, 4), (5, 6), (6, 1),
(6, 2), (6, 3), (6, 4), and (6, 5) ,
which has thirty elements giving a probability P(F) = 30
36
= 5
6
.
The event EF consist of the event where at least one die is a six and the other two die have
different numbers. The elements of this set are given by
(1, 6), (2, 6), (3, 6), (4, 6), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5) ,
which has ten elements so P(EF) = 10
36
= 5
18
. With these two results we have
P(E|F) =
P(EF)
P(F)
=
(5/18)
(5/6)
=
6
18
=
1
3
.
Problem 2 (more fair dice)
Let Ei be the event that the sum of the two dice is i. Let F6 denote the event that the first
die is a six. Then we want to compute P(F6|Ei) for i = 2, 3, · · · , 12. This expression is given
by
P(F6 ∩ Ei)
P(Ei)
.
From Problem 24 from Chapter 2 we know the values of P(Ei) for i = 2, 3, · · · , 12. Thus we
only need to compute the events F6 ∩ Ei for each i. We have (if φ is the empty set)
F6 ∩ E2 = φ, F6 ∩ E4 = φ, F6 ∩ E6 = φ , F6 ∩ E3 = φ, F6 ∩ E5 = φ, F6 ∩ E7 = {6, 1} ,
F6 ∩ E8 = {6, 2}, F6 ∩ E9 = {6, 3}, F6 ∩ E10 = {6, 4} , F6 ∩ E11 = {6, 5}, and
F6 ∩ E12 = {6, 6} ,
Thus if F6 ∩ Ei = φ then P(F6 ∩ Ei) = 0, while if F6 ∩ Ei 6= φ, then P(F6 ∩ Ei) = 1
36
So we
get
P(F6|E2) = 0, P(F6|E3) = 0, P(F6|E4) = 0, P(F6|E5) = 0, P(F6|E6) = 0 .

along with
P(F6|E7) =
1/36
P(E7)
=
1/36
6/36
=
1
6
P(F6|E8) =
1/36
P(E8)
=
1/36
5/36
=
1
5
P(F6|E9) =
1/36
P(E9)
=
1/36
4/36
=
1
4
P(F6|E10) =
1/36
P(E10)
=
1/36
3/36
=
1
3
P(F6|E11) =
1/36
P(E11)
=
1/36
2/36
=
1
2
P(F6|E12) =
1/36
P(E12)
=
1/36
1/36
= 1 .
Problem 3 (hands of bridge)
Equation 2.1 in the book is
p(E|F) =
p(EF)
p(F)
.
To use this equation for this problem, let E be the event that East has three spades and F
be the event that the combined North-South pair has eight spades. Then
P(F) =

13
8

39
18

52
26
.
This can be reasoned as follows. We have thirteen total spades from which we should pick
eight to the give the North-South pair (the rest will go to the East-West pair). This gives
the factor

13
8

. We then have 52 − 13 = 39 other cards (non-spades) from which to pick
the remaining 26 − 8 = 18 cards to make the required total of 26 cards for the North-South
pair. This gives the factor

39
18

. The product of these two expressions gives the total
number of ways we can obtain the stated condition. This product is divided the number of
ways to select 26 cards from the deck of 52 total cards. When we evaluate the above fraction
we find P(F) = 9102/56243 = 0.161833.
Now the joint event EF means that East has three spades and North-South has eight spades
so that West must have 13 − 3 − 8 = 2 spades. Thus to evaluate P(EF) the approach we
take is to enumerate the required number and type of cards to East and then do the same
for West. For each player we do this in two parts, first the number of spade cards and then

the number of non-spade cards. Using this logic we find that
P(EF) =

13
3

39
10

10
2

52 − 13 − 10
11

52
13

39
13
.
This can be reasoned as follows. The first factor

13
3

in the numerator is the number of
ways we can select three required spades for East. The second factor

39
10

is the number
of ways we can select the remaining 13 − 3 = 10 non-spade cards for East. The third factor

10
2

is the number of ways we can select the required two spade cards for West. We then
have 52 − 13 − 10 remaining possible non-spade cards from which we need to draw 11 to
complete the hand of West. This gives the factor

52 − 13 − 10
11

. The denominator is
the number of ways we can draw East and West’s hands without any restrictions. When we
evaluate the above fraction we find P(EF) = 2397/43675 = 0.054883.
With these two results we see that P(E|F) = 39/115 = 0.339130, the same as in the back
of the book. See the Matlab/Octave file chap 3 prob 3.m for the evaluation of the above
fractions.
Problem 4 (at least one six)
This is solved in the same way as in problem number 2. In solving we will let E be the event
that at least one of the pair of dice lands on a 6 and “X = i” be shorthand for the event the
sum of the two dice is i. Then we desire to compute
p(E|X = i) =
P(E, X = i)
p(X = i)
.
We begin by computing p(X = i) for i = 2, 3, 4, · · · , 12. We find that
p(X = 2) =
1
36
, p(X = 8) =
5
36
p(X = 3) =
2
36
, p(X = 9) =
4
36
p(X = 4) =
3
36
, p(X = 10) =
3
36
p(X = 5) =
4
36
, p(X = 11) =
2
36
p(X = 6) =
5
36
, p(X = 12) =
1
36
p(X = 7) =
6
36
.

We next compute p(E, X = i) for i = 2, 3, 4, · · · , 12 we find that
p(E, X = 2) = 0 , p(E, X = 8) =
2
36
p(E, X = 3) = 0 , p(E, X = 9) =
2
36
p(E, X = 4) = 0 , p(E, X = 10) =
2
36
p(E, X = 5) = 0 , p(E, X = 11) =
2
36
p(E, X = 6) = 0 , p(E, X = 12) =
1
36
p(E, X = 7) =
2
36
.
Finally computing our conditional probabilities we find that
P(E|X = 2) = p(E|X = 3) = p(E|X = 4) = p(E|X = 5) = p(E|X = 6) = 0 .
and
p(E|X = 7) =
1
3
, p(E|X = 10) =
2
3
p(E|X = 8) =
2
5
, p(E|X = 11) =
2
2
= 1
p(E|X = 9) =
1
2
, p(E|X = 12) =
1
1
= 1 .
Problem 5 (the first two selected are white)
We have that
P =

6
2

9
2

15
4

is the probability of drawing two white balls and two black balls independently of the order
of the draws. Since we are concerned with the probability of an ordered sequence of draws
we should enumerate these. Let W by the event that the first two balls are white and B
the event that the second two balls are black. Then we desire the probability P(W ∩ B) =
P(W) P(B|W). Now
P(W) =

6
2

15
2
=
15
105
≈ 0.152
and
P(B|W) =

9
2

13
2
=
36
78
≈ 0.461

so that P(W ∩ B) = 0.0659 = 6
91
.
Problem 6 (exactly three white balls)
Let F be the event that the first and third drawn balls are white and let E be the event that
the sample contains exactly three white balls. Then we desire to compute P(F|E) = P (F ∩E)
P (E)
.
Working the without replacement we have that
P(E) =

8
3

·

4
1

12
4
=
224
495
.
and P(F ∩E) is the probability that our sample has three white balls and the first and third
balls are white. To calculate this we can explicitly enumerate the possibilities in F ∩ E as
{(W, W, W, B), (W, B, W, W)}, showing that
P(F ∩ E) =
2

12
4
.
Given these two results we then have that
P(F|E) =
2

8
3

·

4
1
=
1
112
.
To work the problem with replacement we have that
P(E) =

4
3

2
3
3
1
3

=
25
34
.
As before we can enumerate the sample in E ∩ F. This set is {(W, W, W, B), (W, B, W, W)},
and has probabilities given by

2
3
3
1
3
+

2
3
3
1
3
=
24
34
.
so the probabilities we are after is
24
34
25
34
=
1
2
.
Problem 7 (the king’s sister)
The two possible children have a sample space given by
{(M, M), (M, F), (F, M), (F, F)} ,

each with probability 1/4. Then if we let E be the event that one child is a male and F be
the event that one child is a female and one child is a male, the probability that we want to
compute is given by
P(F|E) =
P(FE)
P(E)
.
Now
P(E) =
1
4
+
1
4
+
1
4
=
3
4
.
and FE consists of the set {(M, F), (F, M)} so
P(FE) =
1
2
,
so that
P(F|E) =
1/2
3/4
=
2
3
.
Problem 8 (two girls)
Let F be the event that both children are girls and E the event that the eldest child is a
girl. Now P(E) = 1
4
+ 1
4
= 1
2
and the event EF has probability 1
4
. Then
P(F|E) =
P(FE)
P(E)
=
1/4
1/2
=
1
2
.
Problem 9 (a white ball from urn A)
Let F be the event that the ball chosen from urn A was white. Let E be the event that
two while balls were chosen. Then the desired probability is P(F|E) = P (F E)
P (E)
. Lets first
calculate P(E) or the probability that two white balls were chosen. This event can happen
in the following mutually exclusive draws
(W, W, R) , (W, R, W) , (R, W, W) .
We can calculate the probabilities of each of these events
• The first draw will happen with probability 2
6
8
12
3
4

= 1
6
• The second draw will happen with probability 1
3
4
12
1
4

= 1
36
• The third draw will happen with probability 4
6
8
12
1
4

= 1
9
so that
P(E) =
1
6
+
1
36
+
1
9
=
11
36
.

Now FE consists of only the events {(W, W, R), (W, R, W)} since now the first draw must
be white. The event FE has probability given by 1
6
+ 1
36
= 7
36
, so that we find
P(F|E) =
7/36
11/36
=
7
11
= 0.636 .
Problem 10 (three spades given that we draw two others)
Let F be the event that the first card selected is a spade and E the event that the second
and third cards are spades. Then we desire to compute P(F|E) = P (F E)
P (E)
. Now P(E) is the
probability that the second and third cards are spades, which equals the union of two events.
The first is event that the first, second, and third cards are spades and the second is the
event that the first card is not a spade while the second and third cards are spades. Note
that this first event is also FE above. Thus we have
P(FE) =
13 · 12 · 11
52 · 51 · 50
Letting G be the event that the first card is not a spade while the second and third cards
are spades, we have that
P(G) =
(52 − 13) · 13 · 12
52 · 51 · 50
=
39 · 13 · 12
52 · 51 · 50
,
so
P(E) =
39 · 13 · 12
52 · 51 · 50
+
13 · 12 · 11
52 · 51 · 50
=
11
39 + 11
=
11
50
= 0.22 .
Problem 11 (probabilities on two cards )
We are told to let B be the event that both cards are aces, As the event that the ace of
spades is chosen and A the event that at least one ace is chosen.
Part (a): We are asked to compute P(B|As). Using the definition of conditional probabil-
ities we have that
P(B|As) =
P(BAs)
P(As)
.
The event BAs is the event that both cards are aces and one is the ace of spades. This event
can be represented by the sample space
{(AD, AS), (AH, AS), (AC, AS)} .
where D, S, H, and C stand for diamonds, spades, hearts, and clubs respectively and the
order of these elements in the set above does not matter. So we see that
P(BAs) =
3

52
2
.

The event As is given by the set {AS, ∗} where ∗ is a wild-card denoting any of the possible
fifty-one other cards besides the ace of spades. Thus we see that
P(As) =
51

52
2
.
These together give that
P(B|As) =
3
51
=
1
17
.
Part (b): We are asked to compute P(B|A). Using the definition of conditional probabilities
we have that
P(B|A) =
P(BA)
P(A)
=
P(B)
P(A)
.
The event B are the hand {(AD, AS), (AD, AH), (AD, · · ·)} and has

4
2

elements i.e.
from the four total aces select two. So that
P(B) =

4
2

52
2
.
The set A is the event that at least one ace is chosen. This is the complement of the set that
no ace is chosen. No ace can be chosen in

48
2

ways so that
P(A) = 1 −

48
2

52
2
=

52
2

−

48
2

52
2
.
This gives for P(B|A) the following
P(B|A) =

4
2

52
2

−

48
2
=
6
198
=
1
33
.
Problem 12 (passing the actuarial exams)
We let Ei be the event that the ith actuarial exam is passed. Then the given probabilities
can be expressed as
P(E1) = 0.9 , P(E2|E1) = 0.8 , P(E3|E1, E2) = 0.7 .

Part (a): The desired probability is given by P(E1E2E3) or conditioning we have
P(E1E2E3) = P(E1)P(E2|E1)P(E3|E1E2) = 0.9 · 0.8 · 0.7 = 0.504 .
Part (b): The desired probability is given by P(Ec
2|(E1E2E3)c
) and can be expressed using
the set identity
(E1E2E3)c
= E1 ∪ (E1Ec
2) ∪ (E1E2Ec
3) ,
are the only ways that one can not pass all three tests i.e. one must fail one of the first three
tests. Note that these sets are mutually independent. Now
P(Ec
2|(E1E2E3)c
) =
P(Ec
2 (E1E2E3)c
)
P((E1E2E3)c)
.
We know how to compute P((E1E2E3)c
) because it is equal to 1 − P(E1E2E3) and we can
compute P(E1E2E3). From the above set identity the event Ec
2 (E1E2E3)c
is composed of
only one set, namely E1Ec
2, since if we don’t pass the second test we don’t take the third
test. We now need to evaluate the probability of this event. We find
P(E1Ec
2) = P(Ec
2|E1)P(E1)
= (1 − P(E2|E1))P(E1)
= (1 − 0.8)(0.9) = 0.18 .
With this the conditional probability sought is given by 0.18
1−0.504
= 0.3629
Problem 13
Define p by p ≡ P(E1E2E3E4). Then by conditioning on the events E1, E1E2, and E1E2E3
we see that p is given by
p = P(E1E2E3E4)
= P(E1)P(E2E3E4|E1)
= P(E1)P(E2|E1)P(E3E4|E1E2)
= P(E1)P(E2|E1)P(E3|E1E2)P(E4|E1E2E3) .

So we need to compute each probability in this product. We have
P(E1) =

4
1

48
12

52
13

P(E2|E1) =

3
1

36
12

39
13

P(E3|E1E2) =

2
1

24
12

26
13

P(E4|E1E2E3) =

1
1

12
12

13
13
= 1 .
so this probability is then given by (when we multiply each of the above expressions)
p = 0.1055 .
See the Matlab file chap 3 prob 13.m for these calculations.
Problem 14
Part (a): We will compute this as a conditional probability since the number of each colored
balls depend on the results from the previous draws. Let Bi be the event that a black ball
is selected on the ith draw and Wi the event that a white ball is selected on the ith draw.
Then the probability we are looking for is given by
P(B1B2W3W4) = P(B1)P(B2|B1)P(W3|B1B2)P(W4|B1B2W3)
=

7
5 + 7

9
5 + 9

5
5 + 11

7
7 + 11

= 0.0455 .
See the Matlab file chap 3 prob 14.m for these calculations.
Part (b): The set discussed is given by the

4
2

= 6 sets given by
(B1, B2, W3, W4) , (B1, W2, B3, W4) , (B1, W2, W3, B4)
(W1, B2, B3, B4) , (W1, B2, W3, B4) , (W1, W2, B3, B4) .
The probabilities of each of these events can be computed as in Part (a) of this problem.
The probability requested is then the sum of the probabilities of all these mutually exclusive
events.

Problem 15 (ectopic pregnancy among smokers )
Let S be the event a woman is a smoker and E the event that a woman has an ectopic preg-
nancy. Then the information given in the problem statement is that P(E|S) = 2P(E|Sc
),
P(S) = 0.32, P(Sc
) = 0.68, and we want to calculate P(S|E). We have using Bayes’ rule
that
P(S|E) =
P(E|S)P(S)
P(E|S)P(S) + P(E|Sc)P(Sc)
=
2P(E|Sc
)(0.32)
2P(E|Sc)(0.32) + P(E|Sc)(0.68)
=
2(0.32)
2(0.32) + 0.68
= 0.4848 .
Problem 16 (surviving a Cesarean birth)
Let C be the event of a Cesarean section birth, let S be the event that the baby survives.
The facts given in the problem are that
P(S) = 0.98 , P(Sc
) = 0.02 , P(C) = 0.15 , P(Cc
) = 0.85 , P(S|C) = 0.96 .
We want to calculate P(S|Cc
). We can compute P(S) by C (the type of birth) as
P(S) = P(S|C)P(C) + P(S|Cc
)P(Cc
) .
Using the information given in the problem into the above we find that
0.98 = 0.96(0.15) + P(S|Cc
)(0.85) ,
or that P(S|Cc
) = 0.983.
Problem 17 (owning pets)
Let D be the event a family owns a dog, and C the event that a family owns a cat. Then
from the numbers given in the problem we have that P(D) = 0.36, P(C) = 0.3, and
P(C|D) = 0.22.
Part (a): We are asked to compute P(CD) = P(C|D)P(D) = 0.22 · 0.36 = 0.0792.
Part (b): We are asked to compute
P(D|C) =
P(C|D)P(D)
P(C)
=
0.22 · (0.36)
0.3
= 0.264 .

Problem 18 (types of voters)
Let I, L, and C be the event that a random person is an independent, liberal, or a conser-
vative respectfully. Let V be the event that a person voted. Then from the problem we are
given that
P(I) = 0.46 , P(L) = 0.3 , P(C) = 0.24 ,
and
P(V |I) = 0.35 , P(V |L) = 0.62 , P(V |C) = 0.58 .
We want to compute P(I|V ), P(L|V ), and P(C|V ) which by Bayes’ rule are given by (for
P(I|V ) for example)
P(I|V ) =
P(V |I)P(I)
P(V )
=
P(V |I)P(I)
P(V |I)P(I) + P(V |L)P(L) + P(V |C)P(C)
.
All desired probabilities will need to calculate P(V ) which we do (as above) by conditioning
on the various types of voters. We find that it is given by
P(V ) = P(V |I)P(I) + P(V |L)P(L) + P(V |C)P(C)
= 0.35(0.46) + 0.62(0.3) + 0.58(0.24) = 0.4862 .
Then the requested conditional probabilities are given by
P(I|V ) =
0.35(0.46)
0.48
= 0.3311
P(L|V ) =
P(V |L)P(L)
P(V )
=
0.62(0.3)
0.4862
= 0.38256
P(C|V ) =
P(V |C)P(C)
P(V )
=
0.58(0.24)
0.4862
= 0.2863 .
Part (d): This is P(V ) which from Part (c) we know to be equal to 0.48.
Problem 19 (attending a smoking success party)
Let M be the event a person who attends the party is male, W the event a person who
attends the party is female, and E the event that a person was smoke free for a year. The
problem gives
P(E|M) = 0.37 , P(M) = 0.62 , P(E|W) = 0.48 , P(W) = 1 − P(M) = 0.38 .
Part (a): We are asked to compute P(W|E) which by Bayes’ rule is given by
P(W|E) =
P(E|W)P(W)
P(E)
=
P(E|W)P(W)
P(E|W)P(W) + P(E|M)P(M)
=
0.48(0.38)
0.48(0.38) + 0.37(0.62)
= 0.442 .

Part (b): For this part we want to compute P(E) which by conditioning on the sex of the
person equals P(E) = P(E|W)P(W) + P(E|M)P(M) = 0.4118.
Problem 20 (majoring in computer science)
Let F be the event that a student is female. Let C be the event that a student is majoring
in computer science. Then we are told that P(F) = 0.52, P(C) = 0.05, and P(FC) = 0.02.
Part (a): We are asked to compute P(F|C) = P (F C)
P (C)
= 0.02
0.05
= 0.4.
Part (b): We are asked to compute P(C|F) = P (F C)
P (F )
= 0.02
0.52
= 0.3846.
Problem 21 (salaries for married workers)
We are given the following joint probabilities
P(W, H) =
212
500
= 0.424
P(W, H) =
198
500
= 0.396
P(W, H) =
36
500
= 0.072
P(W, H) =
54
500
= 0.108 .
Where the notation W is the event that the wife makes less than 25, 000, W is the event
that the wife makes more than 25, 000, H and H are the events that the husband makes
less than or more than 25, 000 respectively.
Part (a): We desire to compute P(H), which we can do by considering all possible situa-
tions involving the wife. We have
P(H) = P(H, W) + P(H, W) =
212
500
+
36
500
= 0.496 .
Part (b): We desire to compute P(W|H) which we do by remembering the definition
of conditional probability. We have P(W|H) = P (W,H)
P (H)
. Since P(H) = 1 − P(H) =
1 − 0.496 = 0.504 using the above we find that P(W|H) = 0.2142 = 3
14
.
Part (c): We have
P(W|H) =
P(W, H)
P(H)
=
0.072
0.496
= 0.145 =
9
62
.

Problem 22 (ordering colored dice)
Part (a): The probability that no two dice land on the same number means that each die
must land on a unique number. To count the number of such possible combinations we see
that there are six choices for the red die, five choices for the blue die, and then four choices for
the yellow die yielding a total of 6·5·4 = 120 choices where each die has a different number.
There are a total of 63
total combinations of all possible die through giving a probability of
120
63
=
5
9
Part (b): We are asked to compute P(B Y R|E) where E is the event that no two dice
land on the same number. From Part (a) above we know that the count of the number of
rolls that satisfy event E is 120. Now the number of rolls that satisfy the event B Y R
can be counted in a manner like Problem 6 from Chapter 1. For example, if R shows a roll
of three then the only possible valid rolls where B Y R for B and Y are B = 1 and
Y = 2. If R shows a four then we have

3
2

= 3 possible choices i.e. either
(B = 1, Y = 2) , (B = 1, Y = 3) , (B = 2, Y = 3) .
for the possible assignments to the two values for the B and Y die. If R = 5 we have

4
2

= 6 possible assignments to B and Y . Finally, if R = 6 we have

5
2

= 10 possible
assignments to B and Y . Thus we find that
P(B Y R|E) =
1 + 3 + 6 + 10
120
=
1
6
Part (c): We see that
P(B Y R) = P(B Y R|E)P(E) + P(B Y R|Ec
)P(Ec
) ,
Since P(B Y R|Ec
) = 0 from the above we have that
P(B Y R) =

1
6

5
9

=
5
54
.
Problem 23 (some urns)
Part (a): Let W be the event that the ball chosen from urn II is white. Then we should
solve this problem by conditioning on the color of the ball drawn from first urn. Specifically
P(W) = P(W|BI = w)P(BI = w) + P(W|BI = r)P(BI = r) .

Here BI = w is the event that the ball drawn from the first urn is white and BI = r is
the event that the the drawn ball is red. We know that P(BI = w) = 1
3
, P(BI = r) = 2
3
,
P(W|BI = w) = 2
3
, and P(W|BI = r) = 1
3
. We then have
P(W) =
2
3
·
1
3
+
1
3
·
2
3
=
2 + 2
9
=
4
9
Part (b): Now we are looking for
P(BI = w|W) =
P(W|BI = w)P(BI = w)
P(W)
.
Since everything is known in the above we can compute this as
P(BI = w|W) =
2
3
1
3

4
9
=
1
2
.
Problem 24 (painted balls in an urn)
Part (a): Let E be the event that both balls are gold and F the event that at least one
ball is gold. The probability we desire to compute is then P(E|F). Using the definition of
conditional probability we have that
P(E|F) =
P(EF)
P(F)
=
P({G, G})
P({G, G}, {G, B}, {B, G})
=
1/4
1/4 + 1/4 + 1/4
=
1
3
Part (b): Since now the balls are mixed together in the urn, the difference between the pair
{G, B} and {B, G} is no longer present. Thus we really have two cases to consider.
• Either both balls are gold or
• One ball is gold and the other is black.
Thus to have a second ball be gold will occur once out of these two choices and our probability
is then 1/2.
Problem 25 (estimating the number of people over fifty)
Let F denote the event that a person is over fifty and denote this probability by p which is
also the number we desire to estimate. Let α1 denote the proportion of the time a person
under fifty spends on the streets and α2 the same proportion for people over fifty. Let S

denote the event that a person (of any age) is found in the streets. Then this event S can be
decomposed into the sets where the person on the streets is less than or greater than fifty as
S = SF ∪ SFc
.
Since the two sets on the right-hand-side of this expression are disjoint we have
P(S) = P(SF) + P(SFc
) .
These sets can be written in terms of S conditional on the persons age F as
P(SF) = P(F)P(S|F) = pP(S|F)
P(SFc
) = P(Fc
)P(S|Fc
) = (1 − p)P(S|F) .
Now by taking measurements of the number/proportion of people over fifty during the day
as suggested by the initial part of this problem we are actually measuring the probability
P(F|S) ,
and not P(F). The expression P(F|S) is related to p and what we desire to measure by
P(F|S) =
P(SF)
P(S)
=
pP(S|F)
pP(S|F) + (1 − p)P(S|Fc)
.
Since we are told that α1 should be the proportion of time someone under the age of fifty
spends in the streets we can express this variable in terms of the above expressions simply
as P(S|Fc
). In the same way P(S|F) = α2. Using this notation we thus have
P(F|S) =
α2p
α2p + α1(1 − p)
=
α2p
α1 + (α2 − α1)p
.
From the above we see that if α1 = α2 we will have P(F|S) = p and we will have actually
measured what we intended to measure.
Problem 26 (colorblindness)
From the problem, assuming that CB represents the event that a person is colorblind, we
are told that
P(CB|M) = 0.05 , and P(CB|W) = 0.0025 .
We are asked to compute P(M|CB), which we will do by using the Bayes’ rule. We find
P(M|CB) =
P(CB|M)P(M)
P(CB)
.
We will begin by computing P(CB) by conditioning on the sex of the person. We have
P(CB) = P(CB|M)P(M) + P(CB|F)P(F)
= 0.05(0.5) + 0.0025(0.5) = 0.02625 .

Then using Bayes’ rule we find that
P(M|CB) =
0.05(0.5)
0.02625
= 0.9523 =
20
21
.
If the population consisted of twice as many males as females we would then have P(M) =
2P(F) giving P(M) = 2
3
and P(F) = 1
3
and our calculation becomes
P(CB) = 0.05

2
3

+ 0.0025

1
3

= 0.03416 .
so that
P(M|CB) =
0.05(2/3)
0.03416
= 0.9756 =
40
41
.
Problem 27 (counting the number of people in each car)
Since we desire to estimate the number of people in a given car, if we choose the first method
we will place too much emphasis on cars that carry a large number of people. For example
if we imagine that a large bus of people arrives then on average we will select more people
from this bus than from cars that only carry one person. This is the same effect as in the
discussion in the book about the number of students counted on various numbers of buses
and would not provide an unbiased estimate. The second method suggested would provide
an unbiased estimate and would be the preferred method.
Another way to see this is to recognize that this problem is testing an understanding of the
ideas of conditional probability. The question asks about the number of people in a car given
that the car is in the company parking lot (the second method). If we start our sampling
by looking at the person level (the first method) we will be counting people who may get
to work by other means (like walk, ride a bicycle, etc.). As far as the number of people in
each car in the parking lot is concerned we are not interested in these later people and they
should not be polled.
Problem 28 (the 21st card)
Part (a): Let F be the event the 20th card is the first ace and let E be the event the 21st
card is the ace of spades. For this part of the problem we want to compute P(E|F). From
the definition of conditional probability this can be written as
p(E|F) =
p(EF)
p(F)
.
Thus we can compute P(E|F) if we can compute P(F) and P(EF). We begin by computing
the value of P(F). To compute this probability we will count the number of ways we can
obtain the special card ordering denoted by event F and then divide this number by the
number of ways we can have all 52 cards ordered with no restrictions on their ordering. This

latter number is given by 52!. To compute the number of card ordering that given rise to
event F consider that in selecting the first card we can select any card that is not an ace
and thus have 52 − 4 = 48 cards to select from. To select the second card we have one less
or 47 cards to select from. Continuing this patter down to the 20th card we have
48 · 47 · 46 · · ·32 · 31 · 30 .
ways to select the cards up to the 20th. For the 20th we have four choices (any one of the
aces). After this card is selected we can select any card from the 52 − 20 = 32 remaining
cards for the 21st card. For the 22nd card we can select any of the 31 remaining cards. Thus
the number of ways to select the remaining block of cards can be done in 32! ways. In total
then we can compute P(F) as
P(F) =
(48 · 47 · 46 · · ·32 · 31 · 30)4(32!)
52!
=
992
54145
.
Next we need to compute P(EF). Since the event EF is similar to the event F but with
exception that the 20th card cannot be the ace of spaces (because the 21st card) the number
of ways we can get the event EF is given by
(48 · 47 · 46 · · ·32 · 31 · 30) · 3 · 1 · (31!) .
Thus the probability P(EF) is given by
P(EF) =
(48 · 47 · 46 · · ·32 · 31 · 30) · 3 · 1 · (31!)
52!
=
93
216580
.
Using these two results we compute
P(E|F) =
992/54145
93/216580
=
3
128
.
As an alternative method to compute these probabilities we can express the events E and
F as boolean combinations of simpler component events Ai, where this component event
describes whether the card at location i in the deck is an ace. The event F defined above
represents the case where the the first 19 cards are not aces while the 20th card is and can
be written in terms of these Ai events as
F = Ac
1 · · · Ac
19A20 .
With this product representation P(F) can be computed by conditioning as
P(F) = P(Ac
1 · · · Ac
19A20) = P(A20|Ac
1 · · · Ac
19)P(Ac
1 · · · Ac
19) . (6)
We can compute the probability the first 19 cards are not aces represented by the expression
P(Ac
1 · · ·Ac
19) by further conditioning on earlier cards as
P(Ac
1 · · · Ac
19) = P(Ac
2Ac
3 · · · Ac
19|Ac
1)P(Ac
1)
= P(Ac
3 · · · Ac
19|Ac
1Ac
2)P(Ac
2|Ac
1)P(Ac
1)
= P(Ac
19|Ac
1Ac
2Ac
3 · · · Ac
18) · · ·P(Ac
3|Ac
1Ac
2)P(Ac
2|Ac
1)P(Ac
1) . (7)

We can now more easily evaluate these probabilities since
P(Ac
1) =
48
52
, P(Ac
2|Ac
1) =
47
51
etc.
Thus changing the order of the product in Equation 7 we find
P(Ac
1 · · ·Ac
19) = P(Ac
1)P(Ac
2|Ac
1)P(Ac
3|Ac
1Ac
2) · · ·P(Ac
19|Ac
1Ac
2Ac
3 · · · Ac
18)
=
48
52
·
47
51
·
46
50
· · ·
30
34
=
8184
54145
.
In the same way we have P(A20|Ac
1 · · · Ac
19) = 4
33
so that using Equation 6 we find
P(F) =
8184
54145
·
4
33
=
992
54145
,
the same result we found earlier.
Next to compute P(EF) we first introduce the event S to denote what type of ace the 20th
is. To do that let S be the event that the 20th ace is the ace of spades. Since using S we
have A20 = S ∪ Sc
we can write the event EF as
EF = Ac
1 · · · Ac
19A20E = Ac
1 · · · Ac
19SE ∪ Ac
1 · · · Ac
19Sc
E ,
and have
P(EF) = P(Ac
1 · · · Ac
19SE) + P(Ac
1 · · ·Ac
19Sc
E) . (8)
To evaluate each of these expressions we can condition like Equation 6 to get
P(Ac
1 · · · Ac
19SE) = P(E|Ac
1 · · · Ac
19S)P(Ac
1 · · · Ac
19S) and
P(Ac
1 · · · Ac
19Sc
E) = P(E|Ac
1 · · · Ac
19Sc
)P(Ac
1 · · · Ac
19Sc
) .
Since S and E cannot both happen P(E|Ac
1 · · · Ac
19S) = 0 and in Equation 8 we are left with
P(EF) = P(Ac
1 · · ·Ac
19Sc
E) = P(E|Ac
1 · · · Ac
19Sc
)P(Ac
1 · · · Ac
19Sc
)
= P(E|Ac
1 · · · Ac
19Sc
)P(Sc
|Ac
1 · · · Ac
19)P(Ac
1 · · · Ac
19)
=
1
32
·
3
33
·
8184
54145
=
93
216580
,
the same result as earlier.
Part (b): As in the first method in Part (a) above for this part let F again be the event
the 20th card is the first ace, but now let E be the event the 21st card is the 2 of clubs. As
before we will solve this problem using definition of conditional probability or
p(E|F) =
p(EF)
p(F)
.
It remains to compute p(EF) in this case since P(F) is the same as previously. Since the
event EF is similar to the event F but with exception that we now know the identity of the
21st card the number of ways we can get the event EF is given by
(47 · 46 · 45 · · ·31 · 30 · 29) · 4 · 1 · (31!) .

Thus the probability P(EF) is given by
P(EF) =
(47 · 46 · 45 · · ·31 · 30 · 29) · 4 · 1 · (31!)
52!
=
18
52037
.
Using these two results we compute
P(E|F) =
18/52037
98/5349
=
29
1536
.
See the Matlab/Octave file chap 3 prob 28.m for the fractional simplifications needed in
this problem.
Problem 29 (used tennis balls)
Let E0, E1, E2, E3 be the event that we select 0, 1, 2, or 3 used tennis balls during our
first draw consisting of three balls. Then let A be the event that when we draw three balls
the second time none of the selected balls have been used. The problem asks us to compute
P(A), which we can compute P(A) by conditioning on the mutually exclusive events Ei for
i = 0, 1, 2, 3 as
P(A) =
3
X
i=0
P(A|Ei)P(Ei) .
Now we can compute the prior probabilities P(Ei) as follows
P(E0) =

6
0

9
3

15
3
, P(E1) =

6
1

9
2

15
3

P(E2) =

6
2

9
1

15
3
, P(E3) =

6
3

9
0

15
3
.
Where the random variable representing the number of selected used tennis balls is a hy-
pergeometric random variable and we have explicitly enumerated these probabilities above.
We can now compute P(A|Ei) for each i. Beginning with P(A|E0) which we recognize as
the probability of event A under the situation where in the first draw of three balls we draw
no used balls initially i.e. we draw all new balls. Since event E0 is assumed to happen with
certainty when we go to draw the second of three balls we have 6 new balls and 9 used balls.
This gives the probability of event A as
P(A|E0) =

9
0

6
3

15
3
.

In the same way we can compute the other probabilities. We find that
P(A|E1) =

8
0

7
3

15
3
, P(A|E2) =

7
0

8
3

15
3
, P(A|E3) =

6
0

9
3

15
3
.
With these results we can calculate P(A). This is done in the Matlab file chap 3 prob 29.m
where we find that P(A) ≈ 0.0893.
Problem 30 (boxes with marbles)
Let B be the event that the drawn ball is black and let X1 (X2) be the event that we select
the first (second) box. Then to calculate P(B) we will condition on the box drawn from as
P(B) = P(B|X1)P(X1) + P(B|X2)P(X2) .
Now P(B|X1) = 1/2, P(B|X2) = 2/3, P(X1) = P(X2) = 1/2 so
P(B) =
1
2

1
2

+
1
2

2
3

=
7
12
.
If we see that the ball is white (i.e. it is not black i.e event Bc
has happened) we now want
to compute that it was drawn from the first box i.e.
P(X1|Bc
) =
P(Bc
|X1)P(X1)
P(Bc|X1)P(X1) + P(Bc|X2)P(X2)
=
3
5
.
Problem 31 (Ms. Aquina’s holiday)
After Ms. Aquina’s tests are completed and the doctor has the results he will flip a coin. If
it lands heads and the results of the tests are good he will call with the good news. If the
results of the test are bad he will not call. If the coin flip lands tails he will not call regardless
of the tests outcome. Lets let B denote the event that Ms. Aquina has cancer and the and
the doctor has bad news. Let G be the event that Ms. Aquina does not have cancer and
the results of the test are good. Finally let C be the event that the doctor calls the house
during the holiday.
Part (a): Now the event that the doctor does not call (i.e. Cc
) will add support to the
hypothesis that Ms. Aquina has cancer (or event B) if and only if it is more likely that
the doctor will not call given that she does have cancer. This is the event Cc
will cause
β ≡ P(B|Cc
) to be greater than α ≡ P(B) if and only if
P(Cc
|B) ≥ P(Cc
|Bc
) = P(Cc
|G) .

From a consideration of all possible outcomes we have that
P(Cc
|B) = 1 ,
since if the results of the tests come back negative (and Ms. Aquina has cancer), the doctor
will not call regardless of the coin flip. We also have that
P(Cc
|G) =
1
2
,
since if the results of the test are good, the doctor will only call if the coin flip lands heads
and not call otherwise. Thus the fact that the doctor does not call adds evidence to the
belief that Ms. Aquina has cancer. Logic similar to this is discussed in the book after the
example of the bridge championship controversy.
Part (b): We want to explicitly find β = P(B|Cc
) using Bayes’ rule. We find that
β =
P(Cc
|B)P(B)
P(Cc)
=
1(α)
(3/4)
=
4
3
α α .
Which explicitly verifies the intuition obtained in Part (a).
Problem 32 (the number of children)
Let C1, C2, C3, C4 be the events that the family has 1, 2, 3, 4 children respectively. Let E be
the evidence that the chosen child is the eldest in the family.
Part (a): We want to compute
P(C1|E) =
P(E|C1)P(C1)
P(E)
.
We will begin by computing P(E). We find that
P(E) =
4
X
i=1
P(E|Ci)P(Ci) = 1(0.1) +
1
2
(0.25) +
1
3
(0.35) +
1
4
(0.3) = 0.4167 ,
so that P(C1|E) = 1(0.1)/0.4167 = 0.24.
Part (b): We want to compute
P(C4|E) =
P(E|C4)P(C4)
P(E)
=
(0.25)(0.3)
0.4167
= 0.18 .
These calculations are done in the file chap 3 prob 32.m.

Problem 33 (English vs. American)
Let E (A) be the event that this man is English (American). Also let L be the evidence
found on the letter. Then we want to compute P(E|L) which we will do with Bayes’ rule.
We find (counting the number of vowels in each word) that
P(E|L) =
P(L|E)P(E)
P(L|E)P(E) + P(L|Ec)P(Ec)
=
(3/6)(0.4)
(3/6)(0.4) + (2/5)(0.6)
=
5
11
.
Problem 34 (some new interpretation of the evidence)
From Example 3f in the book we had that
P(G|C) =
P(GC)
P(C)
=
P(C|G)P(G)
P(C|G)P(G) + P(C|Gc)P(Gc)
.
But now we are told P(C|G) = 0.9, since we are assuming that if we are guilty we will have
the given characteristic with 90% certainty. Thus we now would compute for P(G|C) the
following
P(G|C) =
0.9(0.6)
0.9(0.6) + 0.2(0.4)
=
27
31
.
Problem 35 (which class is superior)
In this problem the superior class is the one that has the larger concentration of good
students. An expert examines a student selected from class A and a student from class B.
To formulate this problem in terms of probabilities lets introduce three events E, F, and
R as follows. Let E be the event class A is the superior class, F be the event the expert
finds the student from class A to be Fair, and R be the event the expert finds the student
from class B to be Poor (P might have been a more intuitive notation to use for this last
event but the letter P conflicts with the notation for probability). Using this notation for
this problem we want to evaluate P(E|FR). Using the definition of conditional probability
we have
P(E|FR) =
P(FR|E)P(E)
P(FR)
=
P(FR|E)P(E)
P(FR|E)P(E) + P(FR|Ec)P(Ec)
.
To evaluate the above, first assume the events E and Ec
are equally likely, that is P(E) =
P(Ec
) = 1
2
. This is reasonable since the labeling of A and B was done randomly and so the
event that the label A was assigned to the superior class would happen with a probability of
1
2
. Next given E (that is A is the superior class) the two events F and R are conditionally
independent. That is
P(FR|E) = P(F|E)P(R|E) ,

and a similar expression when the event FR is conditioned on Ec
. This states that given A
is the superior class, a student selected from one class is Good, Fair, or Poor independent of
a student selected from the other class being Good, Fair or Poor.
To evaluate these probabilities we reason as follows. If we are given the event E then A is
the superior class and thus has 10 Fair students, so P(F|E) = 10
30
, while B is not the superior
class and has 15 Poor students giving P(R|E) = 15
30
. If we are given Ec
then A is not the
superior class so P(F|Ec
) = 5
30
and P(R|Ec
) = 10
30
. Using all of these results we have
P(E|FR) =
P(F|E)P(R|E)P(E)
P(F|E)P(R|E)P(E) + P(F|Ec)P(R|Ec)P(Ec)
=
P(F|E)P(R|E)
P(F|E)P(R|E) + P(F|Ec)P(R|Ec)
=
(10/30)(15/30)
(10/30)(15/30) + (10/30)(5/30)
=
3
4
.
Problem 36 (resignations from store C)
To solve this problem lets begin by introducing several events. Let A be the event a person
works for company A, B be the event a person works for company B, and C be the event a
person works for company C. Finally let W be the event a person is female (a woman). We
desire to find P(C|W). Using the definition of conditional probability we have
P(C|W) =
P(CW)
P(W)
=
P(W|C)P(C)
P(W|A)P(A) + P(W|B)P(B) + P(W|C)P(C)
. (9)
Since 50, 75, and 100 people work for companies A, B, and C, respectively the total number
of workers is 50 + 75 + 100 = 225 and the individual probabilities of A, B, or C is given by
P(A) =
50
225
=
2
9
, P(B) =
75
225
=
1
3
, and P(C) =
100
225
=
4
9
.
We are also told that .5, .6, and .7 are the percentages of the female employees of the
companies A, B, C, respectively. Thus
P(W|A) = 0.5 , P(W|B) = 0.6 , and P(W|C) = 0.7 .
Using these results in Equation 9 we get
P(C|W) =
(0.7)(4/9)
(0.5)(2/9) + (0.6)(1/3) + (0.7)(4/9)
=
1
2
.
See the Matlab/Octave file chap 3 prob 36.m for the fractional simplifications needed in
this problem.

Problem 37 (gambling with a fair coin)
Let F denote the event that the gambler is observing results from a fair coin. Also let O1,
O2, and O3 denote the three observations made during our experiment. We will assume that
before any observations are made the probability that we have selected the fair coin is 1/2.
Part (a): We desire to compute P(F|O1) or the probability we are looking at a fair coin
given the first observation. This can be computed using Bayes’ theorem. We have
P(F|O1) =
P(O1|F)P(F)
P(O1|F)P(F) + P(O1|Fc)P(Fc)
=
1
2
1
2

1
2
1
2

+ 1 1
2
=
1
3
.
Part (b): With the second observation and using the “posteriori’s become priors” during a
recursive update we now have
P(F|O2, O1) =
P(O2|F, O1)P(F|O1)
P(O2|F, O1)P(F|O1) + P(O2|Fc, O1)P(Fc|O1)
=
1
2
1
3

1
2
1
3

+ 1 2
3
=
1
5
.
Part (c): In this case because the two-headed coin cannot land tails we can immediately
conclude that we have selected the fair coin. This result can also be obtained using Bayes’
theorem as we have in the other two parts of this problem. Specifically we have
P(F|O3, O2, O1) =
P(O3|F, O2, O1)P(F|O2, O1)
P(O3|F, O2, O1)P(F|O2, O1) + P(O3|Fc, O2, O1)P(Fc|O2, O1)
=
1
2
1
5

1
2
1
5

+ 0
= 1 .
Verifying what we know must be true.
Problem 38 (drawing white balls)
Let W and B represent the events of drawing a white ball or a black respectively, and let
H and T denote the event of obtaining a head or a tail when we flip the coin. As stated in
the problem when the outcome of the coin flip is heads (event H) a ball is selected from urn
A. This urn has 5 white and 7 black balls. Thus P(W|H) = 5
12
. Similarly, when the coin
flip results in tails a ball is selected from urn B, which has 3 white and 12 black balls. Thus
P(W|T) = 3
15
. We would like to compute P(T|W). Using Bayes’ formula we have
P(T|W) =
P(W|T)P(T)
P(W)
=
P(W|T)P(T)
P(W|T)P(T) + P(W|H)P(H)
=
3
15
1
2

3
15
1
2

+ 5
12
1
2
=
12
37
.

Problem 39 (having accidents)
From example 3a in the book where A1 is the event that a person has an accident during
the first year we recall that P(A1) = 0.26. In this problem we are asked to find P(A2|Ac
1).
We can find this probability by conditioning on whether or not the person is accident prone
(event A). We have
P(A2|Ac
1) =
P(A2Ac
1)
P(Ac
1)
=
P(A2Ac
1|A)P(A) + P(A2Ac
1|Ac
)P(Ac
)
P(Ac
1)
.
We assume that A2 and A1 are conditionally independent given A and thus have
P(A2Ac
1|A) = P(A2|A)P(Ac
1|A) and P(A2Ac
1|Ac
) = P(A2|Ac
)P(Ac
1|Ac
) . (10)
With these simplifications and using the numbers from example 3a we can evaluate P(A2|Ac
1).
We thus find
P(A2|Ac
1) =
P(A2|A)P(Ac
1|A)P(A) + P(A2|Ac
)P(Ac
1|Ac
)P(Ac
)
P(Ac
1)
=
0.4(1 − 0.4)(0.3) + 0.2(1 − 0.2)(0.7)
1 − 0.26
=
46
185
.
Note that instead of assuming conditional independence to simplify probabilities such as
P(A2Ac
1|A) appearing in Equations 10 we could also simply condition on earlier events by
writing this expression as P(A2|Ac
1, A)P(Ac
1|A). The numerical values used to evaluate this
expression would be the same as presented above.
Problem 40 (selecting k white balls)
For this problem we draw balls from an urn that starts with 5 white and 7 red balls and on
each draw we replace each drawn ball with one of the same color as the one drawn. Then to
solve the requested problem let Wk denote the event that a white ball was selected during the
kth draw and Rk denote the even that a red ball was selected on the kth draw for k = 1, 2, 3.
We then can decompose each of the higher level events (the number of white balls) in terms
of the component events Wk and Rk as follows
Part (a): To get 0 white balls requires the event R1R2R3. To compute this probability we
use conditioning to find
P(R1R2R3) = P(R1)P(R2R3|R1) = P(R1)P(R2|R1)P(R3|R1R2)
=
7
12
·
8
13
·
9
14
=
3
13
.
Part (b): We can represent drawing only 1 white ball by the following event
W1R2R3 ∪ R1W2R3 ∪ R1R2W3 .

As in Part (a) by conditioning we have that the probability of the above event is given by
P(W1R2R3 ∪ R1W2R3 ∪ R1R2W3) = P(W1R2R3) + P(R1W2R3) + P(R1R2W3)
= P(W1)P(R2|W1)P(R3|W1R2)
+ P(R1)P(W2|R1)P(R3|R1W2)
+ P(R1)P(R2|R1)P(W3|R1R2)
=
5
12
·
7
13
·
8
14
+
7
12
·
5
13
·
8
14
+
7
12
·
8
13
·
5
14
=
5
13
.
Part (c): We can draw 3 white balls in only one way W1W2W3. Using the above logic as
in Part (a) we have that the probability of this event given by
P(W1W2W3) = P(W1)P(W2|W1)P(W3|W1W2)
=
5
12
6
13
7
14
=
5
52
.
Part (d): We can draw two white balls in the following way
R1W2W3 ∪ W1R2W3 ∪ W1W2R3 .
Again, using the same logic as in Part (a) we have that the probability of the above event
given by
P(R1W2W3 ∪ W1R2W3 ∪ W1W2R3) = P(R1W2W3) + P(W1R2W3) + P(W1W2R3)
= P(R1)P(W2|R1)P(W3|R1W2)
+ P(W1)P(R2|W1)P(W3|W1R2)
+ P(W1)P(W2|W1)P(R3|W1W2)
=
7
12
·
5
13
·
6
14
+
5
12
·
7
13
·
6
14
+
5
12
·
6
13
·
7
14
=
15
52
.
Problem 41 (drawing the same ace)
We want to compute if the second card drawn is an ace. Denoting this event by E and
following the hint lets compute P(E) by conditioning on whether we select the original ace
drawn from the first deck. Let this event by A0. Then we have
P(E) = P(E|A0)P(A0) + P(E|Ac
0)P(Ac
0) .
Now P(A0) = 1
27
since it is one of the twenty seven cards in this second stack and P(Ac
0) = 26
27
and P(E|A0) = 1. Using these values we get
P(E) = 1 ·
1
27
+
26
27
· P(E|Ac
0) .

Thus it remains to compute the expression P(E|Ac
0). Since under the event Ac
0 we know
that we do not draw the original ace, this probability is related to how the original deck of
cards was split. In that case the current half deck could have 3, 2, 1 or 0 aces in it. For each
of these cases we have probabilities given by 3
26
, 2
26
, 1
26
, and 0
26
of drawing an ace if we have
three aces, two aces, one ace, and no aces respectively in our second half deck. Conditioning
on the number of aces in this half deck we have (using D3, D2 and D1 as notation for the
events that this half deck has 3, 2 or 1 aces in it) we obtain
P(E|Ac
0) = P(E|D3, Ac
0)P(D3|Ac
0) + P(E|D2, Ac
0)P(D2|Ac
0) + P(E|D1, Ac
0)P(D1|Ac
0) .
Since one of the aces was found to be in the first pile, the second pile contains k = 1, 2, 3
aces with probability
P(Dk) =

3
k

48
26 − k

51
26
for k = 1, 2, 3 ,
Evaluating the above expression these numbers become
P(E|D3, Ac
0) =

3
3

52 − 4
26 − 3

51
26
=
104
833
P(E|D2, Ac
0) =

3
2

52 − 4
26 − 2

51
26
=
325
833
P(E|D1, Ac
0) =

3
1

52 − 4
26 − 1

51
26
=
314
833
.
We can now evaluate P(E|Ac
0) the probability of selecting an ace, given that it must be one
of the original 26 cards in the second pile, as
P(E|Ac
0) =
3
X
k=1
P(EDk|Ac
0) =
3
X
k=1
P(Dk|Ac
0)P(E|Dk, Ac
0)
=
3
X
k=1
k
26





3
k

48
26 − k

51
26




 =
1
17
,
when we perform the required summation. We can also reason that P(E|Ac
0) = 1
17
in another
way. This probability is equivalent to the case where we have simply removed one ace from
the deck and recognized that the second card drawn could be any one of the remaining 51
cards (three of which remain as aces). Thinking like this would give P(E|Ac
0) = 3
51
= 1
17
the

same result as argued above. The fact that cards are in separate piles is irrelevant. Any one
of the 51 cards could be in any position in either pile.
We can now finally evaluate P(E) we have
P(E) = 1 ·
1
27
+
26
27
·
1
17
=
43
459
,
the same as in the back of the book. See the Matlab file chap 3 prob 41.m for these calcu-
lations.
Problem 42 (special cakes)
Let R be the event that the special cake will rise correctly. Then from the problem statement
we are told that P(R|A) = 0.98, P(R|B) = 0.97, and P(R|C) = 0.95, with the prior
information of P(A) = 0.5, P(B) = 0.3, and P(C) = 0.2. Then this problem asks for
P(A|Rc
). Using Bayes’ rule we have
P(A|Rc
) =
P(Rc
|A)P(A)
P(Rc)
,
where P(Rc
) is given by conditioning on A, B, or C as
P(Rc
) = P(Rc
|A)P(A) + P(Rc
|B)P(B) + P(Rc
|C)P(C)
= 0.02(0.5) + 0.03(0.3) + 0.05(0.2) = 0.029 ,
so that P(A|Rc
) is given by
P(A|Rc
) =
0.02(0.5)
0.029
= 0.344 .
Problem 43 (three coins in a box)
Let C1, C2, C3 be the event that the first, second, and third coin is chosen and flipped.
Then let H be the event that the flipped coin showed heads. Then we would like to evaluate
P(C1|H). Using Bayes’ rule we have
P(C1|H) =
P(H|C1)P(C1)
P(H)
.
We compute P(H) first. We find conditioning on the the coin selected that
P(H) =
3
X
i=1
P(H|Ci)P(Ci) =
1
3
3
X
i=1
P(H|Ci)
=
1
3

1 +
1
2
+
1
3

=
11
18
.
Then P(C1|H) is given by
P(C1|H) =
1(1/3)
(11/18)
=
6
11
.

Problem 44 (a prisoners’ dilemma)
I will argue that this problem is similar to the so called “Monty Hall Problem” and because
of this connection the probability of execution of the prisoner stays at 1/3 instead of 1/2.
See [1] for a nice discussion of the Monty Hall Problem. The probabilities that do change,
however, are the probabilities of the other two prisoners. The probability of execution of the
prisoner to be set free falls to 0 while the probability of the other prisoner increases to 2/3.
To show the similarity of this problem to the Monty Hall Problem, think of the three prisoners
as surrogates for Monty Hall’s three doors. Think of execution as the “prize” that is hidden
“behind” one of the prisoners. Finally, the prisoner that the guard admits to freeing is
equivalent to Monty Hall opening up a door in which he knows does not contain this “prize”.
In the Monty Hall Problem the initial probabilities associated with each door are 1/3 but
once a non-selected door has been opened the probability of having selected the correct door
does not increase from 1/3 to 1/2. The opening of the other door is irrelevant to your odds of
winning if you keep your selection. The remaining door has a probability of 2/3 of containing
the prize.
Following the analogy the jailer revealing a prisoner that can go free is Monty opening a
door known not to contain the prize. By symmetry to Monty Hall, A’s probability of being
executed must remain at 1/3 and not increase to 1/2.
A common error in logic is to argue as follows. Before asking his question the probability of
event A (A is to be executed) is P(A) = 1/3. If prisoner A is told that B (or C) is to be set
free then we need to compute P(A|Bc
). Where A, B, and C are the events that prisoner A,
B, or C is to be executed respectively. Now from Bayes’ rule
P(A|Bc
) =
P(Bc
|A)P(A)
P(Bc)
.
We have that P(Bc
) is given by
P(Bc
) = P(Bc
|A)P(A) + P(Bc
|B)P(B) + P(Bc
|C)P(C) =
1
3
+ 0 +
1
3
=
2
3
.
So the above probability then becomes
P(A|Bc
) =
1(1/3)
2/3
=
1
2

1
3
.
Thus the probability that prisoner A will be executed has increased as claimed by the jailer.
While there is nothing wrong with above logic, the problem with it is that it is not answering
the real question that we want the answer to. This question is: what is the probability that
A will be executed given the statement that the jailer makes. Now the jailer has only two
choices for what he can say; either B or C will be set free. Lets compute P(A|JB) where
JB is the event that the jailer says that prisoner B will be set free. We expect by symmetry

that P(A|JB) = P(A|JC). We then have using Bayes’ rule
P(A|JB) =
P(JB|A)P(A)
P(JB)
=
P(JB|A)P(A)
P(JB|A)P(A) + P(JB|B)P(B) + P(JB|C)P(C)
=
(1/2)(1/3)
(1/2)(1/3) + 0 + 1(1/3)
=
1
3
.
In performing this computation we have used the facts that
P(JB|A) =
1
2
,
since the jailer has two choices he can say if A is to be executed,
P(JB|B) = 0 ,
since the jailer cannot say that B will be set free if B is to be executed, and
P(JB|C) = 1 ,
since in this case prisoner C is to be executed so the jailer cannot say C and being unable
to say that prisoner A is to be set free he must say that prisoner B is to be set free. Thus
we see that regardless to what the jailer says the probability that A is to executed stays the
same.
Problem 45 (is it the fifth coin?)
Let Ci be the event that the ith coin was selected to be flipped. Since any coin is equally
likely we have P(Ci) = 1
10
for all i. Let H be the event that the flipped coin shows heads,
then we want to compute P(C5|H). From Bayes’ rule we have
P(C5|H) =
P(H|C5)P(C5)
P(H)
.
We compute P(H) by conditioning on the selected coin Ci we have
P(H) =
10
X
i=1
P(H|Ci)P(Ci)
=
10
X
i=1
i
10

1
10

=
1
100
10
X
i=1
i
=
1
100

10(10 + 1)
2

=
11
20
.
So that
P(C5|H) =
(5/10)(1/10)
(11/20)
=
1
11
.

Problem 46 (one accident means its more likely that you will have another)
Consider the expression P(A2|A1). By the definition of conditional probability this can be
expressed as
P(A2|A1) =
P(A1, A2)
P(A1)
,
so the desired expression to show is then equivalent to the following
P(A1, A2)
P(A1)
P(A1) ,
or P(A1, A2) P(A1)2
. Considering first the expression P(A1) by conditioning on the sex
of the policy holder we have
P(A1) = P(A1|M)P(M) + P(A1|W)P(W) = pmα + pf (1 − α) .
where M is the event the policy holder is male and W is the event that the policy holder is
female. In the same way we have for the joint probability P(A1, A2) that
P(A1, A2) = P(A1, A2|M)P(M) + P(A1, A2|W)P(W) .
Assuming that A1 and A2 are independent given the specification of the policy holders sex
we have that
P(A1, A2|M) = P(A1|M)P(A2|M) ,
the same expression holds for the event W. Using this in the expression for P(A1, A2) above
we obtain
P(A1, A2) = P(A1|M)P(A2|M)P(M) + P(A1|W)P(A2|W)P(W)
= p2
mα + p2
f (1 − α) .
We now look to see if P(A1, A2) P(A1)2
. Computing the expression P(A1, A2) − P(A1)2
,
(which we hope to be able to show is always positive) we have that
P(A1, A2) − P(A1)2
= p2
mα + p2
f (1 − α) − (pmα + pf (1 − α))2
= p2
mα + p2
f (1 − α) − p2
mα2
− 2pmpf α(1 − α) − p2
f (1 − α)2
= p2
mα(1 − α) + p2
f (1 − α)α − 2pmpf α(1 − α)
= α(1 − α)(p2
m + p2
f − 2pmpf )
= α(1 − α)(pm − pf )2
.
Note that this is always positive. Thus we have shown that P(A1|A2) P(A1). In words,
this means that given that we have an accident in the first year this information will increase
the probability that we will have an accident in the second year to a value greater than we
would have without the knowledge of the accident during year one (A1).

Problem 47 (the probability on which die was rolled)
Let X be the the random variable that specifies the number on the die roll i.e. the integer
1, 2, 3, · · · , 6. Let W be the event that all the balls drawn are white. Then we want to
evaluate P(W), which can be computed by conditioning on the value of X. Thus we have
P(W) =
6
X
i=1
P{W|X = i}P(X = i)
Since P{X = i} = 1/6 for every i, we need only to compute P{W|X = i}. We have that
P{W|X = 1} =
5
15
≈ 0.33
P{W|X = 2} =

5
15

4
14

≈ 0.095
P{W|X = 3} =

5
15

4
14

3
13

≈ 0.022
P{W|X = 4} =

5
15

4
14

3
13

2
12

≈ 0.0036
P{W|X = 5} =

5
15

4
14

3
13

2
12

1
11

≈ 0.0003
P{W|X = 6} = 0
Then we have
P(W) =
1
6
(0.33 + 0.95 + 0.022 + 0.0036 + 0.0003) = 0.0756 .
If all the balls selected are white then the probability our die showed a three was
P{X = 3|W} =
P{W|X = 3}P(X = 3)
P(W)
= 0.048 .
Problem 48 (which cabinet did we select)
This question is the same as asking what is the probability we select cabinet A given that a
silver coin is seen on our draw. Then we want to compute P(A|S) = P (S|A)P (A)
P (S)
. Now
P(S) = P(S|A)P(A) + P(S|B)P(B) = 1

1
2

+

1
2

1
2

=
3
4
Thus
P(A|S) =
1(1/2)
(3/4)
=
2
3
.

Problem 49 (prostate cancer)
Let C be the event that man has cancer and A (for antigen) the event of taking an elevated
PSA measurement. Then in the problem we are given
P(A|Cc
) = 0.135
P(A|C) = 0.268 ,
and in addition we have P(C) = 0.7.
Part (a): We want to evaluate P(C|A) or
P(C|A) =
P(A|C)P(C)
P(A)
=
P(A|C)P(C)
P(A|C)P(C) + P(A|Cc)P(Cc)
=
(0.268)(0.7)
(0.268)(0.7) + (0.135)(0.3)
= 0.822 .
Part (b): We want to evaluate P(C|Ac
) or
P(C|Ac
) =
P(Ac
|C)P(C)
P(Ac)
=
(1 − 0.268)(0.7)
1 − 0.228
= 0.633 .
If the prior probability of cancer changes (i.e. P(C) = 0.3) then the above formulas yield
P(C|A) = 0.459
P(C|Ac
) = 0.266 .
Problem 50 (assigning probabilities of risk)
Let G, A, B be the events that a person is of good risk, an average risk, or a bad risk
respectively. Then in the problem we are told that (if E denotes the event that an accident
occurs)
P(E|G) = 0.05
P(E|A) = 0.15
P(E|B) = 0.3
In addition the a priori assumptions on the proportion of people that are good, average and
bad risks are given by P(G) = 0.2, P(A) = 0.5, and P(B) = 0.3. Then in this problem we
are asked to compute P(E) or the probability that an accident will happen. This can be

computed by conditioning on the probability of a person having an accident from among the
three types, i.e.
P(E) = P(E|G)P(G) + P(E|A)P(A) + P(E|B)P(B)
= 0.05(0.2) + (0.15)(0.5) + (0.3)(0.3) = 0.175 .
If a person had no accident in a given year we want to compute P(G|Ec
) or
P(G|Ec
) =
P(Ec
|G)P(G)
P(Ec)
=
(1 − P(E|G))P(G)
1 − P(E)
=
(1 − 0.05)(0.2)
1 − 0.175
=
38
165
also to compute P(A|Ec
) we have
P(A|Ec
) =
P(Ec
|A)P(A)
P(Ec)
=
(1 − P(E|A))P(A)
1 − P(E)
=
(1 − 0.15)(0.5)
1 − 0.175
=
17
33
Problem 51 (letters of recommendation)
Let Rs, Rm, and Rw be the event that our worker receives a strong, moderate, or weak
recommendation respectively. Let J be the event that our applicant gets the job. Then the
problem specifies
P(J|Rs) = 0.8
P(J|Rm) = 0.4
P(J|Rw) = 0.1 ,
with priors on the type of recommendation given by
P(Rs) = 0.7
P(Rm) = 0.2
P(Rw) = 0.1 ,
Part (a): We are asked to compute P(J) which by conditioning on the type of recommen-
dation received is
P(J) = P(J|Rs)P(Rs) + P(J|Rm)P(Rm) + P(J|Rw)P(Rw)
= 0.8(0.7) + (0.4)(0.2) + (0.1)(0.1) = 0.65 =
13
20
.

Part (b): Given the event J is held true then we are asked to compute the following
P(Rs|J) =
P(J|Rs)P(Rs)
P(J)
=
(0.8)(0.7)
(0.65)
=
56
65
P(Rm|J) =
P(J|Rm)P(Rm)
P(J)
=
(0.4)(0.2)
(0.65)
=
8
65
P(Rw|J) =
P(J|Rw)P(Rw)
P(J)
=
(0.1)(0.1)
(0.65)
=
1
65
Note that this last probability can also be calculated as P(Rw|J) = 1−P(Rw|J)−P(Rw|J).
Part (c): For this we are asked to compute
P(Rs|Jc
) =
P(Jc
|Rs)P(Rs)
P(Jc)
=
(1 − 0.8)(0.7)
(0.35)
=
2
5
P(Rm|Jc
) =
P(Jc
|Rm)P(Rm)
P(Jc)
=
(1 − 0.4)(0.2)
(0.35)
=
12
35
P(Rw|Jc
) =
P(Jc
|Rw)P(Rw)
P(Jc)
=
(1 − 0.1)(0.1)
(0.35)
=
9
35
.
Problem 52 (college acceptance)
Let M, T, W, R, F, and S correspond to the events that mail comes on Monday, Tuesday,
Wednesday, Thursday, Friday, or Saturday (or later) respectively. Let A be the event that
our student is accepted.
Part (a): To compute P(M) we can condition on whether or not the student is accepted as
P(M) = P(M|A)P(A) + P(M|Ac
)P(Ac
) = 0.15(0.6) + 0.05(0.4) = 0.11 .
Part (b): We desire to compute P(T|Mc
). Using the definition of conditional probability
we find that (again conditioning P(T) on whether she is accepted or not)
P(T|Mc
) =
P(T, Mc
)
P(Mc)
=
P(T)
1 − P(M)
=
P(T|A)P(A) + P(T|Ac
)P(Ac
)
1 − P(M)
=
0.2(0.6) + 0.1(0.4)
1 − 0.11
=
16
89
.
Part (c): We want to calculate P(A|Mc
, Tc
, Wc
). Again using the definition of conditional
probability (twice) we have that
P(A|Mc
, Tc
, Wc
) =
P(A, Mc
, Tc
, Wc
)
P(Mc, Tc, Wc)
=
P(Mc
, Tc
, Wc
|A)P(A)
P(Mc, Tc, Wc)
.

To evaluate terms like P(Mc
, Tc
, Wc
|A), and P(Mc
, Tc
, Wc
|Ac
), lets compute the probability
that mail will come on Saturday or later given that she is accepted or not. Using the fact
that P(·|A) and P(·|Ac
) are both probability densities and must sum to one over their first
argument we calculate that
P(S|A) = 1 − 0.15 − 0.2 − 0.25 − 0.15 − 0.1 = 0.15
P(S|Ac
) = 1 − 0.05 − 0.1 − 0.1 − 0.15 − 0.2 = 0.4 .
With this result we can calculate that
P(Mc
, Tc
, Wc
|A) = P(R|A) + P(F|A) + P(S|A) = 0.15 + 0.1 + 0.15 = 0.4
P(Mc
, Tc
, Wc
|Ac
) = P(R|Ac
) + P(F|Ac
) + P(S|Ac
) = 0.15 + 0.2 + 0.4 = 0.75 .
Also we can compute P(Mc
, Tc
, Wc
) by conditioning on whether she is accepted or not. We
find
P(Mc
, Tc
, Wc
) = P(Mc
, Tc
, Wc
|A)P(A) + P(Mc
, Tc
, Wc
|Ac
)P(Ac
)
= 0.4(0.6) + 0.75(0.4) = 0.54 .
Now we finally have all of the components we need to compute what we were asked to. We
find that
P(A|Mc
, Tc
, Wc
) =
P(Mc
, Tc
, Wc
|A)P(A)
P(Mc, Tc, Wc)
=
0.4(0.6)
0.54
=
4
9
.
Part (d): We are asked to compute P(A|R) which using Bayes’ rule gives
P(A|R) =
P(R|A)P(A)
P(R)
.
To compute this lets begin by computing P(R) again obtained by conditioning on whether
our student is accepted or not. We find
P(R) = P(R|A)P(A) + P(R|Ac
)P(Ac
) = 0.15(0.6) + 0.15(0.4) = 0.15 .
So that our desired probability is given by
P(A|R) =
0.15(0.6)
0.15
=
3
5
.
Part (e): We want to calculate P(A|S). Using Bayes’ rule gives
P(A|S) =
P(S|A)P(A)
P(S)
.
To compute this, lets begin by computing P(S) again obtained by conditioning on whether
our student is accepted or not. We find
P(S) = P(S|A)P(A) + P(S|Ac
)P(Ac
) = 0.15(0.6) + 0.4(0.4) = 0.25 .
So that our desired probability is given by
P(A|S) =
0.15(0.6)
0.25
=
9
25
.

Problem 53 (the functioning of a parallel system)
With n components a parallel system will be working if at least one component is working.
Let Hi be the event that the component i for i = 1, 2, 3, · · · , n is working. Let F be the
event that the entire system is functioning. We want to compute P(H1|F). We have
P(H1|F) =
P(F|H1)P(H1)
P(F)
.
Now P(F|H1) = 1 since if the first component is working the system is functioning. In
addition, P(F) = 1 − 1
2
n
since to be not functioning all components must not be working.
Finally P(H1) = 1/2. Thus our probability is
P(H1|F) =
1/2
1 − (1/2)n .
Problem 54 (independence of E and F)
Part (a): These two events would be independent. The fact that one person has blue eyes
and another unrelated person has blue eyes are in no way related.
Part (b): These two events seem unrelated to each other and would be modeled as inde-
pendent.
Part (c): As height and weigh are related, I would think that these two events are not
independent.
Part (d): Since the United States is in the western hemisphere these two two events are
related and they are not independent.
Part (e): Since rain one day would change the probability of rain on other days I would
say that these events are related and therefore not independent.
Problem 55 (independence in class)
Let S be a random variable denoting the sex of the randomly selected person. The S can
take on the values m for male and f for female. Let C be a random variable representing
denoting the class of the chosen student. The C can take on the values f for freshman and
s for sophomore. We want to select the number of sophomore girls such that the random
variables S and C are independent. Let n denote the number of sophomore girls. Then

counting up the number of students that satisfy each requirement we have
P(S = m) =
10
16 + n
P(S = f) =
6 + n
16 + n
P(C = f) =
10
16 + n
P(C = s) =
6 + n
16 + n
.
The joint density can also be computed and are given by
P(S = m, C = f) =
4
16 + n
P(S = m, C = s) =
6
16 + n
P(S = f, C = f) =
6
16 + n
P(S = f, C = s) =
n
16 + n
.
Then to be independent we must have P(C, S) = P(S)P(C) for all possible C and S values.
Considering the point case where (S = m, C = f) we have that n must satisfy
P(S = m, C = f) = P(S = m)P(C = f)
4
16 + n
=

10
16 + n

10
16 + n

which when we solve for n gives n = 9. Now one should check that this value of n works for
all other equalities that must be true, for example one needs to check that when n = 9 the
following are true
P(S = m, C = s) = P(S = m)P(C = s)
P(S = f, C = f) = P(S = f)P(C = f)
P(S = f, C = s) = P(S = f)P(C = s) .
As these can be shown to be true, n = 9 is the correct answer.
Problem 56 (is the nth coupon new?)
Let Ci be the identity of the n coupon and Ai the event that after collecting n − 1 coupons
at least one coupon of type i exists. Then the event that the nth coupon is new if we obtain
one of type i is the event Ac
i ∩ Ci, which is the the event that the ith coupon is not in the
first n − 1 coupons i.e. Ac
i and that the nth coupon is not the ith one. Then if Ei is the
event that the ith coupon is new we have
P(Ei) = P(Ac
i ∩ Ci) = P(Ac
i |Ci)P(Ci) = P(Ac
i )pi = (1 − P(Ai))pi ,

where from Example 4 i from the book we have that P(Ai) = 1−(1−pi)n−1
so the probability
of a new coupon being of type i is (1 − (1 − pi)n−1
)pi, so the probability of a new coupon (at
all) is given by
P(E) =
m
X
i=1
P(Ei) =
m
X
i=1
(1 − (1 − pi)n−1
)pi .
Problem 57 (the price path of a stock)
For this problem it helps to draw a diagram of the stocks path v.s. time for the various
situations.
Part (a): To be the same price in two days the stock can go up and then back down or
down and then back up. Giving a total probability of 2p(1 − p).
Part (b): To go up only one unit in three steps we must go up twice and down once. We
can have the single down day happen on any of the three days. Thus, the three possible
paths are (with +1 denoting a day where the stock goes up and −1 denoting a day where
the stock goes down) given by
(+1, +1, −1) , (+1, −1, +1) , (−1, +1, +1) ,
each with probability p2
(1 − p). Thus since each path is mutually exclusive we have a total
probability of 3p2
(1 − p).
Part (c): When we count the number of paths where we go up on the first day (two) and
divide by the total number of paths (three) we get the probability 2
3
.
Problem 58 (generating fair flips with a biased coin)
Part (a): Consider pairs of flips. Let E be the event that a pair of flips returns (H, T)
and let F be the event that the pair of flips returns (T, H). From the discussion on Page 93
Example 4h the event E will occur first with probability
P(E)
P(E) + P(F)
.
Now P(E) = p(1 − p) and P(F) = (1 − p)p, so the probability of obtaining event E and
declaring tails before the event F (from which we would declare heads) would be
p(1 − p)
2p(1 − p)
=
1
2
.
In the same way we will have the event F occur before the event E with probability 1
2
. Thus
we have an equally likely chance of obtaining heads or tails. Note: its important to note that

the procedure described is effectively working with ordered pairs of flips, we flip two coins
and only make a decision after looking at both coins and the order in which they come out.
Part (b): Lets compute the probability of declaring heads under this procedure. Assume
we are considering a sequence of coin flips. Let H be the event that we declare a head.
Then conditioning on the outcome of the previous two flips we have with Pf and Cf random
variables denoting the previous and the current flip respectively that
P(H) = P(H|Pf = T, Cf = T)P{Pf = T, Cf = T}
+ P(H|Pf = T, Cf = H)P{Pf = T, Cf = H}
+ P(H|Pf = H, Cf = T)P{Pf = H, Cf = T}
+ P(H|Pf = H, Cf = H)P{Pf = H, Cf = H} .
Now since
P{Pf = T, Cf = T} = 0 , P{Pf = T, Cf = H} = 1
P{Pf = H, Cf = T} = 0 , P{Pf = H, Cf = H} = 0 .
we see that P(H) = P(Pf = T, Cf = H) = (1 − p)p 6= 1
2
. In the same way P(T) = p(1 − p).
Thus this procedure would produce a head or a tail with equal probability but this probability
would not be 1/2.
Problem 59 (the first four outcomes)
Part (a): This probability would be p4
.
Part (b): This probability would be (1 − p)p3
.
Part (c): Given two mutually exclusive events E and F the probability that E occurs before
F is given by
P(E)
P(E) + P(F)
.
Denoting E by the event that we obtain a T, H, H, H pattern and F the event that we obtain
a H, H, H, H pattern the above becomes
p3
(1 − p)
p4 + p3(1 − p)
=
1 − p
p + (1 − p)
= 1 − p .
Problem 60 (the color of your eyes)
Since Smith’s sister has blue eyes and this is a recessive trait, both of Smith’s parents must
have the gene for blue eyes. Let R denote the gene for brown eyes and L denote the gene
for blue eyes (these are the second letters in the words brown and blue respectively). Then

Smith will have a gene makeup possibly given by (R, R), (R, L), (L, R), where the left gene
is the one received from his mother and the right gene is the one received from his father.
Part (a): With the gene makeup given above we see that in two cases from three total
Smith will have a blue gene. Thus this probability is 2/3.
Part (b): Since Smith’s wife has blue eyes, Smith’s child will receive a L gene from his
mother. The probability Smith’s first child will have blue eyes is then dependent on what
gene they receive from Smith. Letting B be the event that Smith’s first child has blue eyes
(and conditioning on the possible genes Smith could give his child) we have
P(B) = 0

1
3

+
1
2

1
3

+
1
2

1
3

=
1
3
.
As stated above, this result is obtained by conditioning on the possible gene makeups of
Smith. For example let (X, Y ) be the notation for the “event” that Smith has a gene make
up given by (X, Y ) then the above can be written symbolically (in terms of events) as
P(B) = P(B|(R, R))P(R, R) + P(B|(R, L))P(R, L) + P(B|(L, R))P(L, R) .
Evaluating each of the above probabilities gives the result already stated.
Part (c): The fact that the first child has brown eyes makes it more likely that Smith has
a genotype, given by (R, R). We compute the probability of this genotype given the event
E (the event that the first child has brown eyes using Bayes’ rule as)
P((R, R)|E) =
P(E|(R, R))P(R, R)
P(E|(R, R))P(R, R) + P(E|(R, L))P(R, L) + P(E|(L, R))P(L, R)
=
1 1
3

1 1
3

+ 1
2
1
3

+ 1
2
1
3

=
1
2
.
In the same way we have for the other possible genotypes that
P((R, L)|E) =
1
2
1
3

2
3
=
1
4
= P((L, R)|E) .
Thus the same calculation as in Part (b), but now conditioning on the fact that the first
child has brown eyes (event E) gives for a probability of the event B2 (that the second child
we have blue eyes)
P(B2|E) = P(B2|(R, R), E)P((R, R)|E) + P(B2|(R, L), E)P((R, L)|E) + P(B2|(L, R), E)P((L, R)|E
= 0

1
2

+
1
2

1
4

+
1
2

1
4

=
1
4
.
This means that the probability that the second child has brown eyes is then
1 − P(B2|E) =
3
4
.

Problem 61 (more recessive traits)
From the information that the two parents are normal but that they produced an albino child
we know that both parents must be carriers of albinism. Their non-albino child can have
any of three possible genotypes each with probability 1/3 given by (A, A), (A, a), (a, A). Lets
denote this parent by P1 and the event that this parent is a carrier for albinism as C1. Note
that P(C1) = 2/3 and P(Cc
1) = 1/3. We are told that the spouse of this person (denoted
P2) is a carrier for albinism.
Part (a): The probability their first offspring is an albino depends on how likely our first
parent is a carrier of albinism. We have (with E1 the event that their first child is an albino)
that
P(E1) = P(E1|C1)P(C1) + P(E1|Cc
1)P(Cc
1) .
Now P(E1|C1) = 1
2
1
2

= 1
4
, since both parents must contribute their albino gene, and
P(E1|Cc
1) = 0 so we have that
P(E1) =
1
4

2
3

=
1
6
.
Part (b): The fact that the first newborn is not an albino changes the probability that the
first parent is a carrier or the value of P(C1). To calculate this we will use Bayes’ rule
P(C1|Ec
1) =
P(Ec
1|C1)P(C1)
P(Ec
1|C1)P(C1) + P(Ec
1|Cc
1)P(Cc
1)
=
3
4
2
3

3
4
2
3

+ 1 1
3

=
3
5
.
so we have that P(Cc
1|Ec
1) = 2
5
, and following the steps in Part (a) we have (with E2 the
event that the couples second child is an albino)
P(E2|Ec
1) = P(E2|Ec
1, C1)P(C1|Ec
1) + P(E2|Ec
1, Cc
1)P(Cc
1|Ec
1)
=
1
4

3
5

=
3
20
.
Problem 62 (target shooting with Barbara and Dianne)
Let H be the event that the duck is “hit”, by either Barbra or Dianne’s shot. Let B and
D be the events that Barbra (respectively Dianne) hit the target. Then the outcome of the
experiment where both Dianne and Barbra fire at the target (assuming that their shots work
independently is)
P(Bc
, Dc
) = (1 − p1)(1 − p2)
P(Bc
, D) = (1 − p1)p2
P(B, Dc
) = p1(1 − p2)
P(B, D) = p1p2 .

Part (a): We desire to compute P(B, D|H) which equals
P(B, D|H) =
P(B, D, H)
P(H)
=
P(B, D)
P(H)
Now P(H) = (1 − p1)p2 + p1(1 − p2) + p1p2 so the above probability becomes
p1p2
(1 − p1)p2 + p1(1 − p2) + p1p2
=
p1p2
p1 + p2 − p1p2
.
Part (b): We desire to compute P(B|H) which equals
P(B|H) = P(B, D|H) + P(B, Dc
|H) .
Since the first term P(B, D|H) has already been computed we only need to compute P(B, Dc
|H).
As before we find it to be
P(B, Dc
|H) =
p1(1 − p2)
(1 − p1)p2 + p1(1 − p2) + p1p2
.
So the total result becomes
P(B|H) =
p1p2 + p1(1 − p2)
(1 − p1)p2 + p1(1 − p2) + p1p2
=
p1
p1 + p2 − p1p2
.
Problem 63 (dueling)
For a given trial while dueling we have the following possible outcomes (events) and their
associated probabilities
• Event I: A is hit and B is not hit. This happens with probability pB(1 − pA).
• Event II: A is not hit and B is hit. This happens with probability pA(1 − pB).
• Event III: A is hit and B is hit. This happens with probability pApB.
• Event IV : A is not hit and B is not hit. This happens with probability (1−pA)(1−pB).
With these definitions we can compute the probabilities of various other events.
Part (a): To solve this we recognize that A is hit if events I and III happen and the
dueling continues if event IV happens. We can compute p(A) (the probability that A is hit)
by conditioning on the outcome of the first duel. We have
p(A) = p(A|I)p(I) + p(A|II)p(II) + p(A|III)p(III) + p(A|IV )p(IV ) .
Now in the case of event IV the dual continues afresh and we see that p(A|IV ) = p(A).
Using this fact and the definitions of events I-IV we have that the above becomes
p(A) = 1 · pB(1 − pA) + 0 · pA(1 − pB) + 1 · pApB + p(A) · (1 − pA)(1 − pB) .

Now solving for p(A) in the above we find that
p(A) =
pB
(1 − (1 − pA)(1 − pB))
.
Part (b): Let D be the event that both duelists are hit. Then to compute this, we can
condition on the outcome of the first dual. Using the same arguments as above we find
p(D) = p(D|I)p(I) + p(D|II)p(II) + p(D|III)p(III) + p(D|IV )p(IV )
= 0 + 0 + 1 · pApB + p(D) · (1 − pA)(1 − pB) .
On solving for P(D) we have
p(D) =
pApB
1 − (1 − pA)(1 − pB)
.
Part (c): Lets begin by computing the probability that the dual ends after one dual. Let G1
be the event that the game ends with more than (or after) one dual. We have, conditioning
on the events I-IV that
p(G1) = 0 + 0 + 0 + 1 · (1 − pA)(1 − pB) = (1 − pA)(1 − pB) .
Now let G2 be the event that the game ends with more than (or after) two duals. Then
p(G2) = (1 − pA)(1 − pB)p(G1) = (1 − pA)2
(1 − pB)2
.
Generalizing this result we have for the probability that the games ends after n duels is
p(Gn) = (1 − pA)n
(1 − pB)n
.
Part (d): Let G1 be the event that the game ends with more than one dual and let A be
the event that A is hit. Then to compute p(G1|Ac
) by conditioning on the first experiment
we have
p(G1|Ac
) = p(G1, I|Ac
)p(I) + p(G1, II|Ac
)p(II)
+ p(G1, III|Ac
)p(III) + p(G1, IV |Ac
)p(IV )
= 0 + 0 + 0 + p(G1, IV |Ac
)(1 − pA)(1 − pB) .
So now we need to evaluate p(G1, IV |Ac
), which we do using the definition of conditional
probability. We find
p(G1, IV |Ac
) =
p(G1, IV, Ac
)
p(Ac)
=
1
p(Ac)
.
Where p(Ac
) is the probability that A is not hit on the first experiment. This can be
computed as
p(A) = pB(1 − pA) + pApB = pB so
p(Ac
) = 1 − pB ,

Woman answers correctly Woman answers incorrectly
Man answers correctly p2
p(1 − p)
Man answers incorrectly (1 − p)p (1 − p)2
Table 5: The possible probabilities of agreement for the couple in Problem 64, Chapter 3.
When asked a question four possible outcomes can occur, corresponding to the correctness
of the mans (woman’s) answer. The first row corresponds to the times when the husband
answers the question correctly, the second row to the times when the husband answers the
question incorrectly. In the same way, the first column corresponds to the times when the
wife is correct and second column to the times when the wife is incorrect.
and the above is then given by
p(G1|Ac
) =
(1 − pA)(1 − pB)
1 − pB
= 1 − pA .
In the same way as before this would generalize to the following (for the event Gn)
p(Gn) = (1 − pA)n
(1 − pB)n−1
Part (e): Let AB be the event that both duelists are hit. Then in the same way as Part (d)
above we see that
p(G1, IV |AB) =
p(G1, IV, AB)
p(AB)
=
1
p(AB)
.
Here p(AB) is the probability that A and B are hit on any given experiment so p(AB) =
pApB, and
p(G1|AB) =
(1 − pA)(1 − pB)
pApB
and in general
p(Gn|AB) =
(1 − pA)n
(1 − pB)n
pApB
.
Problem 64 (game show strategies)
Part (a): Since each person has probability p of getting the correct answer, either one
selected to represent the couple will answer correctly with probability p.
Part (b): To compute the probability that the couple answers correctly under this strategy
we will condition our probability on the “agreement” matrix in Table 5, i.e. the possible
combinations of outcomes the couple may encounter when asked a question that they both
answer. Lets define E be the event that the couple answers correctly, and let Cm (Cw) be
the events that the man (women) answers the question correctly. We find that
P(E) = P(E|Cm, Cw)P(Cm, Cw) + P(E|Cm, Cc
w)P(Cm, Cc
w)
+ P(E|Cc
m, Cw)P(Cc
m, Cw) + P(E|Cc
m, Cc
w)P(Cc
m, Cc
w) .

Now P(E|Cc
m, Cc
w) = 0 since both the man and the woman agree but they both answer
the question incorrectly. In that case the couple would return the incorrect answer to the
question. In the same way we have that P(E|Cm, Cw) = 1. Following the strategy of flipping
a coin when the couple answers disagree we note that P(E|Cm, Cc
w) = P(E|Cc
m, Cw) = 1/2,
so that the above probability when using this strategy becomes
P(E) = 1 · p2
+
1
2
p(1 − p) +
1
2
(1 − p)p = p ,
where in computing this result we have used the joint probabilities found in Table 5 to
evaluate terms like P(Cm, Cc
w). Note that this result is the same as in Part (a) of this
problem showing that there is no benefit to using this strategy.
Problem 65 (how accurate are we when we agree/disagree)
Part (a): We want to compute (using the notation from the previous problem)
P(E|(Cm, Cw) ∪ (Cc
m, Cc
w)) .
Defining the event A to be equal to (Cm, Cw) ∪ (Cc
m, Cc
w). We see that this is equal to
P(E|(Cm, Cw) ∪ (Cc
m, Cc
w)) =
P(E, A)
P(A)
=
p2
p2 + (1 − p)2
=
0.36
0.36 + 0.16
=
9
13
.
Part (b): We want to compute P(E|(Cc
m, Cw)∪(Cm, Cc
w)), but in the second strategy above
if the couple disagrees they flip a fair coin to decide. Thus this probability is equal to 1/2.
Problem 66 (relay circuits)
Part (a): Let E be the event that current flows from A to B. Then
P(E) = P(E|5 Closed)p5
= p(1 and 2 closed or 3 and 4 closed|5 closed)p5
= (p1p2 + p3p4)p5 .
Part (b): Conditioning on relay 3. Let Ci be the event the ith relay is closed. Then
P(E) = P(E|C3)P(C3) + P(E|Cc
3)P(Cc
3)
= (p1p4 + p1p5 + p2p5 + p2p4)p3 + (p1p4 + p2p5)(1 − p3) .
Both of these can be checked by considering the entire joint distribution and eliminating
combinations that don’t allow current to flow. For example for Part (a) we have (conditioned

on switch five being closed) that
1 = p1p2p3p4 + (1 − p1)p2p3p4 + p1(1 − p2)p3p4 + p1p2(1 − p3)p4
+ p1p2p3(1 − p4) + (1 − p1)(1 − p2)p3p4 + (1 − p1)p2(1 − p3)p4
+ (1 − p1)p2p3(1 − p4) + p1(1 − p2)(1 − p3)p4 + p1(1 − p2)p3(1 − p4)
+ p1p2(1 − p3)(1 − p4) + (1 − p1)(1 − p2)(1 − p3)p4 + (1 − p1)(1 − p2)p3(1 − p4)
+ (1 − p1)p2(1 − p3)(1 − p4) + p1(1 − p2)(1 − p3)(1 − p4)
+ (1 − p1)(1 − p2)(1 − p3)(1 − p4) .
This explicit enumeration is possible is possible because these are Bernoulli random variables
(they can be on or off) and thus there are 2|c|
total elements in the joint distribution. From
the above enumeration we find (eliminating the non-functioning combinations) that
P(E|C5) = p1p2p3p4 + (1 − p1)p2p3p4 + p1(1 − p2)p3p4
+ p1p2(1 − p3)p4 + (1 − p1)(1 − p2)p3p4 + p1p2(1 − p3)(1 − p4)
= p2p3p4 + (1 − p2)p3p4 + p1p2(1 − p3)
= p3p4 + p1p2(1 − p3) .
Problem 67 (k-out-of-n systems)
Part (a): We must have two or more of the four components functioning so we can have
(with E the event that we have a functioning system) that
P(E) = p1p2p3p4 + (1 − p1)p2p3p4 + p1(1 − p2)p3p4 + p1p2(1 − p3)p4
+ p1p2p3(1 − p4) + (1 − p1)(1 − p2)p3p4 + (1 − p1)p2(1 − p3)p4
+ (1 − p1)p2p3(1 − p4) + p1(1 − p2)(1 − p3)p4 + p1(1 − p2)p3(1 − p4)
+ p1p2(1 − p3)(1 − p4) .
Problem 68 (is the relay open?)
For this problem let Ci be the event the ith relay is open and let E be the event current
flows from A to B. We can write the event E in terms of the events Ci as
E = (C1C2 ∪ C3C4)C5 .
The probability we want to evaluate is P((C1C2)c
|E) = 1 − P(C1C2|E). Now from the
definition of conditional probability we have
P(C1C2|E) =
P(C1C2E)
P(E)
.
Consider C1C2E as an “set” using the expression for E above it can be written as
C1C2E = C1C2C5 ∪ C1C2C3C4C5 = C1C2C3C4C5 ,

P1 P2 C = P1 C = P2
a,a a,a 1 1
a,a a,A 1/2 1/2
a,a A,A 0 0
a,A a,a 1/2 1/2
a,A a,A 1/2 1/2
a,A A,A 1/2 1/2
A,A a,a 0 0
A,A a,A 1/2 1/2
A,A A,A 1 1
Table 6: The probability of various matching genotypes, for the child denoted by C and the
two parents P1 and P2. The notations C = P1 and C = P2 means the the child’s genotype
matches that of the first and second parent respectively.
since C1C2C5 is a larger set than C1C2C3C4C5 in other words C1C2C5 ⊃ C1C2C3C4C5. Thus
P(C1C2E) = P(C1C2C3C4C5) = p1p2p3p4p5 .
Next using the relationship between the probability of the union of events we have
P(E) = P((C1C2 ∪ C3C4)C5) = p5P(C1C2 ∪ C3C4)
= p5(P(C1C2) + P(C3C4) − P(C1C2C3C4))
= p5(p1p2 + p3p4 − p1p2p3p4) .
Using these two expressions we have
P(C1C2|E) =
p1p2p3p4
p1p2 + p3p4 − p1p2p3p4
,
so that the desired probability is given by
P((C1C2)c
|E) = 1 − P(C1C2|E) =
p1p2 + p3p4 − 2p1p2p3p4
p1p2 + p3p4 − p1p2p3p4
.
Problem 69 (genotypes and phenotypes)
For this problem lets first consider the simplified case where we compute the probability
that a child receives various genotypes/phenotypes when crossing one single gene pair from
each parent. In Table 6 we list the probability of a child having genotypes that match the
first parent and second parent. These probabilities can be computed by considering the
possible genes that a given parent can give to his/her offspring. An example of this type of
subcalculation that goes into producing the entries in the second row of Table 6 is given in
Table 7. In this table we see the first parent has the gene pair aa and the second parent has
the gene pair Aa (or aA since they are equivalent). See the table caption for more details.
In the same way in Table 8 we list the probability of a child having phenotypes that match
(or not) the two parents. As each pair of genes is independent of the others by using the
two Tables 6 and 8 we can now answer the questions for this problem.

A a
a aA aa
a aA aa
Table 7: Example of the potential genotypes (and phenotypes) of the offspring produced
from mating the first parent with an aa genotype and a second parent with an Aa genotype.
We see that in 2 of 4 possible cases we get the gene pair aA and in 2 of the 4 possible cases
we get the gene pair aa. This gives the probability of 1/2 for either genotype aa or aA, and
probabilities of 1/2 for the recessive phenotype a and the dominant phenotype A.
Part (a): Using Table 6 and independence we see that to get a child’s genotype that matches
the first parent will happen with probability of
1
2
·
1
2
·
1
2
·
1
2
·
1
2
=
1
25
=
1
32
.
Using Table 8 and independence we see that to get a child’s phenotype that matches the
first parent will happen with probability of
1
2
·
3
4
·
1
2
·
3
4
·
1
2
=
9
27
=
9
128
.
Part (b): Using the Tables 6 and 8 we have the probability that a child’s genotype and
phenotype matches the second parent will happen with
1
2
·
1
2
·
1
2
·
1
2
·
1
2
=
1
25
=
1
32
,
and
1
2
·
3
4
·
1
2
·
3
4
·
1
2
=
9
27
=
9
128
.
Part (c): To match genotypes with either parent we must exactly match the first or the
second parent. Thus using the results from Part (a) and (b) above we get the probability of
matching either parents genotype is given by
1
32
+
1
32
=
1
16
.
In the same way to match phenotypes means that our phenotype must match the first or
the second child. Thus again using the results from Part (a) and (b) above we have the
probability of matching either parents phenotype is given by
9
128
+
9
128
=
9
64
.
Part (d): If we desire the probability we don’t match either parent, this is the complement
of the probability we do match one of the parents. Thus using the result from Part (c) above
we have that the probability of matching neither parents genotype is given by
1 −
1
16
=
15
16
,

P1 P2 C = P1 C = P2
a,a a,a 1 1
a,a a,A 1/2 1/2
a,a A,A 0 1
a,A a,a 1/2 1/2
a,A a,A 3/4 3/4
a,A A,A 1 1
A,A a,a 1 0
A,A a,A 1 1
A,A A,A 1 1
Table 8: The probability of various matching phenotypes, for the child denoted by C and
the two parents P1 and P2. The notations here match the same ones given in Table 6 but is
for phenotypes rather than genotypes.
and phenotype is given by
1 −
9
64
=
55
64
.
Problem 70 (hemophilia and the queen)
Let C be the event that the queen is a carrier of the gene for hemophilia. We are told that
P(C) = 0.5. Let Hi be the event that the i-th prince has hemophilia. The we observe the
event Hc
1Hc
2Hc
3 and we want to compute P(C|Hc
1Hc
2Hc
3). Using Bayes’ rule we have that
P(C|Hc
1Hc
2Hc
3) =
P(Hc
1Hc
2Hc
3|C)P(C)
P(Hc
1Hc
2Hc
3|C)P(C) + P(Hc
1Hc
2Hc
3|Cc)P(Cc)
.
Now
P(Hc
1Hc
2Hc
3|C) = P(Hc
1|C)P(Hc
2|C)P(Hc
3|C) .
By the independence of the birth of the princes. Now P(Hc
i |C) = 0.5 so that the above is
given by
P(Hc
1Hc
2Hc
3|C) = (0.5)3
=
1
8
.
Also P(Hc
1Hc
2Hc
3|Cc
) = 1 so the above probability becomes
P(C|Hc
1Hc
2Hc
3) =
(0.5)3
(0.5)
(0.5)3(0.5) + 1(0.5)
=
1
9
.
In the next part of this problem (below) we will need the complement of this probability or
P(Cc
|Hc
1Hc
2Hc
3) = 1 − P(C|Hc
1Hc
2Hc
3) =
8
9
.
If the queen has a fourth prince, then we want to compute P(H4|Hc
1Hc
2Hc
3). Let A be the
event Hc
1Hc
2Hc
3 (so that we don’t have to keep writing this) then conditioning on whether

Probability Win Loss Total Wins Total Losses
1
2
3
= 1
8
0 3 87 75
3 1
2
3
= 3
8
1 2 88 74
3 1
2
3
= 3
8
2 1 89 73
1
2
3
= 1
8
3 0 90 72
Table 9: The win/loss record for the Atlanta Braves each of the four total possible outcomes
when they play the San Diego Padres.
S.F.G. Total Wins S.F.G. Total Losses L.A.D. Total Wins L.A.D. Total Losses
86 76 89 73
87 75 88 74
88 74 87 75
89 73 86 76
Table 10: The total win/loss record for both the San Francisco Giants (S.F.G) and the Los
Angeles Dodgers (L.A.D.). The first row corresponds to the San Francisco Giants winning
no games while the Los Angeles Dodgers win three games. The number of wins going to the
San Francisco Giants increases as we move down the rows of the table, until we reach the
third row where the Giants have won three games and the Dodgers none.
the queen is a carrier, we see that the probability we seek is given by
P(H4|A) = P(H4|C, A)P(C|A) + P(H4|Cc
, A)P(Cc
|A)
= P(H4|C)P(C|A) + P(H4|Cc
)P(Cc
|A)
=
1
2

1
9

=
1
18
.
Problem 71 (winning the western division)
We are asked to compute the probabilities that each of the given team wins the western
division. We will assume that the team with the largest total number of wins will be the
division winner. We are also told that each team is equally likely to win each game it plays.
We can take this information to mean that each team wins each game it plays with probability
1/2. We begin to solve this problem, by considering the three games that the Atlanta Braves
play against the San Diego Padres. In Table 9 we enumerate all of the possible outcomes,
i.e. the total number of wins or losses that can occur to the Atlanta Braves during these
three games, along with the probability that each occurs.
We can construct the same type of a table for the San Francisco Giants when they play the
Los Angeles Dodgers. In Table 10 we list all of the possible total win/loss records for both
the San Francisco Giants and the Los Angeles Dodgers. Since the probabilities are the same
as listed in Table 9 the table does not explicitly enumerate these probabilities.
From these results (and assuming that the the team with the most wins will win the division)

1/8 3/8 3/8 1/8
1/8 D D G G
3/8 D B/D B/G G
3/8 B/D B B B/G
1/8 B B B B
Table 11: The possible division winners depending on the outcome of the three games that
each team must play. The rows (from top to bottom) correspond to the Atlanta Braves
winning more and more games (from the three that they play). The columns (from left to
right) correspond to the San Francisco Giants winning more and more games (from the three
they play). Note that as the Giants win more games the Dodgers must loose more games.
Ties are determined by the presence of two symbols at a given location.
we can construct a table which represents for each of the possible wins/losses combination
above, which team will be the division winner. Define the events B, G, and D to be the
events that the Braves, Giants, and Los Angles Dodgers win the western division. Then in
Table 11 we summarize the results of the two tables above where for the first row assumes
that the Atlanta Braves win none of their games and the last row assumes that the Atlanta
Braves win all of their games. In the same way the first column corresponds to the case
when the San Francisco Giants win none of their games and the last column corresponds to
the case when they win all of their games.
In anytime that two teams tie each team has a 1/2 of a chance of winning the tie-breaking
game that they play next. Using this result and the probabilities derived above we can
evaluate the individual probabilities that each team wins. We find that
P(D) =
1
8

1
8
+
3
8

+
3
8

1
8
+
1
2
·
3
8

+
3
8

1
2
·
1
8

=
13
64
P(G) =
1
8

3
8
+
1
8

+
3
8

1
2
·
3
8
+
1
8

+
3
8

1
2
·
1
8

=
13
64
P(B) =
3
8

1
2
·
3
8
+
1
2
·
3
8

+
3
8

1
2
·
1
8
+
3
8
+
3
8
+
1
2
·
1
8

+
1
8
(1) =
19
32
Note that these probabilities to add to one as they should. The calculations for this problems
are chap 3 prob 71.m.
Problem 72 (town council vote)
We can solve this first part of the problem in two ways. In the first, we select a member of
the steering committee (SC) to consider. Without loss of generality let this person be the
first member of the steering committee or i = 1 for i = 1, 2, 3. Let Vi be the event that
the ith person on the steering committee votes for the given piece of legislation. Then the
event V c
i is the event the i member votes against. Now all possible voting for a given piece
of legislation are
V1V2V3 , V c
1 V2V3 , V1V c
2 V3 , V1V2V c
3 , V1V c
2 V 2
3 , V c
1 V2V c
3 , V c
1 V c
2 V3 , V c
1 V c
2 V c
3 .

Since each event Vi is independent with probability p the probability of each of the events
above can be easily calculated. From the above, in only the events
V1V c
2 V3 , V1V2V c
3 , V c
1 V2V c
3 , V c
1 V c
2 V3 ,
will changing the vote of the i = 1 member change the total outcome. Summing the proba-
bility of these four events we find the probability we seek given by
p2
(1−p)+p2
(1−p)+p(1−p)2
+p(1−p)2
= 2p2
(1−p)+2p(1−p)2
= 2p(1−p)[p+1−p] = 2p(1−p) .
As a second way to work this problem let E be the event that the total vote outcome from the
steering committee will be different if the selected member changes his vote. Lets compute
P(E) by conditioning on whether V , our member voted for the legislation, or V c
he did not.
We have
P(E) = P(E|V )P(V ) + P(E|V c
)P(V c
) = P(E|V )p + P(E|V c
)(1 − p) .
Now to determine P(E|V ) the event that changing from a “yes” vote to a “no” vote will
change the outcome of the total decision we note that in order for that to be true we need to
have one “yes” vote and one “no” vote from the other members. In that case if we change
from “yes” to “no” the legislation will be rejected. Having one “yes” and one “no” vote
happens with probability
2
1

p(1 − p) .
Now to determine P(E|V c
) we reason the same way. This is the event that changing from
a “no” vote to a “yes” vote will change the outcome of the total decision. In order for that
to be true, we need to have one “no” vote and one “yes” vote from the other members. In
that case if we change from “no” to “yes” the legislation will be accepted. Having one “no”
and one “yes” vote happens (again) with probability

2
1

(1 − p)p .
Thus, summing the two results above we find
P(E) = 2p2
(1 − p) + 2(1 − p)2
p = 2p(1 − p) ,
the same as before.
When we move to the case with seven councilmen we assume that there is no guarantee
that the members of the original steering committee will vote the same way as earlier. Then
again to evaluate P(E) we condition on V and we have
P(E) = P(E|V )p + P(E|V c
)(1 − p) .
Evaluating P(E|V ), since there are 7 total people we need to have 4 total “yes” and 3 total
“no” thus removing the fact that V has occurred we need 6 “yes” and 3 “no” for a probability
of
P(E|V ) =

6
3

p3
(1 − p)3
.

The calculation of P(E|V c
) is the same. We need 4 total “no” and 3 total “yes” thus
removing the fact that V c
has occurred (we voted “no”) we need 6 “no” and 3 “yes” for a
probability of
P(E|V c
) =

6
3

(1 − p)3
p3
.
Thus we find since 6
3

= 20 that
P(E) = 20p4
(1 − p)3
+ 20p3
(1 − p)4
= 20p3
(1 − p)3
.
Problem 73 (5 children)
Part (a): To have all of the same type of children means that they are all girls or all boys
and will happen with a probability
1
2
5
+
1
2
5
=
1
32
+
1
32
=
1
16
.
Part (b): To first have 3 boys and then 2 girls will happen with probability
1
2
5
=
1
32
.
Part (c): To have exactly 3 boys (independent of their ordering) will happen with proba-
bility
5
3
1
2
31
2
5
=
10
32
.
Part (d): To have the first 2 children be girls (independent of what the other children are)
will happen with probability 1
2
2
=
1
4
Part (e): To have at least one girl (not all boys) is the complement of the even of having
no girls, or of having all boys. Thus
1 −
1
2
5
= 1 −
1
32
=
31
32
.
Problem 74 (our dice sum to 9 or 6)
From Equation 1 we see that the probability player A rolls a 9 is given by pA = 1
9
and the
probability player B rolls a 6 is given by pB = 5
36
. We can compute the probability the
games stops after n total rolls by reasoning as follows

• Since A starts the probability we stop with only one roll is pA.
• The probability we stop at the second roll is (1 − pA)pB.
• The probability we stop at the third roll is (1 − pA)(1 − pB)pA.
• The probability we stop at the fourth roll is
(1 − pA)(1 − pB)(1 − pA)pB = (1 − pA)2
(1 − pB)pB .
• The probability we stop at the fifth roll is (1 − pA)2
(1 − pB)2
pA.
• The probability we stop at the sixth roll is (1 − pA)3
(1 − pB)2
pB.
From the above special cases, it looks like the probability we stop after an odd number say,
2n − 1 rolls for n ≥ 1 is given by
(1 − pA)n−1
(1 − pB)n−1
pA ,
and we stop after an even number of rolls say 2n for n ≥ 1 is
(1 − pA)n
(1 − pB)n−1
pB .
The final roll is made by A with probability we roll 1, 3, 5, or an odd number of times.
We can evaluate the probability that A wins then as the sum of the elemental probabilities
above. We find
P{A wins} = pA + (1 − pA)(1 − pB)pA + · · · + (1 − pA)n−1
(1 − pB)n−1
pA + · · ·
=
∞
X
n=1
pA(1 − pA)n−1
(1 − pB)n−1
= pA
∞
X
n=0
(1 − pA)n
(1 − pB)n
= pA

1
1 − (1 − pA)(1 − pB)

.
When we use the values of pA and pB stated at the beginning of this problem we find
P{A wins} = 9
19
.
As another way to work this problem, let E be the event the last roll is made by player A
and let Ai be the event that A wins on round i for i = 1, 3, 5, · · · and let Bi be the event B
wins on round i for i = 2, 4, 6, · · ·. Then the event E (A wins) is the union of disjoint events
(much as above) as
E = A1 ∪ Ac
1Bc
2A3 ∪ Ac
1Bc
2Ac
3Bc
4A5 ∪ Ac
1Bc
2Ac
3Bc
4Ac
5Bc
6A7 ∪ . . .
Note that we can write the above in a way that emphasizes whether A wins on the first trial
(or not) in the following way
E = A1 ∪ Ac
1Bc
2 [A3 ∪ Ac
3Bc
4A5 ∪ Ac
3Bc
4Ac
5Bc
6A7 . . . ] . (11)
Notice that the term in brackets or
A3 ∪ Ac
3Bc
4A5 ∪ Ac
3Bc
4Ac
5Bc
6A7 . . . ,

is the event that A wins given that he did not win on the first roll and that B did not win
on the second roll. The probability of this even is the same as that of E since when A and
B do not win we are starting the game anew. Thus we can evaluate this as
P(Ac
1Bc
2 [A3 ∪ Ac
3Bc
4A5 ∪ Ac
3Bc
4Ac
5Bc
6A7 . . . ]) = P(A3 ∪ Ac
3Bc
4A5 ∪ Ac
3Bc
4Ac
5Bc
6A7 . . . |Ac
1Bc
2)P(Ac
1Bc
2)
= P(E)P(Ac
1Bc
2) .
Using Equation 11 we thus have shown
P(E) = P(A1) + P(Ac
1Bc
2)P(E) ,
or solving for P(E) in the above we get
P(E) =
P(A1)
1 − P(Ac
1Bc
2)
=
P(A1)
1 − P(Ac
1)P(Bc
2)
=
pA
1 − (1 − pA)(1 − pB)
,
the same expression as before.
Problem 75 (the eldest son)
Part (a): Let assume we have N families each of which can have four possible birth orderings
of their two children
BB , BG , GB , GG .
Here the notation B stands for boy child and G stands for girl child, thus the combined event
BG means that the family has a boy first followed by a girl and the same for the others.
From the N total families that have two children there are N
4
of them have two sons, N
2
of
them have one son, and N
4
of them have no sons. The total number of boys is then
2

N
4

+ 1

N
2

= N .
Since in the birth ordering BB there is only one eldest sons, while in BG and GB there is
also one eldest son the total number of eldest sons is
N
4
+
N
2
=
3N
4
.
Thus the fraction of all sons that are an eldest son is
3
4
N
N
=
3
4
.
Part (b): As in the first part, each family can have three children from the following possible
birth order
BBB , BBG , BGB , GBB , BGG , GBG , GGB , GGG .
From this we see that

• N
8
of the families have 3 sons of which only 1 is the eldest.
• 3N
8
of the families have 2 sons of which only 1 is the eldest.
• 3N
8
of the families have 1 son (who is also the oldest).
• N
8
of the families have 0 sons.
The total number of sons then is
3

N
8

+ 2

3N
8

+

3N
8

=
3N
2
.
The total number of eldest sons is
N
8
+
3N
8
+
3N
8
=
7N
8
.
The fraction of all sons that are eldest is
7N
8
3N
2
=
7
12
.
Problem 76 (mutually exclusive events)
If E and F are mutually exclusive events in an experiment, then P(E ∪ F) = P(E) + P(F).
We desire to compute the probability that E occurs before F, which we will denote by p. To
compute p we condition on the three mutually exclusive events E, F, or (E ∪ F)c
. This last
event are all the outcomes not in E or F. Letting the event A be the event that E occurs
before F, we have that
p = P(A|E)P(E) + P(A|F)P(F) + P(A|(E ∪ F)c
)P((E ∪ F)c
) .
Now
P(A|E) = 1
P(A|F) = 0
P(A|(E ∪ F)c
) = p ,
since if neither E or F happen the next experiment will have E before F (and thus event A
with probability p). Thus we have that
p = P(E) + pP((E ∪ F)c
)
= P(E) + p(1 − P(E ∪ F))
= P(E) + p(1 − P(E) − P(F)) .
Solving for p gives
p =
P(E)
P(E) + P(F)
,
as we were to show.

Problem 77 (running independent trials)
Part (a): Since the trials are independent, knowledge of the third experiment does not give
us any information about the outcome of the first experiment. Thus it could be any of the
3 choices (equally likely) so the probability we obtained a value of 1 is thus 1/3.
Part (b): Again by independence, we have a 1/3 chance in the first trial of getting a 1 and
a 1/3 chance of getting a 1 on the second trial. Thus we get a 1/9 chance of the first two
trials outputting 1’s.
Problem 78 (play till someone wins)
Part (a): According to the problem description whoever wins the last game of these four is
the winner and in the previous three games must have won once more than the other player.
Let E be the event that the game ends after only four games. Let EA be the event that
the winner of the last game is A. The event that B wins the last game is then Ec
A. Then
conditioning on who wins the last game we get
P(E) = P(E|EA)P(EA) + P(E|Ec
A)P(Ec
A)
= P(E|EA)p + P(E|Ec
A)(1 − p) .
To calculate P(E|EA) we note that in the three games before the last one is played because
the rules state we declare a winner if A or B gets two games ahead when we consider all the
possible ways we can assign winners to these three games:
AAA , AAB , ABA , BAA , ABB , BAB , BBA , BBB ,
we see that several of the orderings are not relevant for us. For example with the win pattern
AAA, AAB, BBA, BBB we would have stopped playing before the third game. For the
win patterns ABB and BAB we would keep playing after the fourth game won by A. Thus
only two orderings ABA and BAA will result in A winning after he wins the fourth game.
Thus
p(E|EA) = 2p2
(1 − p) .
By symmetry of A and B and p and 1 − p we thus have
P(E|Ec
A) = 2(1 − p)2
p ,
and the probability we want is given by
P(E) = 2p3
(1 − p) + 2(1 − p)3
p = 2p(1 − p)(p2
+ (1 − p)2
) .
Part (b): Consider the first two games. If we stop and declare a total winner then they
must be A1A2 or B1B2. If we don’t stop they must be A1B2 or B1A2. If we don’t stop, then
since A and B both have won one game they are “tied” and its like the game is started all

over again. Thus if we let E be the event that A wins the match we can evaluate P(E) by
conditioning on the result of the first two games. We find
P(E) = P(E|A1A2)P(A1A2) + P(E|B1B2)P(B1B2)
+ P(E|A1B2)P(A1B2) + P(E|B1A2)P(B1A2) .
Note that
P(E|A1A2) = 1
P(E|B1B2) = 0
P(E|A1B2) = P(E|B1A2) = P(E) ,
as in the last two cases the match “starts over”. Using independence we then have
P(E) = p2
+ 2p(1 − p)P(E) .
Solving for P(E) in the above we get
P(E) =
p2
1 − 2p(1 − p)
=
p2
(1 − p + p)2 − 2p(1 − p)
=
p2
(1 − p)2 + p2
.
Problem 79 (2 7’s before 6 even numbers)
In the problem statement we take the statement that we get 2 7’s before 6 even numbers
to not require 2 consecutive 7’s or 6 consecutive even numbers. What is required is that
the second occurrence of a 7 is before the sixth occurrence of an even number in the total
sequence of trials. We will evaluate this probability by considering what happens on the first
trial. Let S1 be the event a seven is rolled on the first trial, let E1 be the event an even
number is rolled on the first trial and let X1 be the event that neither a 7 or an even number
is rolled on the first trial. Finally, let Ri,j be the event we roll i 7’s before j even numbers
for i ≥ 0 and j ≥ 0. We will compute the probability of the event Ri,j by conditioning on
what happens on the first trial as
P(Ri,j) = P(Ri,j|S1)P(S1) + P(Ri,j|E1)P(E1) + P(Ri,j|X1)P(X1) . (12)
Using Equation 1 the probabilities of the “one step” events S1, E1, and X1 are given by
P(S1) =
1
6
≡ s , P(E1) =
1
2
≡ e , P(X1) = 1 −
1
6
−
1
2
=
1
3
≡ x ,
where these probabilities are constant on each subsequent trial. The probabilities of the
conditional events when i 1 and j 1 are given by
P(Ri,j|S1) = P(Ri−1,j)
P(Ri,j|E1) = P(Ri,j−1)
P(Ri,j|X1) = P(Ri,j) .

With all of this Equation 12 above becomes
P(Ri,j) = sP(Ri−1,j) + eP(Ri,j−1) + xP(Ri,j) for i 1 , j 1 . (13)
For this problem we want to evaluate P(R2,6) which using Equation 13 with i = 2 gives
P(R2,j) = sP(R1,j) + eP(R2,j−1) + xP(R2,j) ,
or solving for P(R2,j) we have
P(R2,j) =
s
1 − x
P(R1,j) +
e
1 − x
P(R2,j−1) . (14)
The first term on the right-hand-side shows that we need to be able to evaluate is P(R1,j)
for j 1. We can compute this expression by taking i = 1 in Equation 12, where we get
P(R1,j) = sP(R1,j|S1) + eP(R1,j|E1) + xP(R1,j|X1)
= s + eP(R1,j−1) + xP(R1,j) ,
where we have simplified by using P(R1,j|S1) = 1 and P(R1,j|E1) = P(R1,j−1). Solving for
P(R1,j) we find
P(R1,j) =
s
1 − x
+
e
1 − x
P(R1,j−1) .
Based on the terms in the above expression lets define η = s
1−x
and ξ = e
1−x
where the above
becomes
P(R1,j) = η + ξP(R1,j−1) .
This can be iterated for j = 2, 3, · · · to give
P(R1,2) = η + ξP(R1,1)
P(R1,3) = η + ξη + ξ2
P(R1,1)
P(R1,4) = η + ξη + ξ2
η + ξ3
P(R1,1)
.
.
.
P(R1,j) = η
j−2
X
k=0
ξk
+ ξj−1
P(R1,1) for j ≥ 2 .
We can find the value of P(R1,1) by taking i = 1 and j = 1 in Equation 12. We find
P(R1,1) = P(R1,1|S1)P(S1) + P(R1,1|E1)P(E1) + P(R1,1|X1)P(X1)
= s + xP(R1,1) .
When we solve the above for P(R1,1) we get P(R1,1) = s
1−x
= η. Thus, using this we see
that the expression for P(R1,j) then becomes
P(R1,j) = η
j−1
X
k=0
ξk
for j ≥ 1 . (15)
Note that for any given value of j ≥ 1 we can evaluate the above sum to compute P(R1,j).
Thus we can consider this a known function of j. Using this, and the expression for P(R2,j)
in Equation 14 we have
P(R2,j) = ηP(R1,j) + ξP(R2,j−1) .

This we can iterate by letting j = 2, 3, · · · and observing the resulting pattern. We find
P(R2,2) = ηP(R1,2) + ξP(R2,1)
P(R2,3) = ηP(R1,3) + ξP(R2,2) = ηP(R1,3) + ξηP(R1,2) + ξ2
P(R2,1)
= η(P(R1,3) + ξP(R1,2)) + ξ2
P(R2,1)
P(R2,4) = ηP(R1,4) + ξηP(R1,3) + ηξ2
P(R1,2) + ξ3
P(R2,1)
= η(P(R1,4) + ξP(R1,3) + ξ2
P(R1,2)) + ξ3
P(R2,1)
.
.
.
P(R2,j) = η
j−2
X
k=0
ξk
P(R1,j−k) + ξj−1
P(R2,1) for j ≥ 2 .
To use this we need to evaluate P(R). Again using Equation 12 we have
P(R2,1) = sP(R1,1) + eP(R2,1|E1) + xP(R2,1) = sP(R1,1) + xP(R2,1) ,
Solving for P(R2,1) we get
P(R2,1) =
s
1 − x
P(R1,1) =

s
1 − x
2
= η2
.
Thus we finally have for P(R2,j) the following expression
P(R2,j) = η
j−2
X
k=0
ξk
P(R1,j−k) + η2
ξj−1
for j ≥ 2 .
From the numbers given for s, e, and x we find
η =
s
1 − x
=
s
s + e
=
1
6
1
6
+ 1
2
=
1
4
, ξ =
e
1 − x
=
e
s + e
=
1
2
1
6
+ 1
2
=
3
4
.
Thus we can evaluate P(R2,6) using the above sum. In the python code chap 3 prob 79.py
we implement the above computation. When we evaluate that code we get the probability
0.555053710938.
Problem 80 (contest playoff)
Part (a): We have that A will play in the first contest with probability one. He will play
in the second contest only if he wins the first contest which happens with probability 1
2
.
He will play in the third contest only if he wins the first two contests which happens with
probability 1
2
2
= 1
4
. In general we have
P(Ai) =

1
2
i
for 1 ≤ i ≤ n .
Part (d): Consider various values for n. Then by considering the game trees for these values
we see that, if n = 1 then 21
people play 1 game. If n = 2 then 23
play 3 games. If n = 3

then 7 games are played. Note in all of these cases the number of games played is 2n
− 1.
Let Gn be the number of games played when we have 2n
players. We can derive a recursive
relationship for this as follows. Since on the first round we pair the 2n
players in the first
game we will play 1
2
2n
games. After the first round is played, half of the players have lost
and no longer need to be considered. We thus have to consider 2n−1
players who will play
Gn−1 games. The total number of games played is the sum of these two numbers. This gives
that
Gn =
1
2
2n
+ Gn−1 = 2n−1
+ Gn−1 .
Using this recursion relationship and the first few value of Gn computed earlier we can prove
by induction that Gn = 2n
− 1.
Problem 81 (the stock market)
We are told that the value of the stock goes up or down 1 point successively and we want
the probability that the value of the stock goes up to 40 before it goes down to 10. Another
way to state this is that since the initial value of the stock is 25, we want the probability the
stock goes up 15 points before it goes down 15 points. This problem is like the gambler’s
ruin problem, discussed in this chapter of the book in Example 4k. In the terminology of
that example we assume that the stock owner and “the stock market” are playing a game
that the stock owner wins with probability p = 0.55 and each player starts with 15 points.
We then want the probability the gambler’s fortune goes up to 30 units before it goes down
to 0 units starting with an initial fortune of 15 units. Let E be the event gambler A (the
stock owner) ends up with all the money when he starts with 15 units and gambler B (the
market) starts with (30 − 15) = 15 units. From Example 4k we have
P(E) =
1 − (q/p)15
1 − (q/p)30
=
1 − (0.45/0.55)15
1 − (0.45/0.55)30
= 0.95302 .
Problem 82 (flipping coins until a tail occurs)
Because Parts (a) and (c) are similar while Parts (b) and (d) are similar, I’ve solved them
grouped in that order.
Part (a): For this part we want the probability that A gets 2 heads in a row before B does.
To find this, let A be the event that A gets 2 heads in a row before B in a game where A
goes first. In the same way let B be the event A gets 2 heads in a row before B in a game
where now B goes first. Note in each event A or B the player A “wins”. These events will
come up naturally when we try to evaluate P(A) by conditioning on the outcome of the first
two flips. To do that we introduce the events that denote what happens on the first two
flips. Let HA
1 , HA
2 , TA
1 , TA
2 be the events A’s coin lands heads up or tails up on the first or
second flips. Let HB
1 , HB
2 , TB
1 , TB
2 be the events B’s coin lands heads up or tails up on the

first or second flips. Then conditioning on what happens in the first few flips we have
P(A) = P(A|HA
1 HA
2 )P(HA
1 HA
2 ) + P(A|HA
1 TA
2 )P(HA
1 TA
2 ) + P(A|TA
1 )P(TA
1 )
= (1)P(HA
1 HA
2 ) + P(B)P(HA
1 TA
2 ) + P(B)P(TA
1 )
= P2
1 + [P1(1 − P1) + (1 − P1)]P(B)
= P2
1 + (1 + P1)(1 − P1)P(B) .
Where we have used expressions like P(A|HA
1 TA
2 ) = P(B) since in the case where A first flips
HA
1 TA
2 (or TA
1 ) the dice go to B and all memory of the number of heads flipped is forgotten.
In the above expression we need to evaluate P(B) we can do this by again conditioning on
the first few flips. We find
P(B) = P(B|HB
1 HB
2 )P(HB
1 HB
2 ) + P(B|HB
1 TB
2 )P(HB
1 TB
2 ) + P(B|TB
1 )P(TB
1 )
= (0)P(HB
1 HB
2 ) + P(A)P(HB
1 TB
2 ) + P(A)P(TB
1 )
= (P2(1 − P2) + (1 − P2))P(A)
= (1 + P2)(1 − P2)P(A) .
Putting this expression into the previous expression derived for P(A) and we get
P(A) = P2
1 + (1 + P1)(1 − P1)(1 + P2)(1 − P2)P(A) .
Solving for P(A) we get
P(A) =
P2
1
1 − (1 + P1)(1 − P1)(1 + P2)(1 − P2)
=
P2
1
1 − (1 − P2
1 )(1 − P2
2 )
.
Part (c): For this part we want the probability that A gets 3 heads in a row before B does.
In the same way as Part (a), let A be the event that A gets 3 heads in a row before B in a
game where A goes first. In the same way let B be the event A gets 3 heads in a row before
B in a game where now B goes first. Again in each event the player A “wins”. Again let
HA
1 , HA
2 , HA
3 , TA
1 , TA
2 , TA
3 be the events the A’s coin lands heads up or tails up on the first,
second or third flips. Let HB
1 , HB
2 , HB
3 , TB
1 , TB
2 , TB
3 be the events the B’s coin lands heads up
or tails up on the first, second or third flips. By conditioning on the first few flips we have
P(A) = P(A|HA
1 HA
2 HA
3 )P(HA
1 HA
2 HA
3 )
+ P(A|HA
1 HA
2 TA
3 )P(HA
1 HA
2 TA
3 ) + P(A|HA
1 TA
2 )P(HA
1 TA
2 ) + P(A|TA
1 )P(TA
1 )
= (1)P(HA
1 HA
2 HA
3 ) + P(HA
1 HA
2 TA
3 )P(B) + P(HA
1 TA
2 )P(B) + P(TA
1 )P(B)
= P3
1 + [P2
1 (1 − P1) + P1(1 − P1) + (1 − P1)]P(B)
= P3
1 + (1 − P1)(P2
1 + P1 + 1)P(B) .
As this involves P(B) in the same way as earlier we compute that now
P(B) = P(B|HB
1 HB
2 HB
3 )P(HB
1 HB
2 HB
3 )
+ P(B|HB
1 HB
2 TB
3 )P(HB
1 HB
2 TB
3 ) + P(B|HB
1 TB
2 )P(HB
1 TB
2 ) + P(B|TB
1 )P(TB
1 )
= (0)P(HB
1 HB
2 HB
3 ) + P(A)P(HB
1 HB
2 TB
3 ) + P(A)P(HB
1 TB
2 ) + P(A)P(TB
1 )
= (P2
2 (1 − P2) + P2(1 − P2) + (1 − P2))P(A)
= (1 − P2)(P2
2 + P2 + 1)P(A) .

Using this in the expression derived for P(A) earlier we have
P(A) = P3
1 + (1 − P1)(P2
1 + P1 + 1)(1 − P2)(P2
2 + P2 + 1)P(A) .
Solving this for P(A) we find
P(A) =
P3
1
1 − (P2
1 + P1 + 1)(1 − P1)(P2
2 + P2 + 1)(1 − P2)
=
P3
1
1 − (1 − P3
1 )(1 − P3
2 )
.
Part (b): For this part we want the probability that A gets a total of 2 heads before B
does. Unlike the previous parts of the problem we now have to remember the number of
heads that a given player has already made. Since that number makes it more or less likely
for him to win the game. The motivation for this solution is that if A starts and gets a head
on the first flip then A continues flipping. We can view the rest of the game from this point
on as a new game where A starts and A wins if A gets a total of 2 − 1 = 1 head before B
gets a total of 2 heads. In the case when A gets tails on the first flip then B starts flipping.
We can view the rest of the game as a new game where B now starts and A wins if A gets a
total of 2 heads before B gets a total of 2 heads. Motivated by this, let A(i, j) be the event
A gets 2 − i heads before B gets 2 − j heads in a game where A starts and in the same way
we let B(i, j) be the event A gets 2 − i heads before B gets 2 − j heads in a game where B
starts. In all cases we have 0 ≤ i ≤ 2. In both of these events A “wins”. Thus the indices
i and j count the number of heads that each player has already received. The problem asks
us to then find P(A(0, 0)). To do this it will also be easier to compute P(B(0, 0)) at the
same time. Then conditioning on the result of the first flip we can compute P(A(i, j)) and
P(B(i, j)) as
P(A(i, j)) = P(A(i + 1, j))P1 + P((B(i, j))(1 − P1)
P(B(i, j)) = P(B(i, j + 1))P2 + P((A(i, j))(1 − P2) . (16)
This is a recursive system where we need to evaluate P(A(i, j)) and P(B(i, j)) at various
value of i and j to get P(A(0, 0)) and P(B(0, 0)). Note that to evaluate each of P(A(i, j))
and P(B(i, j)) using the above we need to know P(A(i + 1, j)) and P(B(i, j + 1)). That
is we need probability values at the neighboring grid points (i + 1, j) and (i, j + 1). This
motivates us to start with the boundary conditions
P(A(2, j)) = 1 for 0 ≤ j ≤ 1
P(B(2, j)) = 1 for 0 ≤ j ≤ 1
P(A(i, 2)) = 0 for 0 ≤ i ≤ 1
P(B(i, 2)) = 0 for 0 ≤ i ≤ 1 .
and work backwards. We do this by using the above equations to solve for P(A(i, j)) and
P(B(i, j)) at (i, j) = (1, 1), then (i, j) = (0, 1), then (i, j) = (1, 0) and finally (i, j) = (0, 0).
To begin, let i = j = 1 in Equation 16 to get
P(A(1, 1)) = P1P(A(2, 1)) + (1 − P1)P(B(1, 1)) = P1 + (1 − P1)P(B(1, 1))
P(B(1, 1)) = P2P(B(1, 2)) + (1 − P2)P(A(1, 1)) = (1 − P2)P(A(1, 1)) .

When we solve this for P(A(1, 1)) and P(B(1, 1)) we get
P(A(1, 1)) =
P1
1 − (1 − P1)(1 − P2)
=
P1
P1 + P2 − P1P2
P(B(1, 1)) =
P1(1 − P2)
1 − (1 − P1)(1 − P2)
=
P1(1 − P2)
P1 + P2 − P1P2
. (17)
Now let i = 1 and j = 0 in Equation 16 and we get
P(A(1, 0)) = P1P(A(2, 0)) + (1 − P1)P(B(1, 0)) = P1 + (1 − P1)P(B(1, 0))
P(B(1, 0)) = P2P(B(1, 1)) + (1 − P2)P(A(1, 0)) .
Since we know the value of P(B(1, 1)) via Equation 17 we can solve the above for P(A(1, 0))
and P(B(1, 0)). Using Mathematica we get
P(A(1, 0)) =
P1(P1(1 − P2)2
+ (2 − P2)P2)
(P1 + P2 − P1P2)2
P(B(1, 0)) =
P1(1 − P2)(P1 + 2P2 − P1P2)
(P1 + P2 − P1P2)2
.
Next let i = 0 and j = 1 in Equation 16 to get
P(A(0, 1)) = P1P(A(1, 1)) + (1 − P1)P(B(0, 1))
P(B(0, 1)) = P2P(B(0, 2)) + (1 − P2)P(A(0, 1)) = (1 − P2)P(A(0, 1)) .
Since we have already evaluated P(A(1, 1)) we can solve the above for P(A(0, 1)) and
P(B(0, 1)). Using Mathematica when we do that we get
P(A(0, 1)) =
P2
1
(P1 + P2 − P1P2)2
P(B(0, 1)) =
P2
1 (1 − P2)
(P1 + P2 − P1P2)2
.
Finally let i = 0 and j = 0 in Equation 16 to get
P(A(0, 0)) = P1P(A(1, 0)) + (1 − P1)P(B(0, 0))
P(B(0, 0)) = P2P(B(0, 1)) + (1 − P2)P(A(0, 0)) .
Since we know expressions for P(A(1, 0)) and P(B(0, 1)) we can solve the above for P(A(0, 0))
and P(B(0, 0)). Where we find
P(A(0, 0)) =
P2
1 ((3 − 2P2)P2 + P1(1 − 3P2 + 2P2
2 ))
(P1(1 − P2) + P2)3
P(B(0, 0)) =
P2
1 (1 − P2)(P1(1 − P2)2
+ (3 − P2)P2)
(P1(1 − P2) + P2)3
.
Part (d): Just as in Part (b) above we now have to keep track of the number of heads that
each player has received as they play the game. We will use the same notation as above. We

now have the boundary conditions
P(A(3, j)) = 1 for 0 ≤ j ≤ 2
P(B(3, j)) = 1 for 0 ≤ j ≤ 2
P(A(i, 3)) = 0 for 0 ≤ i ≤ 2
P(B(i, 3)) = 0 for 0 ≤ i ≤ 2 .
with the same recursion relationship given by Equation 16 and work backwards to derive
an expression for P(A(0, 0)) and P(B(0, 0)). We will outline the calculation that we will
perform before we will write out recurrence relationships, as before, such that we will always
be able to solve for P(A(·, ·)) and P(B(·, ·)) at each step. We write out the recurrence
relationship for the grid point (i, j) in the following orders: (2, 2), (1, 2), (2, 1), (0, 2), (1, 1),
(2, 1), (0, 1), (1, 0), and finally (0, 0) which will be the desired result. The ordering of these
points is starting from the upper right corner of the (i, j) plane and working to the lower
left corner towards (0, 0) diagonally. In this manner we have the correct boundary equations
and previous solutions in place to solve for the next values of P(A(·, ·)) and P(B(·, ·)). This
procedure can be implemented in Mathematica where we obtain
P3
1 (P2
2 (10 − 12P2 + 3P2
2 ) + P2
1 (−1 + P2)2
(1 − 3P2 + 3P2
2 ) + P1P2(5 − 20P2 + 21P2
2 − 6P3
2 ))
(P1 + P2 − P1P2)5
,
for P(A(0, 0)) and
P3
1 (1 − P2)(P2
1 (−1 + P2)4
+ P2
2 (10 − 8P2 + P2
2 ) + P1P2(5 − 15P2 + 12P2
2 − 2P3
2 ))
(P1 + P2 − P1P2)5
,
for P(B(0, 0)). The algebra for this problem is worked in the Mathematica file chap 3 prob 82.nb.
Problem 83 (red and white dice)
Let H be the event the coin lands heads up and we select the die A. From this Hc
is then
the event that the coin lands tails up and we select die B. Let Rn be the event a red face is
showing on the nth roll of the die.
Part (a): We can compute P(Rn) by conditioning on the result of the coin flip. We have
P(Rn) = P(H)P(Rn|H) + P(Hc
)P(Rn|Hc
)
=

1
2

4
6

+

1
2

2
6

=
1
2
,
when we simplify.
Part (b): We want to compute P(R3|R1, R2). We will do that using the definition of
conditional probability or
P(R3|R1R2) =
P(R1R2R3)
P(R1R2)
.

We will evaluate P(R1R2) and P(R1R2R3) by conditioning on the result of the first coin flip.
We have
P(R1R2) = P(R1R2|H)P(H) + P(R1R2|Hc
)P(Hc
)
= P(R1|H)P(R2|H)P(H) + P(R1|Hc
)P(R2|Hc
)P(Hc
) .
and the same for P(R1R2R3) or
P(R1R2R3) = P(R1R2R3|H)P(H) + P(R1R2R3|Hc
)P(Hc
)
= P(H)P(R1|H)P(R2|H)P(R3|H) + P(Hc
)P(R1|H)P(R2|H)P(R3|Hc
) .
Thus we get for the probability we want
P(R3|R1R2) =
P(H)P(R1|H)P(R2|H)P(R3|H) + P(Hc
)P(R1|H)P(R2|H)P(R3|Hc
)
P(H)P(R1|H)P(R2|H) + P(Hc)P(R1|Hc)P(R2|Hc)
=
1
2
4
6
3
+
1
2
2
6
3
1
2
4
6
2
+
1
2
2
6
2 =
2
3
3
+
1
3
3
2
3
2
+
1
3
2 =
8
27
+
1
27
4
9
+
1
9
=
9
27
5
9
=
3
5
.
Part (c): For this part we want P(H|R1R2). We have
P(H|R1R2) =
P(HR1R2)
P(R1R2)
=
P(H)P(R1|H)P(R2|H)
P(H)P(R1|H)P(R2|H) + P(Hc)P(R1|Hc)P(R2|Hc)
=
4
9
5
9
=
4
5
.
Problem 84 (4 white balls in an urn)
Some definitions that will be used in both parts of this problem. Let i be the index of
the trial where one ball is drawn. Depending on what happens during the game A will be
drawing on the trials i = 1, 4, 7, 10, . . ., B will be drawing on the trials i = 2, 5, 8, 11, . . . and
C will be drawing on the trials i = 3, 6, 9, 12, . . .. Let Wi be the event that a white ball is
selected on the ith draw (by whoever is drawing at the time). Finally, let A be the event A
draws the first white ball and therefore wins. The same for the events B and C.
Part (a): In this case there is no memory as we place and the players keep drawing balls
until there is a winner. We can compute the probability that A wins by conditioning on the
result of the first set of draws. We find
P(A) = P(A|W1)P(W1) + P(A|Wc
1 Wc
2 Wc
3 )P(Wc
1 Wc
2 Wc
3 )
= P(W1)P(A|W1) + P(A|Wc
1 Wc
2 Wc
3 )P(Wc
1 )P(Wc
2 )P(Wc
3 )
= P(W1)(1) + P(A)P(Wc
1 )P(Wc
2 )P(Wc
3 ) .

We can solve the above for P(A) where we find
P(A) =
P(W1)
1 − P(Wc
1 )P(Wc
2 )P(Wc
3 )
=
4
12
1 −
8
12
3 =
1
3
1 −
2
3
3 =
9
19
.
We can evaluate P(B) in the same way. We have
P(B) = P(B|Wc
1 W2)P(Wc
1 W2) + P(Wc
4 B|Wc
1 Wc
2 Wc
3 )P(Wc
1 Wc
2 Wc
3 )
= P(B|Wc
1 W2)P(Wc
1 )P(W2) + P(Wc
4 B|Wc
1 Wc
2 Wc
3 )P(Wc
1 )P(Wc
2 )P(Wc
3 )
= P(Wc
1 )P(W2)(1) + P(Wc
1 )P(Wc
2 )P(Wc
3 )P(B) .
Solving for P(B) we get
P(B) =
P(Wc
1 )P(W2)
1 − P(Wc
1 )P(Wc
2 )P(Wc
3 )
=
(
8
12
)(
4
12
)
1 − (
8
12
)3
=
(
2
3
)(
1
3
)
1 − (
2
3
)3
=
6
19
.
For P(C) we have
P(C) = P(C|Wc
1 Wc
2 W3)P(Wc
1 Wc
2 W3) + P(Wc
4 Wc
5 C|Wc
1 Wc
2 Wc
3 )P(Wc
1 Wc
2 Wc
3 )
= P(C|Wc
1 Wc
2 W3)P(Wc
1 )P(Wc
2 )P(W3) + P(Wc
4 Wc
5 C|Wc
1 Wc
2 Wc
3 )P(Wc
1 )P(Wc
2 )P(Wc
3 )
= P(Wc
1 )P(Wc
2 )P(W3)(1) + P(Wc
1 )P(Wc
2 )P(Wc
3 )P(C) .
Solving for P(C) we get
P(C) =
P(Wc
1 )P(Wc
2 )P(W3)
1 − (P(Wc
1 )P(Wc
2 )P(Wc
3 ))
=
(
8
12
)2
(
4
12
)
1 − (
8
12
)3)
=
(
2
3
)2
(
1
3
)
1 − (
2
3
)3)
=
4
19
We can check that P(A) + P(B) + P(C) = 1 using the above numbers as it should.
Part (b): In this case we remove balls as each draw is made. Because of this the probabilities
change after each ball is removed. In this case the game cannot run forever since the longest
it can run would be the case where the white balls are drawn after all of the others. In other
words we can draw at most 8 balls before getting a white ball. To compute the probability
that A wins we can write this event as
W1 ∪ Wc
1 Wc
2 Wc
3 W4 ∪ Wc
1 Wc
2 Wc
3 Wc
4 Wc
5 Wc
6 W7 .
Note we don’t have the possibility of W8 since if A does not draw a white ball in the first
7 draws there is no way he can (some other player will draw it). Using the product rule of
probability to evaluate the above union of independent events we have
P(A) = P(W1) + P(Wc
1 Wc
2 Wc
3 W4) + P(Wc
1 Wc
2 Wc
3 Wc
4 Wc
5 Wc
6 W7)
= P(W1) + P(Wc
1 )P(Wc
2 |Wc
1 )P(Wc
3 |Wc
1 Wc
2 )P(W4|Wc
1 Wc
2 Wc
3 )
+ P(Wc
1 )P(Wc
2 |Wc
1 )P(Wc
3 |Wc
1 Wc
2 )P(Wc
4 |Wc
1 Wc
2 Wc
3 )
× P(Wc
5 |Wc
1 Wc
2 Wc
3 Wc
4 )P(Wc
6 |Wc
1 Wc
2 Wc
3 Wc
4 Wc
5 )P(W7|Wc
1 Wc
2 Wc
3 Wc
4 Wc
5 Wc
6 )
=
4
12
+ (
8
12
)(
7
11
)(
6
10
)(
4
9
) + (
8
12
)(
7
11
)(
6
10
)(
5
9
)(
4
8
)(
3
7
)(
4
6
) =
7
15
.

To compute the probability that B wins we write this event as
Wc
1 W2 ∪ Wc
1 Wc
2 Wc
3 Wc
4 W5 ∪ Wc
1 Wc
2 Wc
3 Wc
4 Wc
5 Wc
6 Wc
7 W8 .
From which we can compute the probability using
P(B) = P(Wc
1 W2 ∪ Wc
1 Wc
2 Wc
3 Wc
4 W5 ∪ Wc
1 Wc
2 Wc
3 Wc
4 Wc
5 Wc
6 Wc
7 W8)
= P(Wc
1 W2) + P(Wc
1 Wc
2 Wc
3 Wc
4 W5) + P(Wc
1 Wc
2 Wc
3 Wc
4 Wc
5 Wc
6 Wc
7 W8)
= P(Wc
1 )P(W2|Wc
1 )
+ P(Wc
1 )P(Wc
2 |Wc
1 )P(Wc
3 |Wc
1 Wc
2 )P(Wc
4 |Wc
1 Wc
2 Wc
3 )P(W5|Wc
1 Wc
2 Wc
3 Wc
4 )
+ P(Wc
1 )P(Wc
2 |Wc
1 )P(Wc
3 |Wc
1 Wc
2 )P(Wc
4 |Wc
1 Wc
2 Wc
3 )P(Wc
5 |Wc
1 Wc
2 Wc
3 Wc
4 )
× P(Wc
6 |Wc
1 Wc
2 Wc
3 Wc
4 Wc
5 )P(Wc
7 |Wc
1 Wc
2 Wc
3 Wc
4 Wc
5 Wc
6 )P(W8|Wc
1 Wc
2 Wc
3 Wc
4 Wc
5 Wc
6 Wc
7 )
= (
8
12
)(
4
11
) + (
8
12
)(
7
11
)(
6
10
)(
5
9
)(
4
8
) + (
8
12
)(
7
11
)(
6
10
)(
5
9
)(
4
8
)(
3
7
)(
2
6
)(
4
5
) =
53
165
.
To compute the probability that C wins we write this event as
Wc
1 Wc
2 W3 ∪ Wc
1 Wc
2 Wc
3 Wc
4 Wc
5 W6 ∪ Wc
1 Wc
2 Wc
3 Wc
4 Wc
5 Wc
6 Wc
7 Wc
8 W9 .
From which we can compute the probability using
P(C) = P(Wc
1 Wc
2 W3 ∪ Wc
1 Wc
2 Wc
3 Wc
4 Wc
5 W6 ∪ Wc
1 Wc
2 Wc
3 Wc
4 Wc
5 Wc
6 Wc
7 Wc
8 W9)
= P(Wc
1 Wc
2 W3) + P(Wc
1 Wc
2 Wc
3 Wc
4 Wc
5 W6) + P(Wc
1 Wc
2 Wc
3 Wc
4 Wc
5 Wc
6 Wc
7 Wc
8 W9)
= P(Wc
1 )P(W2|Wc
1 )P(W3|Wc
1 Wc
2 )
+ P(Wc
1 )P(Wc
2 |Wc
1 )P(Wc
3 |Wc
1 Wc
2 )
× P(Wc
4 |Wc
1 Wc
2 Wc
3 )P(Wc
5 |Wc
1 Wc
2 Wc
3 Wc
4 )P(W6|Wc
1 Wc
2 Wc
3 Wc
4 Wc
5 )
+ P(Wc
1 )P(Wc
2 |Wc
1 )P(Wc
3 |Wc
1 Wc
2 )
× P(Wc
4 |Wc
1 Wc
2 Wc
3 )P(Wc
5 |Wc
1 Wc
2 Wc
3 Wc
4 )P(Wc
6 |Wc
1 Wc
2 Wc
3 Wc
4 Wc
5 )
× P(Wc
7 |Wc
1 Wc
2 Wc
3 Wc
4 Wc
5 Wc
6 )P(Wc
8 |Wc
1 Wc
2 Wc
3 Wc
4 Wc
5 Wc
6 Wc
7 )P(W9|Wc
1 Wc
2 Wc
3 Wc
4 Wc
5 Wc
6 Wc
7 Wc
8 )
= (
8
12
)(
7
11
)(
4
10
) + (
8
12
)(
7
11
)(
6
10
)(
5
9
)(
4
8
)(
4
7
) + (
8
12
)(
7
11
)(
6
10
)(
5
9
)(
4
8
)(
3
7
)(
2
6
)(
1
5
)(1) =
7
33
.
We can again check that P(A) + P(B) + P(C) = 1 using the above numbers as it should.
Problem 85 (4 white balls in 3 urns)
Part (a): This is the same as Problem 84 Part (a).
Part (b): First note that each tail A flips reduces the number of non-white balls by one.
Thus we can have at most 8 non-white flips before we must get a white ball. Using the same

notation from Problem 84 we have that the event that A wins given by
A = W1 ∪ Wc
1 Wc
2 Wc
3 W4 ∪ Wc
1 Wc
2 Wc
3 Wc
4 Wc
5 Wc
6 W7 . . .
= W1 ∪
3
Y
i=1
Wc
i
!
W4 ∪
6
Y
i=1
Wc
i
!
W7 ∪
9
Y
i=1
Wc
i
!
W10
∪
12
Y
i=1
Wc
i
!
W13 ∪
15
Y
i=1
Wc
i
!
W16 ∪
18
Y
i=1
Wc
i
!
W19
∪
21
Y
i=1
Wc
i
!
W22 ∪
24
Y
i=1
Wc
i
!
W25
= ∪8
n=0
3n
Y
i=1
Wc
i
!
W3n+1 .
Which are the events that A wins by flipping no tails, one tail, two tails etc. We are using
the convention that
Q0
i=1 · = 1. Thus the probability A wins is given by
P(A) =
4
12
+
8
12
3 4
11

+
8
12
3 7
11
3 4
10

+ . . .
=
8
X
n=0
n
Y
k=1

8 − (k − 1)
12 − (k − 1)
3
!
4
12 − n

=
8
X
n=0
n
Y
k=1

9 − k
13 − k
3
!
4
12 − n

.
The event B wins is given by
B = Wc
1 W2 ∪ Wc
1 Wc
2 Wc
3 Wc
4 W5 ∪ Wc
1 Wc
2 Wc
3 Wc
4 Wc
5 Wc
6 Wc
7 W8 ∪ · · · ∪
3n+1
Y
i=1
Wc
i
!
W3n+2 ∪ . . . .
Which are the events that B wins by flipping no tails, one tail, two tails etc. Since A can
fail to draw a white ball at most 8 times, the last set in the above union can be when n = 7.
Thus we have
B = ∪7
n=0
3n+1
Y
i=1
Wc
i
!
W3n+2 .
Thus the probability B wins is given by
P(B) =
8
12
4
12

+
8
12
3 7
11
4
11

+
8
12
3 7
11
3 6
10
4
10

+ . . .
=
7
X
n=0
n
Y
k=1

9 − k
13 − k
3
!
8 − n
12 − n

4
12 − n

.
Finally, the event C wins is given by
C = Wc
1 Wc
2 W3 ∪ Wc
1 Wc
2 Wc
3 Wc
4 Wc
5 W6 ∪ Wc
1 Wc
2 Wc
3 Wc
4 Wc
5 Wc
6 Wc
7 Wc
8 W9 ∪ · · ·
= ∪7
n=0
3n+2
Y
i=1
Wc
i
!
W3n+3 .

Which are the events that C wins by flipping no tails, one tail, two tails etc. Thus the
probability C wins is given by
P(C) =
8
12
2 4
12

+
8
12
3 7
11
2 4
11

+
8
12
3 7
11
3 6
10
3 4
10

+ . . .
=
7
X
n=0
n
Y
k=1

9 − k
13 − k
3
!
8 − n
12 − n
2
4
12 − n

.
We evaluate all of these expressions in the python code chap 3 prob 85.py where we get
the values P(A) = 0.48058, P(B) = 0.3139, and P(C) = 0.20543. These numbers satisfy
P(A) + P(B) + P(C) = 1. as they should.
Problem 86 (A ⊂ B)
Part (a): First note that there are n
i

subsets with i elements. There are a total of 2n
subsets of S. Thus
P(N(B) = i) =
n
i

2n
.
To evaluate P(A ⊂ B|N(B) = i) note that originally the set A can be any of the 2n
subsets
of S. Since B has i elements to have A ⊂ B means that all the elements of A must actually
also be elements of B (of which there are i). Thus A must be one of the 2i
subsets of B and
we have
P(A ⊂ B|N(B) = i) =
2i
2n
.
Using these two results we now have
P(E) =
n
X
i=0
P(A ⊂ B|N(B) = i)P(N(B) = i) =
n
X
i=0
n
i

2n
2i
2n

=
1
2n
1
2n
n
X
i=0

n
i

2i
=
1
2n
1
2n

1 + 2
n
=
3
4
n
.
Part (b): Note that AB = ∅ is equivalent to the statement A ⊂ Bc
. Since Bc
is also a
subset of S. From the previous part we have
P(A ⊂ Bc
) =
3
4
n
.
Problem 87 (Laplace’s rule of succession I)
As shown in example 5e, where Ci is the event we draw the ith coin 0 ≤ i ≤ k and Fn is the
event that the first n flips give head we have
P(Ci) =
1
k + 1
, and P(Fn|Ci) =
i
k
n
,

Thus the conditional probability requested is then
P(Ci|Fn) =
P(CiFn)
P(Fn)
=
P(Ci)P(Fn|Ci)
Pk
j=0 P(Cj)P(Fn|Cj)
=
1
k + 1
i
k
n
Pk
j=0
1
k + 1
j
k
n
=
i
k
n
Pk
j=0
j
k
n
.
Problem 88 (Laplace’s rule of succession II)
The outcomes of the successive flips are not independent. The can be argued by noting the
fact that we get a head on the first flip make it more likely that we drew a coin that was
biased towards landing heads. We can show that by showing that P(H1H2) 6= P(H1)P(H2).
Again conditioning on the initial coin selected we have
P(H1H2) =
k
X
i=0
P(H1H2Ci) =
k
X
i=0
P(Ci)P(H1H2|Ci)
=
k
X
i=0
P(Ci)P(H1|Ci)P(H2|Ci)
=
k
X
i=0
1
k + 1
i
k
2
=
1
k + 1
1
k2
k
X
i=0
i2
=
1
k + 1
1
k2

1
6
k(k + 1)(2k + 1)

=
2k + 1
6k
.
The probability of a single head is
P(H1) =
k
X
i=0
P(H1Ci) =
k
X
i=0
P(Ci)P(H1|Ci)
=
k
X
i=0
1
k + 1
i
k

=
1
k + 1
1
k
k
X
i=0
i =
1
k + 1
1
k
(k + 1)k
2
=
1
2
.
Notice that P(H1)P(H2) =
1
4
6= P(H1H2) and the events are not independent.
Problem 89 (3 judges vote guilty)
As suggested from the book let Ei be the event that judge i votes guilty for i = 1, 2, 3. Let
G be the event the person is actually guilty. Then the problem tells us that P(Ei|G) = .7,
P(Ei|Gc
) = .2, and P(G) = .7.

Part (a): We want to evaluate P(E3|E1E2) which we do by the definition of conditional
independence as
P(E3|E1E2) =
P(E1E2E3)
P(E1E2)
.
We can evaluate each of the probabilities above by conditioning on whether or not the person
is guilty. That is
P(E1E2) = P(E1E2|G)P(G) + P(E1E2|Gc
)P(Gc
)
= P(E1|G)P(E2|G)P(G) + P(E1|Gc
)P(E2|Gc
)P(Gc
)
= (0.7)(0.7)2
+ (0.3)(0.2)2
,
and
P(E1E2E3) = P(E1E2E3|G)P(G) + P(E1E2E3|Gc
)P(Gc
)
= P(E1|G)P(E2|G)P(E3|G)P(G) + P(E1|Gc
)P(E2|Gc
)P(E3|Gc
)P(Gc
)
= (0.7)(0.7)3
+ (0.3)(0.2)3
.
Thus
P(E3|E1E2) =
(0.7)(0.7)3
+ (0.3)(0.2)3
(0.7)(0.7)2 + (0.3)(0.2)2
=
0.2401 + 0.0024
0.343 + 0.012
=
0.2425
0.355
=
97
142
.
Part (b): One way to interpret the problem is to say that we are asked for P(E3|E1Ec
2)
and then this part can be worked just like the previous one. As above we would need to
evaluate P(E1Ec
2E3) and P(E1Ec
2) and then take their ratio. If, instead, we interpret the
problem to be: compute the probability that judge 3 voted guilty given that one of the two
previous judges voted guilty. In this case we don’t know which of the two previous judge’s
voted guilty and which did not. Thus the event we are conditioning on is E1Ec
2 ∪ Ec
1E2. We
thus have
P(E3|E1Ec
2 ∪ Ec
1E2) =
P(E1Ec
2E3 ∪ Ec
1E2E3)
P(E1Ec
2 ∪ Ec
1E2)
=
P(E1Ec
2E3) + P(Ec
1E2E3)
P(E1Ec
2) + P(Ec
1E2)
.
We now calculate each of these probabilities in tern
P(E1Ec
2E3) = P(G)P(E1Ec
2E3|G) + P(Gc
)P(E1Ec
2E3|Gc
)
= P(G)P(E1|G)P(Ec
2|G)P(E3|G) + P(Gc
)P(E1|Gc
)P(Ec
2|Gc
)P(E3|Gc
)
= (0.7)(0.7)(1 − 0.7)(0.7) + (0.3)(0.2)(1 − 0.2)(0.2) .
We do the same thing for P(Ec
1E2E3) and find
P(Ec
1E2E3) = (0.7)(1 − 0.7)(0.7)2
+ (0.3)(0.2)(1 − 0.2)(0.2)2
.
Now for P(E1Ec
2) we have
P(E1Ec
2) = P(G)P(E1Ec
2|G) + P(Gc
)P(E1Ec
2|Gc
)
= P(G)P(E1|G)P(Ec
2|G) + P(Gc
)P(E1|Gc
)P(Ec
2|Gc
)
= (0.7)(0.7)(1 − 0.7) + (0.3)(0.2)(1 − 0.2) .

We do the same thing for P(E1E2) and find
P(Ec
1E2) = (0.7)(1 − 0.7)(0.7) + (0.3)(1 − 0.2)(0.2) .
Thus we get
P(E3|E1Ec
2 ∪ Ec
1E2) =
2[(0.7)(0.7)2
(1 − 0.7) + (0.3)(0.2)2
(1 − 0.2)]
2[(0.7)(0.7)(1 − 0.7) + (0.3)(0.2)(1 − 0.2)]
=
0.1029 + 0.0096
0.147 + 0.048
=
0.1125
0.195
=
15
26
.
Part (c): We want to evaluate P(E3|Ec
1Ec
2) which we also do by the definition of conditional
independence from
P(E3|Ec
1Ec
2) =
P(Ec
1Ec
2E3)
P(Ec
1Ec
2)
.
As before
P(Ec
1Ec
2E3) = P(G)P(Ec
1Ec
2E3|G) + P(Gc
)P(Ec
1Ec
2E3|Gc
)
= P(G)P(Ec
1|G)P(Ec
2|G)P(E3|G) + P(Gc
)P(Ec
1|Gc
)P(Ec
2|Gc
)P(E3|Gc
)
= (0.7)(1 − 0.7)2
(0.7) + (0.3)(1 − 0.2)2
(0.2) ,
and
P(Ec
1Ec
2) = P(G)P(Ec
1Ec
2|G) + P(Gc
)P(Ec
1Ec
2|Gc
)
= P(G)P(Ec
1|G)P(Ec
2|G) + P(Gc
)P(Ec
1|Gc
)P(Ec
2|Gc
)
= (0.7)(1 − 0.7)2
+ (0.3)(1 − 0.2)2
.
Thus we have
P(E3|E1E2) =
(0.7)(1 − 0.7)2
(0.7) + (0.3)(1 − 0.2)2
(0.2)
(0.7)(1 − 0.7)2 + (0.3)(1 − 0.2)2
=
0.0441 + 0.0384
0.063 + 0.192
=
0.0825
0.255
=
33
102
.
Problem 90 (n trials, 3 outcomes)
We want to evaluate the probability of the event E defined such that
outcomes 1 and 2 both occur at least once.
This is equivalent to the event
outcome 1 occurs at least once and outcome 2 occurs at least once.
The event is also equal to the contra-positive of the above statement which is
not (1 never occurs or 2 never occurs).
This last event is in tern is equivalent to the event
not (0 or 2 always occur or 0 or 1 always occur).

We will evaluate the probability of this last event and then from it derive the probability of
the event of interest or E. Let Ui, Vi, Wi be the events outcome 0,1,2 occurs on the ith trial.
From the above we have argued that
E =
n

i=1
(Ui ∪ Wi) ∪
n

i=1
(Ui ∪ Vi)
c
.
We first find P(Ec
) given by
P(Ec
) = P
n

i=1
(Ui ∪ Wi) ∪
n

i=1
(Ui ∪ Vi)

= P
n

i=1
(Ui ∪ Wi)

+ P
n

i=1
(Ui ∪ Vi)

− P
n

i=1
(Ui ∪ Wi)
n

i=1
(Ui ∪ Vi)

= P
n

i=1
(Ui ∪ Wi)

+ P
n

i=1
(Ui ∪ Vi)

− P
n

i=1
(Ui ∪ Wi)(Ui ∪ Vi)

= P
n

i=1
(Ui ∪ Wi)

+ P
n

i=1
(Ui ∪ Vi)

− P
n

i=1
Ui

=
n
Y
i=1
P(Ui ∪ Wi) +
n
Y
i=1
P(Ui ∪ Vi) −
n
Y
i=1
P(Ui)
= (p0 + p2)n
+ (p0 + p1)n
− pn
0 .
Thus using this result we have for P(E) that
P(E) = P(
n

i=1
(Ui ∪ Wi) ∪
n

i=1
(Ui ∪ Vi)
c
)
= 1 − P
n

i=1
(Ui ∪ Wi) ∪
n

i=1
(Ui ∪ Vi)

= 1 − (p0 + p2)n
− (p0 + p1)n
+ pn
0 .
Problem 1 (conditioning on more information)
We have
P(A ∩ B|A) =
P(A ∩ B ∩ A)
P(A)
=
P(A ∩ B)
P(A)
.
and
P(A ∩ B|A ∪ B) =
P((A ∩ B) ∩ (A ∪ B))
P(A ∪ B)
=
P(A ∩ B)
P(A ∪ B)
.
But since A ∪ B ⊃ A, the probabilities P(A ∪ B) ≥ P(A), so
P(A ∩ B)
P(A)
≥
P(A ∩ B)
P(A ∪ B)

giving
P(A ∩ B|A) ≥ P(A ∩ B|A ∪ B) ,
the desired result.
Problem 2 (simplifying conditional expressions)
Using the definition of conditional probability we can compute
P(A|B) =
P(A ∩ B)
P(B)
=
P(A)
P(B)
.
since A ⊂ B. In words P(A|B) is the amount of A in B. For P(A|¬B) we have
P(A|¬B) =
P(A ∩ ¬B)
P(¬B)
=
P(φ)
P(¬B)
= 0 .
Since if A ⊂ B then A ∩ ¬B is empty or in words given that ¬B occurred and A ⊂ B, A
cannot have occurred and therefore has zero probability. For P(B|A) we have
P(B|A) =
P(A ∩ B)
P(A)
=
P(A)
P(A)
= 1 ,
or in words since A occurs and B contains A, B must have occurred giving probability one.
For P(B|¬A) we have
P(B|¬A) =
P(B ∩ ¬A)
P(¬A)
,
which cannot be simplified further.
Problem 3 (biased selection of the first born)
We define n1 to be the number of families with one child, n2 the number of families with two
children, and in general nk to be the number of families with k children. In this problem
we want to compare two different methods for selecting children. In the first method, M1,
we pick one of the m families and then randomly choose a child from that family. In the
second method, M2, we directly pick one of the
Pk
i=1 ini children randomly. Let E be the
event that a first born child is chosen. Then the question seeks to prove that
P(E|M1) P(E|M2) .
We will solve this problem by conditioning no the number of families with i children. For
example under M1 we have (dropping the conditioning on M1 for notational simplicity) that
P(E) =
k
X
i=1
P(E|Fi)P(Fi) ,

where Fi is the event that the chosen family has i children. This later probability is given
by
P(Fi) =
ni
m
,
for we have ni families with i children from m total families. Also
P(E|Fi) =
1
i
,
since the event Fi means that our chosen family has i children and the event E means that
we select the first born, which can be done in 1
i
ways. In total then we have under M1 the
following for P(E)
P(E) =
k
X
i=1
P(E|Fi)P(Fi) =
k
X
i=1
1
i
ni
m

=
1
m
k
X
i=1
ni
i
.
Now under the second method again P(E) =
Pk
i=1 P(E|Fi)P(Fi) but under the second
method P(Fi) is the probability we have selected a family with i children and is given by
ini
Pk
i=1 ini
,
since ini is are the number of children from families with i children and the denominator is
the total number of children. Now P(E|Fi) is still the probability of having selected a family
with ith children we select the first born child. This is
ni
ini
=
1
i
,
since we have ini total children from the families with i children and ni of them are first
born. Thus under the second method we have
P(E) =
k
X
i=1

1
i

ini
Pk
l=1 lnl
!
=
1
Pk
l=1 lnl

k
X
i=1
ni .
Then our claim that P(E|M1) P(E|M2) is equivalent to the statement that
1
m
k
X
i=1
ni
i
≥
Pk
i=1 ni
Pk
i=1 ini
or remembering that m =
Pk
i=1 ni that
k
X
i=1
ini
! k
X
j=1
nj
j
!
≥
k
X
i=1
ni
! k
X
j=1
nj
!
.
To show that this is true (and that all earlier results are true) expand each expression. First
the left hand side LHS, we obtain
LHS = (n1 + 2n2 + 3n3 + · · · + knk)

n1 +
n2
2
+
n3
3
+ · · · +
nk
k

= n2
1 +
n1n2
2
+
n1n3
3
+
n1n4
4
+ · · · +
n1nk
k
+ 2n2n1 + n2
2 +
2n2n3
3
+ · · · +
2n2nk
k
+ · · ·
+ knkn1 +
knkn2
2
+ · · · + n2
k .

By grouping the polynomial terms we find that the above is equivalent to
LHS = n2
1 +

1
2
+ 2

n1n2 +

1
3
+ 3

n1n3 + · · · +

1
k
+ k

n1nk
+ n2
2 +

2
3
+
3
2

n2n3 +

4
2
+
2
4

n2n4 + · · · +

2
k
+
k
2

n2nk
+ n2
3 +

3
4
+
4
3

n3n4 + · · · +

3
k
+
k
3

n3nk + · · ·
Thus in general the ninj term has a coefficient given by 1 if i = j and by
i
j
+
j
i
,
if i j ≤ k. While the expansion of the right hand side RHS will have a coefficient given by
1 if i = j and 2 if i j ≤ k. Thus a sufficient condition for the left hand side to be greater
than the right hand side is for
i
j
+
j
i
2 when i 6= j and i j ≤ k .
By multiplying by the product ij, we have the above equivalent to
i2
+ j2
2ij ,
which in turn is equivalent to
i2
− 2ij + j2
0 ,
or
(i − j)2
0 ,
which we know to be true for all i and j. Because (using reversible transformations) we have
reduced our desired inequality to one that we know to be true we have shown the desired
identity.
Problem 4 (fuzzy searching for balls in a box)
Let Ei be the event that the ball is present in box i. Let Si be the event that the search of
box i yields a success or “finds” the ball. Then the statement of the problem tells us that
P(Ei) = Pi and
P(Si|Ej) =

αi j = i
0 j 6= i
.
We desire to compute P(Ej|Sc
i ) which by Bayes’ rule is equal to
P(Ej|Sc
i ) =
P(Sc
i |Ej)P(Ej)
P(Sc
i )
=
(1 − P(Si|Ej))P(Ej)
1 − P(Si)
.
Lets begin by computing P(Si). We have
P(Si) =
n
X
k=1
P(Si|Ek)P(Ek) = αipi .

Using the above we can compute our desired expression. We find
P(Ej|Sc
i ) =
(
(1−αi)pi
1−αipi
j = i
pj
1−αipi
j 6= i
,
which is the desired result.
Problem 5 (negative information)
An event F is said to carry negative information about an event E and is written F ց E if
P(E|F) ≤ P(E).
Part (a): If F ց E then P(E|F) ≤ P(E) so that using Bayes’ rule for P(E|F) we see that
this is equivalent to the expression
P(F|E)P(E)
P(F)
≤ P(E) ,
and assuming P(E) 6= 0. This implies that P(F|E) ≤ P(F) so that E ց F.
Problem 6 (the union of independent events)
We recognize that E1 ∪ E2 ∪ · · · En as the event that at least one of the events Ei occurs.
Consider the event ¬E1 ∩ ¬E2 ∩ ¬E3 · · · ∩ ¬En = ¬(E1 ∪ E2 ∪ E3 ∪ · · · ∪ En), which is the
event that none of the Ei occur. Then we have
P(E1 ∪ E2 ∪ · · · ∪ En) = 1 − P(¬(E1 ∪ E2 ∪ · · · ∪ En)) = 1 − P(¬E1 ∩ ¬E2 ∩ · · · ¬En) .
As a lemma for this problem assume we have only two independent events E1 and E2 and
consider
P(¬E1 ∩ ¬E2) = P(¬(E1 ∪ E2)) = 1 − P(E1 ∪ E2)
= 1 − (P(E1) + P(E2) − P(E1 ∩ E2))
= 1 − P(E1) − P(E2) + P(E1)P(E2)
= (1 − P(E1))(1 − P(E2)) = P(¬E1)P(¬E2) ,
using the independence of the sets E1 and E2. This result shows that for independent events
the product rule works for the negation of the sets as well as the direct sets themselves. Thus
we have for the problem at hand that
P(E1 ∪ E2 ∪ · · · En) = 1 − P(¬E1 ∩ ¬E2 ∩ · · ·¬En)
= 1 − P(¬E1)P(¬E2)P(¬E3) · · ·P(¬En)
= 1 − (1 − P(E1))(1 − P(E2))(1 − P(E3)) · · ·(1 − P(En)) ,
the required result.

Problem 7 (extinct fish)
Part (a): We desire to compute Pw the probability that the last ball drawn is white. This
probability will be
Pw =
n
n + m
,
because we have n white balls that can be selected from n+m total balls that can be placed
in the last spot.
Part (b): Let R be the event that the red fish species are the first species to become extinct.
Then following the hint we write P(R) as
P(R) = P(R|Gl)P(Gl) + P(R|Bl)P(Bl) .
Here Gl is the event that the green fish species are the last fish species to become extinct
and Bl the event that the blue fish species are the last fish species to become extinct. Now
we conclude that
P(Gl) =
g
r + b + g
,
and
P(Bl) =
b
r + b + g
.
We can see these by considering the blue fish as an example. If the blue fish are the last ones
extinct then we have b possible blue fish to select from the r + b + g total number of fish to
be the last fish. Now we need to compute the conditional probabilities P(R|Gl). This can
be thought of as the event that the red fish go extinct and then the blue fish. This is the
same type of experiment as in Part (a) of this problem in that we must have a blue fish go
extinct (i.e. a draw with a blue fish last). This can happen with probability
b
r + b + g − 1
,
where the denominator is one less than r + b + g since the last fish drawn must be a green
fish by the required conditioning. In the same way we have that
P(R|Bl) =
g
r + b + g − 1
.
So that the total probability P(R) is then given by
P(R) =

b
r + b + g − 1

g
r + b + g

+

g
r + b + g − 1

b
r + b + g

=
2bg
(r + b + g − 1)(r + b + g)
.

Problem 8 (some inequalities)
Part (a): If P(A|C) P(B|C) and P(A|Cc
) P(B|Cc
), then consider P(A) which by
conditioning on C and Cc
becomes
P(A) = P(A|C)P(C) + P(A|Cc
)P(Cc
)
P(B|C)P(C) + P(B|Cc
)P(Cc
) = P(B) .
Where the second line follows from the given inequalities.
Part (b): Following the hint, let C be the event that the sum of the pair of dice is 10,
A the event that the first die lands on a 6 and B the event that the second die lands a 6.
Then P(A|C) = 1
3
, and P(A|Cc
) = 5
36−3
= 5
33
. So that P(A|C) P(A|Cc
) as expected.
Now P(B|C) and P(B|Cc
) will have the same probabilities as for A. Finally, we see that
P(A ∩ B|C) = 0, while P(A ∩ B|Cc
) = 1
33
0. So we have found an example where
P(A ∩ B|C) P(A ∩ B|Cc
) and a counter example has been found.
Problem 9 (pairwise independence)
Let A be the event that the first toss lands heads and let B be the event that the second
toss lands heads, and finally let C be the event that both lands on the same side. Now
P(A, B) = 1
4
, and P(A) = P(B) = 1
2
, so A and B are independent. Now
P(A, C) = P(C|A)P(A) =
1
2

1
2

=
1
4
.
but P(C) = 1
2
so P(A, C) = P(A)P(C) and A and C are independent. Finally
P(B, C) = P(C|B)P(B) =
1
4
,
so again B and C are independent. Thus A, B, and C are pairwise independent but for
three sets to be fully independent we must have in addition that
P(A, B, C) = P(A)P(B)P(C) .
The right hand side of this expression is 1
2
3
while the left hand side is the event that
both tosses land heads and so P(A, B, C) = 1
4
6= P(A)P(B)P(C) and the three sets are not
independent.
Problem 10 (pairwise independence does not imply independence)
Let Ai,j be the event that person i and j have the same birthday. We desire to show that
these events are pairwise independent. That is the two events Ai,j and Ar,s are independent

but the totality of all

n
2

events are not independent. Now
P(Ai,j) = P(Ar,s) =
1
365
,
since for the specification of either one persons birthday the probability that the other person
will have that birthday is 1/365. Now
P(Ai,j ∩ Ar,s) = P(Ai,j|Ar,s)P(Ar,s) =

1
365

1
365

=
1
3652
.
This is because P(Ai,j|Ar,s) = P(Ai,j) i.e. the fact that people r and s have the same
birthday has no effect on whether the event Ai,j is true. This is true even if one of the people
in the pairs (i, j) and (r, s) is the same. When we consider the intersection of all the sets
Ai,j, the situation changes. This is because the event ∩(i,j)Ai,j (where the intersection is
over all pairs (i, j)) is the event that every pair of people have the same birthday, i.e. that
everyone has the same birthday. This will happen with probability

1
365
n−1
,
while if the events Ai,j were independent the required probability would be
Y
(i,j)
P(Ai,j) =

1
365



n
2


=

1
365
n(n−1)
2
.
Since

n
2

6= n − 1, these two results are not equal and the totality of events Ai,j are not
independent.
Problem 11 (at least one head)
The probability that we obtain at least on head is one minus the probability that we obtain
all tails in n flips. Thus this probability is 1 − (1 − p)n
. If this is to be made greater than 1
2
we have
1 − (1 − p)n

1
2
,
or solving for n we have n ln(1/2)
ln(1−p)
, so since the this expression can be non-integer take n
we need to take the next integer larger than or equal to this number. That is take n to be
n =

ln(1/2)
ln(1 − p)

.

Problem 12 (an infinite sequence of flips)
Let ai be the probability that the ith coin lands heads. The consider the random variable
N, specifying the location where the first head occurs. This problem then is like a geometric
random variable where we want to determine the first time a success occurs. Then we have
for a distribution of P{N} the following
P{N = n} = an
n−1
Y
i=1
(1 − ai) .
This states that the first n − 1 flips must land tails and the last flip (the nth) then lands
heads. Then when we add this probability up for n = 1, 2, 3, · · · , ∞ i.e.
∞
X
n=1

an
n−1
Y
i=1
(1 − ai)
#
,
is the probability that a head occurs somewhere in the infinite sequence of flips. The other
possibility would be for a head to never appear. This will happen with a probability of
∞
Y
i=1
(1 − ai) .
Together these two expressions consist of all possible outcomes and therefore must sum to
one. Thus we have proven the identity
∞
X
n=1

an
n−1
Y
i=1
(1 − ai)
#
+
∞
Y
i=1
(1 − ai) = 1 ,
or the desired result.
Problem 13 (winning by flipping)
Let Pn,m be the probability that A who starts the game accumulates n head before B
accumulates m heads. We can evaluate this probability by conditioning on the outcome of
the first flip made by A. If this flip lands heads, then A has to get n−1 more flips before B’s
obtains m. If this flip lands tails then B obtains control of the coin and will receive m flips
before A receives n with probability Pm,n by the implicit symmetry in the problem. Thus A
will accumulate the correct number of heads with probability 1 − Pm,n. Putting these two
outcomes together (since they are the mutually exclusive and exhaustive) we have
Pn,m = pPn−1,m + (1 − p)(1 − Pm,n) ,
or the desired result.

Problem 14 (gambling against the rich)
Let Pi be the probability you eventually go broke when your initial fortune is i. Then
conditioning on the result of the first wager we see that Pi satisfies the following difference
equation
Pi = pPi+1 + (1 − p)Pi−1 .
This simply states that the probability you go broke when you have a fortune of i is p times
Pi+1 if you win the first wager (since if you win the first wager you now have i + 1 as your
fortune) plus 1 − p times Pi−1 if you loose the first wager (since if you loose the first wager
you will have i − 1 as your fortune). To solve this difference equation we recognize that its
solution must be given in terms of a constant raised to the ith power i.e. αi
. Using the
ansatz that Pi = αi
and inserting this into the above equation we find that α must satisfy
the following
pα2
− α + (1 − p) = 0 .
Using the quadratic equation to solve this equation for α we find α given by
α =
1 ±
p
1 − 4p(1 − p)
2p
=
1 ±
p
(2p − 1)2
2p
=
1 ± (2p − 1)
2p
.
Taking the plus sign gives α+
= 1, while taking the minus sign in the above gives α−
= q
p
.
Now the general solution to this difference equation is then given by
Pi = C1 + C2

q
p
i
for i ≥ 0 .
Problem 15 (n trials and r successes)
The event that we want the probability of is the sum of independent events where in each
of these events we have r − 1 success in the n − 1 trials and the n trial is also a success (and
the last one needed). Based on combining a binomial probability with this last success, the
probability we seek is then

n − 1
r − 1

pr−1
(1 − p)n−1−(r−1)

p =

n − 1
r − 1

pr
(1 − p)n−r
.
Recall that the problem of the points is the situation where we obtain a success with prob-
ability p and we want the probability we have n successes before m failures. To place the
problem of the points in the framework of this problem we can consider extending the num-
ber of trials we perform in such a way that we always perform n + m games. Since the
probability we want to compute is the probability that from these total n + m games we get
n success before m failures this event must be one of the following independent events
• We get our n successes in the first n games and the remaining m games are all failures.
• We get our n successes in the first n+ 1 games (so have one failure) and the remaining
m − 1 games are all failures.

• We get our n successes in the first n + 2 games (and thus have two failures) and the
remaining m − 2 are all failures.
• etc.
• We get our n successes in the first n + m − 2 games (and thus have m − 2 failures) and
the remaining 2 games are all failures
• We get our n successes in the first n + m − 1 games (and thus have m − 1 failures) and
the remaining 1 game is a failure
We have computed the probability of each of these events the first part of this problem.
Thus by adding the contributions from each independent event we have
n+m−1
X
i=n

i − 1
n − 1

pn
(1 − p)i−n
.
Note: I’m not sure how to show that this is equivalent to the Fermat’s solution which is
n+m−1
X
i=n

m + n − 1
i

pi
(1 − p)m+n−1−i
,
if anyone knows how to show this please contact me.
Problem 16 (the probability of an even number of successes)
Let Pn be the probability that n Bernoulli trials result in an even number of successes. Then
the given difference equation can be obtained by conditioning on the result of the first trial
as follows. If the first trial is a success then we have n − 1 trials to go and to obtain an even
total number of tosses we want the number of successes in this n − 1 trials to be odd This
occurs with probability 1 − Pn−1. If the first trial is a failure then we have n − 1 trials to go
and to obtain an even total number of tosses we want the number of successes in this n − 1
trials to be even This occurs with probability Pn−1. Thus we find that
Pn = p(1 − Pn−1) + (1 − p)Pn−1 for n ≥ 1 .
Some special point cases are easily computed. We have by assumption that P0 = 1, and
P1 = q since with only one trial, this trial must be a failure to get a total even number of
successes. Given this difference equation and a potential solution we can verify that this
solution satisfies our equation and therefore know that it is a solution. One can easily check
that the given Pn satisfies P0 = 1 and P1 = q. In addition, for the given assumed solution
we have that
Pn−1 =
1 + (1 − 2p)n−1
2
,

From which we find (using this expression in the right hand side of the difference equation
above)
p(1 − Pn−1) + (1 − p)Pn−1 = p + (1 − 2p)Pn−1
= p + (1 − 2p)

1 + (1 − 2p)n−1
2

= p +
1 − 2p
2
+
(1 − 2p)n
2
=
1
2
+
(1 − 2p)n
2
= Pn .
Showing that Pn is a solution the given difference equation.
Problem 17 (odd number of successes)
Let Si be the event the ith trial results in success and let En be the event the number of
successes in n total trials is odd.
Part (a): The probability P(E1) is the probability the first trial is a success so we have
P(E1) = P(S1) =
1
2(1) + 1
=
1
3
.
The probability P(E2) is the union of the independent events where the first trial is a success
and the second trial is a failure or the first trial is a failure and the second trial is a success.
Thus
E2 = S1Sc
2 ∪ Sc
1S2 so
P(E2) = P(S1Sc
2) + P(Sc
1S2) = (
1
3
)(
4
5
) + (
2
3
)(
1
5
) =
6
(3)(5)
=
2
5
The probability P(E3) is the union of the independent events where we have three total
successes or one success and two failures. Thus
E3 = S1S2S3 ∪ S1Sc
2Sc
3 ∪ Sc
1S2Sc
3 ∪ Sc
1S2S3 so
P(E3) = P(S1S2S3) + P(S1Sc
2Sc
3) + P(Sc
1S2Sc
3) + P(Sc
1Sc
2S3)
= (
1
3
)(
1
5
)(
1
7
) + (
1
3
)(
4
5
)(
6
7
) + (
2
3
)(
1
5
)(
6
7
) + (
2
3
)(
4
5
)(
1
7
) =
45
(3)(5)(7)
=
3
7
.
In the same way the probability P(E4) is the union of the independent events where we have
three successes with one failure or one success with three failures. Thus the the probability
P(E4) is given by
P(E4) = P(Sc
1S2S3S4) + P(S1Sc
2S3S4) + P(S1S2Sc
3S4) + P(S1S2S3Sc
4)
+ P(S1Sc
2Sc
3Sc
4) + P(Sc
1S2Sc
3Sc
4) + P(Sc
1Sc
2S3Sc
4) + P(Sc
1Sc
2Sc
3S4)
= (
2
3
)(
1
5
)(
1
7
)(
1
9
) + (
1
3
)(
4
5
)(
1
7
)(
1
9
) + (
1
3
)(
1
5
)(
6
7
)(
1
9
) + (
1
3
)(
1
5
)(
1
7
)(
8
9
)
+ (
1
3
)(
4
5
)(
6
7
)(
8
9
) + (
2
3
)(
1
5
)(
6
7
)(
8
9
) + (
2
3
)(
4
5
)(
1
7
)(
8
9
) + (
2
3
)(
4
5
)(
6
7
)(
1
9
)
=
420
(3)(5)(7)(9)
=
4
9
.

Finally, the probability P(E5) is the union of the independent events where we have all
success, three successes with two failures or one success with four failures. Thus the the
probability P(E5) is given by
P(E5) = P(S1S2S3S4S5)
+ P(Sc
1Sc
2S3S4S5) + P(Sc
1S2Sc
3S4S5) + P(Sc
1S2S3Sc
4S5)
+ P(Sc
1S2S3S4Sc
5) + P(S1Sc
2Sc
3S4S5) + P(S1Sc
2S3Sc
4S5)
+ P(S1Sc
2S3S4Sc
5) + P(S1S2Sc
3Sc
4S5) + P(S1S2Sc
3S4Sc
5)
+ P(S1S2S3Sc
4Sc
5) + P(S1Sc
2Sc
3Sc
4Sc
5) + P(Sc
1S2Sc
3Sc
4Sc
5)
+ P(Sc
1Sc
2S3Sc
4Sc
5) + P(Sc
1Sc
2Sc
3S4Sc
5) + P(Sc
1Sc
2Sc
3Sc
4S5)
= (
1
3
)(
1
5
)(
1
7
)(
1
9
)(
1
11
)
+ (
2
3
)(
4
5
)(
1
7
)(
1
9
)(
1
11
) + (
2
3
)(
1
5
)(
6
7
)(
1
9
)(
1
11
) + (
2
3
)(
1
5
)(
1
7
)(
8
9
)(
1
11
)
+ (
2
3
)(
1
5
)(
1
7
)(
1
9
)(
10
11
) + (
1
3
)(
4
5
)(
6
7
)(
1
9
)(
1
11
) + (
1
3
)(
4
5
)(
1
7
)(
8
9
)(
1
11
)
+ (
1
3
)(
4
5
)(
1
7
)(
1
9
)(
10
11
) + (
1
3
)(
1
5
)(
6
7
)(
8
9
)(
1
11
) + (
1
3
)(
1
5
)(
6
7
)(
1
9
)(
10
11
)
+ (
1
3
)(
1
5
)(
1
7
)(
8
9
)(
10
11
) + (
1
3
)(
4
5
)(
6
7
)(
8
9
)(
10
11
) + (
2
3
)(
1
5
)(
6
7
)(
8
9
)(
10
11
)
+ (
2
3
)(
4
5
)(
1
7
)(
8
9
)(
10
11
) + (
2
3
)(
4
5
)(
6
7
)(
1
9
)(
10
11
) + (
2
3
)(
4
5
)(
6
7
)(
8
9
)(
1
11
)
=
4725
(3)(5)(7)(9)(11)
=
5
11
.
Part (b): From the above expressions we hypothesize that
Pn ≡ P(En) =
n
2n + 1
.
Part (c): Now Pn is the probability that we have an odd number of successes in n trials
and Pn−1 must be the probability we have an odd number of success in n − 1 trials. We
can compute Pn in terms of Pn−1 by conditioning on the result of the last trial. If the result
of the nth trial is success, the number of successes in the total n trials will be odd if and
only if the number of successes in the first n − 1 trials is even. This last event happens with
probability 1 − Pn−1. At the same time if the result of the nth trial is failure, the number
of successes in n trials will be odd if and only if the number of successes in the first n − 1
trials is odd. Thus we have just argued that
Pn =

1
2n + 1

(1 − Pn−1) +

1 −
1
2n + 1

Pn−1
=
1
2n + 1

(1 − Pn−1) +
2n
2n + 1

Pn−1 (18)
=
1
2n + 1
+
2n − 1
2n + 1

Pn−1 .

Part (d): For this part we want to show that Pn = n
2n+1
is a solution to the above difference
equation. Note that
Pn−1 =
n − 1
2(n − 1) + 1
=
n − 1
2n − 1
.
Thus the right-hand-side of Equation 18 gives
RHS =
1
2n + 1

1 −
n − 1
2n − 1

+
2n
2n + 1
n − 1
2n − 1

=
1
2n + 1
−
1
2n + 1
n − 1
2n − 1

+
2n
2n + 1
n − 1
2n − 1

=
2n − 1
(2n + 1)(2n − 1)
−
n − 1
(2n + 1)(2n − 1)
+
2n(n − 1)
(2n + 1)(2n − 1)
=
n + 2n2
− 2n
(2n + 1)(2n − 1)
=
n(2n − 1)
(2n + 1)(2n − 1)
=
n
2n + 1
,
which is the same as the functional form for Pn showing the equivalence.
Problem 18 (3 consecutive heads)
Let Qn be the probability that in n tosses of a fair coin no run of three consecutive heads
appears. Since when n = 0, 1, 2 is is not possible to flip three heads we have Q0 = Q1 =
Q2 = 1. We can compute Qn by conditioning on the result of the first few flips. If the first
flip is a tail then we cannot have this flip as the beginning of a sequence of three consecutive
heads and the probability that no run appears is the probability that none appear in the
remaining n−1 flips which is the value Qn−1. If the first flip is in fact a head then we need to
consider what happens in the next flips. If the second flip is then a tail this first flip cannot
be in a run of heads and the probability that no run of three heads occur in the next n − 2
flips is Qn−2. If the second flip is a head then again we need to look at the next flip. If is
a tail the same logic above applies and probability that no run of three heads occur in the
next n−3 flips is Qn−3. If in fact the third flip is a head then we do have a run of heads and
the probability of no run of heads is 0. When we combine the above with the probability of
the individual flips, we can summarize this discussion as the probability of interest Qn can
be decomposed into the following independent events and their probabilities:
• T · · · with a probability of 1
2
Qn−1
• HT · · · with a probability of 1
4
Qn−2
• HHT · · · with a probability of 1
8
Qn−3
• HHH · · · with a probability of 0.
Here the · · · notation represents unrealized coin flips. Thus we can add these probabilities
to get Qn and we have shown
Qn =
1
2
Qn−1 +
1
4
Qn−2 +
1
8
Qn−3 .

Given the initial conditions on the first three values of Qn of can use the above to compute
Q8.
Problem 19 (the n-game gambler’s ruin)
With the probability p that A wins (and in a slightly different notation) by conditioning on
the outcome of the ith flip we have
P(n, i) = pP(n − 1, i + 1) + (1 − p)P(n − 1, i − 1) .
Next note that if i = N, then gambler A has all the money and must certainly win and
we have P(n, N) = 1. In the same way if i = 0, gambler B has all the money and A must
certainly loose and we have P(n, 0) = 0. If n = 0, then they stop playing and A cannot win.
Thus we take P(0, i) = 0 for 0 ≤ i N. To find the probability of interest let N = 5, i = 3,
n = 7. Using the above relationship we find some results we will need later
P(1, 3) = pP(0, 4) + qP(0, 2) = 0
P(1, 1) = pP(0, 2) + qP(0, 0) = 0
P(2, 4) = pP(1, 5) + qP(1, 3) = p
P(2, 2) = pP(1, 3) + qP(1, 1) = 0
P(3, 3) = pP(2, 4) + qP(2, 2) = p2
P(3, 1) = pP(2, 2) + qP(2, 0) = 0
P(4, 4) = pP(3, 5) + qP(3, 3) = p + qp2
P(4, 2) = pP(3, 3) + qP(3, 1) = p3
,
and also
P(5, 3) = pP(4, 4) + qP(4, 2) = p(p + qp2
) + qp3
= p2
+ 2qp3
P(5, 1) = pP(4, 2) + qP(4, 0) = p4
P(6, 4) = pP(5, 5) + qP(5, 3) = p + q(p2
+ 2qp3
) = p + qp2
+ 2q2
p3
P(6, 2) = pP(5, 3) + qP(5, 1) = p(p2
+ 2qp3
) + qp4
= p3
+ 3qp4
P(7, 3) = pP(6, 4) + qP(6, 2)
= p(p + qp2
+ 2q2
p3
) + q(p3
+ 3qp4
) = p2
+ 2qp3
+ 5q2
p4
.
Problem 20 (white and black balls, 2 urns)
With probability α a ball is chosen from the first urn. All subsequent selections are made
based on the fact that if a black ball is drawn we “switch” urns and begin to draw from the

other/alternative urn. Let αn be the probability that the nth ball is drawn from the first
urn. We take α1 = α. The to calculate αn+1 we can condition on whether the nth ball was
drawn from the first urn or not. If it was, then with probability p we would draw a while
ball from that urn and we would not have switch urns. Thus ball n + 1 is from urn number
1. If it was not, then with probability 1 − p′
we would have drawn a black ball from the
second urn and would have to switch urns on the n + 1st draw to the first urn. Thus
αn+1 = pαn + (1 − p′
)(1 − αn) = αn(p + p′
− 1) + 1 − p′
. (19)
For n ≥ 1 and with α1 = α. The solution to this recursion relationship is given by
αn =
1 − p′
2 − p − p′
+

α −
1 − p′
2 − p − p′

(p + p′
− 1)n−1
. (20)
To prove this not that when n = 1 we get α1 = α as we should in the above. Assume the
functional form for αn given by Equation 20 holds up to some n. Then put that expression
into the right-hand-side of Equation 19. We get
RHS =

1 − p′
2 − p − p′
+

α −
1 − p′
2 − p − p′

(p + p′
− 1)n−1

(p + p′
− 1) + 1 − p′
=
1 − p′
2 − p − p′
(p + p′
− 1) +

α −
1 − p′
2 − p − p′

(p + p′
− 1)n
+ 1 − p′
= (1 − p′
)

p + p′
− 1
2 − p − p′
+ 1

+

α −
1 − p′
2 − p − p′

(p + p′
− 1)n
= (1 − p′
)

p + p′
− 1 + 2 − p − p′
2 − p − p′

+

α −
1 − p′
2 − p − p′

(p + p′
− 1)n
=
1 − p′
2 − p − p′
+

α −
1 − p′
2 − p − p′

(p + p′
− 1)n
,
which is Equation 20 evaluated at n + 1 proving the expression.
Next Let Pn be the probability that the nth ball selected is white. This depends on the urn
from which we are selecting from. We have
Pn = pαn + p′
(1 − αn) = p′
+ (p − p′
)αn for n ≥ 2 .
From what we know about αn given in Equation 20 we have a solution for Pn.
To calculate the requested limits note that if 0 p, p′
1, then
0 p + p′
2, thus − 1 p + p′
− 1 1 .
Using this we have
lim
n→∞
(p + p′
− 1)n
= 0 ,
and we have
lim
n→∞
αn =
1 − p′
2 − p − p′
=
1 − p′
(1 − p) + (1 − p′)
.
For Pn we have
lim
n→∞
Pn = p′
+
1 − p′
(1 − p) + (1 − p′)
(p − p′
) =
p(1 − p′
) + p′
(1 − p)
(1 − p) + (1 − p′)
.

Problem 21(counting votes)
Part (a): We are told for this problem that candidate A receives n votes and candidate B
receives m votes where we assume that n m. Lets represent the each sequential vote by a
sequence of A’s and B’s. The number of AB sequences is given by
(n + m)!
n! m!
(21)
We will assume all sequences are equally likely. Let E(n, m) be the event A is always ahead
in the counting of votes and Pn,m the events probability.
To calculate P2,1 the only sequence where A leads B for all counts is {AAB}. Using Equa-
tion 21 we then have P2,1 = 1
3
.
To calculate P3,1 the sequences where A leads B for all counts are: {AAAB, AABA}. Using
Equation 21 we then have P3,1 = 2
4
= 1
2
.
To calculate P4,1 the sequences where A leads B for all counts are: {AAAAB, AAABA, AABAA}.
Using Equation 21 we then have P4,1 = 3
5
.
To calculate P3,2 the sequences where A leads B for all counts are: {AAABB, AABAB}.
Using Equation 21 we have (3+2)!
3!2!
= 10 and thus P3,2 = 1
5
.
To calculate P4,2 the sequences where A leads B for all counts are:
{AAAABB, AAABAB, AAABBA, AABAAB, AABABA} .
Using Equation 21 we have (4+2)!
4!2!
= 15 and thus P4,2 = 5
15
= 1
3
.
To calculate P4,3 the sequences where A leads B for all counts are:
{AAAABBB, AAABABB, AAABBAB, AABAABB, AABABAB}
Using Equation 21 again we find P4,3 = 5
35
= 1
7
.
Part (b): For Pn,1 note that from Equation 21 there are (n+1)!
n! 1!
= n + 1 sequences. Each
sequence has one vote for B. Each of these sequences start with the characters AA, AB or
B. Note that the sequences that start with B or AB will not have A leading in votes for
all time, while for the sequences that start with AA the number of A votes will always be
larger than the number of votes for B by at least 1. The number of sequences starting with
AA (since we have specified that the first two votes are for A we have n + 1 − 2 votes to yet
place) is
(n + 1 − 2)!
(n − 2)! 1!
= n − 1 .
Thus Pn,1 = n−1
n+1
.

For Pn,2 note that there are (n+2)!
n! 2!
= (n+2)(n+1)
2
sequences of A and B’s and each sequence
will have two B’s. All our sequences start with either AAA, AABA, AABB, AB or B. The
only of these sequences where A is leading for all counts are AAA and AABA. We thus
need to count the number of sequences starting with each of these prefixes. The number of
sequences starting with AAA (since now we have n + 2 − 3 votes to distribute) is
(n + 2 − 3)!
(n − 3)!2!
=
(n − 1)(n − 2)
2
,
while the number of sequences starting with AABA (since now we have n + 2 − 4 votes to
distributed) is
(n + 2 − 4)!
(n − 3)!1!
= n − 2 .
The total number of sequences where A is leading in votes for all time is then the sum of
these two events or
(n − 1)(n − 2)
2
+ n − 2 =
(n + 1)(n − 2)
2
Using this we find
Pn,2 =
(n+1)(n−2)
2
(n+2)(n+1)
2
=
n − 2
n + 2
.
Part (c): It looks like the pattern for Pn,m is Pn,m = n−m
n+m
.
Part (d): By conditioning on who received the last vote (either A or B) we find
Pn,m =

n
n + m

Pn−1,m +

m
n + m

Pn,m−1 . (22)
Part (e): We will show that the fraction Pn,m = n−m
n+m
satisfies the right-hand-side of Equa-
tion 22 and when simplified gives the left-hand-side of the equation.
RHS =
n
n + m
n − 1 − m
n − 1 + m

+
m
n + m
n − m + 1
n + m − 1

=
n2
− n − nm + mn − m2
+ m
(n + m)(n + m − 1)
=
(n − m)(n + m − 1)
(n + m)(n + m − 1)
=
n − m
n + m
= LHS .
Problem 22 (is it rainy or dry?)
Let Dn be the event the weather is dry on day n and the weather (either wet or dry) tomorrow
will be the same as the weather today with probability p (so that it is different weather with
a probability of 1 − p). This means that
P(Dn|Dn−1) = P(Dc
n|Dc
n−1) = p .
We are told that the weather is dry on day 0 which means that P(D0) = 1, and let n ≥ 1.
Then it can become dry on day n in two ways. If it was dry on the day n − 1 and it stays

dry, or it was wet on day n − 1 and it became dry. Conditioning on what the weather was
yesterday (dry or wet) and the necessary transition we have
P(Dn) = P(Dn−1)P(Dn|Dn−1) + P(Dc
n−1)P(Dn|Dc
n−1)
= P(Dn−1)p + P(Dc
n−1)(1 − p)
= P(Dn−1)p + (1 − P(Dn−1))(1 − p)
= P(Dn−1)(2p − 1) + (1 − p) for n ≥ 1 .
We next want to show that the solution to the above recurrence is given by
P(Dn) =
1
2
+
1
2
(2p − 1)n
, n ≥ 0 . (23)
We will show that by induction on n. First for n = 0 we have
P(D0) =
1
2
+
1
2
(2p − 1)0
= 1 ,
as it should. Let n ≥ 1 and assume that Equation 23 is true for n − 1. We then have
P(Dn) = P(Dn−1)(2p − 1) + (1 − p)
=
1
2
+
1
2
(2p − 1)n−1

(2p − 1) + (1 − p)
=
1
2
(2p − 1) +
1
2
(2p − 1)n
+ (1 − p) =
1
2
+
1
2
(2p − 1)n
,
showing that Equation 23 is true for n as well.
Problem 24 (round robin tournaments)
In this problem we specify an integer k and then ask whether it is possible for every set of k
players to have there exist a member from the other n − k players that beat these k players
when competing against these k. To show that this is possible if the given inequality is true,
we follow the hint. In the hint we enumerate the

n
k

sets of k players and let Bi be the
event that no of the other n − k contestant beats every one of the k players in the i set of
k. Then P(∪iBi) is the probability that at least one of the subsets of size k has no external
player that beats everyone. Then 1 − P(∪iBi) is the probability that every subset of size k
has an external player that beats everyone. Since this is the event we want to be possible
we desire that
1 − P(∪iBi) 0 ,
or equivalently
P(∪iBi) 1 .
Now Boole’s inequality states that P(∪iBi) ≤
P
i P(Bi), so if we pick our k such that
P
i P(Bi) 1, we will necessarily have P(∪iBi) 1 possible. Thus we will focus on ensuring
that
P
i P(Bi) 1.

Lets now focus on evaluating P(Bi). Since this is the probability that no contestant from
outside the ith cluster beats all players inside, we can evaluate it by considering a particular
player outside the k member set. Denote the other player by X. Then X would beat all k
members with probability 1
2
k
, and thus with probability 1 − 1
2
k
does not beat all players
in this set. As the set Bi, requires that all n − k players not beat the k players in this ith
set, each of the n − k exterior players must fail at beating the k players and we have
P(Bi) = 1 −

1
2
k
!n−k
.
Now P(Bi) is in fact independent of i (there is no reason it should depend on the particular
subset of players) we can factor this result out of the sum above and simply multiply by the
number of terms in the sum which is

n
k

giving the requirement for possibility of

n
k

1 −

1
2
k
!n−k
1 ,
as was desired to be shown.
Problem 25 (a direct proof of conditioning)
Consider P(E|F) which is equal to P(E, G|F) + P(E, Gc
|F), since the events (E, G) and
(E, Gc
) are independent. Now these component events can be simplified as
P(E, G|F) = P(E|F, G)P(G|F)
P(E, Gc
|F) = P(E|F, Gc
)P(Gc
|F) ,
and the above becomes
P(E|F) = P(E|F, G)P(G|F) + P(E|F, Gc
)P(Gc
|F) ,
as expected.
Problem 26 (conditional independence)
Equations 5.11 and 5.12 from the book are two equivalent statements of conditional inde-
pendence. Namely E1 and E2 are conditionally independent given F occurs if
P(E1|E2, F) = P(E1|F) , (24)
or equivalently
P(E1E2|F) = P(E1|F)P(E2|F) . (25)
To prove this equivalence, consider the left-hand-side of Equation 25. We have when we use
Equation 24 in the second step that
P(E1E2|F) = P(E1|E2F)P(E2|F) = P(E1|F)P(E2|F) .
Proving the equivalence.

Problem 27 (extension of conditional independence)
A set {Ei : i ∈ I} is said to be conditionally independent, given F, if for every finite subset
of events {Eik
: 1 ≤ k ≤ n} with two or more members (n ≥ 2) we have
P
n

k=1
Eik
| F
!
=
n
Y
k=1
P(Eik
| F)
Problem 28 (does independence imply conditional independence)
This statement is false and can be shown with the following example. Consider the event
where we toss two fair coins. The outcomes are thus {HH, HT, TH, TT}. Then let H1 be
the event the first coin lands heads up and let H2 be the event the second coin lands heads
up. Then we have the events H1, H2, and H1H2 given by
H1 = {HH, HT} , H2 = {HH, TH} , H1H2 = {HH} .
Thus we have P(H1) = P(H2) = 1
2
, and P(H1H2) = 1
4
. Now let F be the event the coins land
the same way. Then this event F is given by F = {HH, TT} and we have H1F = {HH},
H2F = {HH}, with H1H2F = {HH}. From these we can calculate probabilities. We have
P(F) = 1
2
, and P(H1F) = P(H2F) = P(H1H2F) = 1
4
. We know that H1 and H2 are
independent events from
P(H1H2) =
1
4
=
1
2
1
2

= P(H1)P(H2) .
With condition probabilities given by
P(H1H2|F) =
P(H1H2F)
P(F)
=
1
4
1
2
=
1
2
P(H1|F) =
P(H1F)
P(F)
=
1
4
1
2
=
1
2
P(H2|F) =
P(H2F)
P(F)
=
1
4
1
2
=
1
2
.
Thus
P(H1H2|F) =
1
2
6=
1
4
= (
1
2
)(
1
2
) = P(H1|F)P(H2|F) ,
and we have that H1 and H2 are not conditionally independent given F even thought they
are independent.
Problem 30 (extensions on Laplace’s rule of succession)
In Laplace’s rule of succession we assume we have k + 1 coins, the ith one of which yields
heads when flipped with probability i
k
. Then from this version of the experiment the first n

flips of the chosen coin results in r heads and n − r tails. Let H denote the event that the
n + 1 flip will land heads. then conditioning on the chosen coin Ci for 0 ≤ i ≤ k we have the
following
P(H|Fn) =
n
X
i=0
P(H|Ci, Fn)P(Ci|Fn)
Then P(H|Ci, Fn) = P(H|Ci) = i
k
and by Bayes’ rule
P(Ci|Fn) =
P(Fn|Ci)P(Ci)
Pn
i=0 P(Fn|Ci)P(Ci)
so since we are told that flipping our coin n times generates r heads and n − r tails we have
that
P(Fn|Ci) =

n
r

i
k
r
1 −
i
k
n−r
,
and that
P(Ci) =
1
k + 1
,
so that P(Ci|Fn) becomes
P(Ci|Fn) =

n
r

i
k
r
1 − i
k
n−r
· 1
k+1

Pn
i=0

n
r

i
k
r
1 − i
k
n−r
· 1
k+1

=
i
k
r
1 − i
k
n−r
Pn
i=0
i
k
r
1 − i
k
n−r ,
so that our probability of a head becomes
P(H|Fn) =
Pn
i=0
i
k
r+1
1 − i
k
n−r
Pn
i=0
i
k
r
1 − i
k
n−r
If k is large then we can write (the integral identity is proven below)
1
k
n
X
i=0

i
k
r
1 −
i
k
n−r
≈
Z 1
0
xr
(1 − x)n−r
dx =
r!(n − r)!
(n + 1)!
.
Thus for large k our probability P(H|Fn) becomes
P(H|Fn) =
(r+1)!(n−r)!
(n+2)!
r!(n−r)!
(n+1)!
=
(r + 1)!
(n + 2)!
·
(n + 1)!
r!
=
r + 1
n + 2
.
Where we have used the identity
Z 1
0
yn
(1 − y)m
dy =
n! m!
(n + m + 1)!
.

To prove this identity we will define C(n, m) to be this integral and use integration by parts
to derive a difference equation for C(n, m). Remembering the integration by parts formula
R
udv = uv −
R
vdu we see that
C(n, m) ≡
Z 1
0
yn
(1 − y)m
dy
= (1 − y)m yn+1
n + 1
1
0
+
Z 1
0
yn+1
n + 1
m(1 − y)m−1
dy
= 0 +
m
n + 1
Z 1
0
yn+1
(1 − y)m−1
dy
=
m
n + 1
C(n + 1, m + 1) .
Using this recurrence relationship we will prove the proposed representation for C by using
mathematical induction. We begin by determining some initial conditions for C(·, ·). We
have that
C(n, 0) =
Z 1
0
yn
dy =
yn+1
n + 1
1
0
=
1
n + 1
.
Note that this incidental equals n!0!
(n+1)!
as it should. Using our recurrence relation derived
above we then find that
C(n, 1) =
1
n + 1
C(n + 1, 0) =
1
(n + 2)(n + 1)
and
C(n, 2) =
2
n + 1
C(n + 1, 1) =
2
(n + 1)(n + 3)(n + 2)
.
Note that these two expressions equal n!1!
(n+2)!
and n!2!
(n+2+1)!
respectively as they should. We
have shown that
C(n, m) =
n!m!
(n + m + 1)!
for m ≤ 2 ,
so to prove this by induction we will assume that it holds in general and prove that it holds
for C(n, m + 1). For the expression using our recurrence relationship (and our induction
hypothesis) we have that
C(n, m + 1) =
m + 1
n + 1
C(n + 1, m) =
m + 1
n + 1

(n + 1)!m!
(n + m + 2)!

=
n!(m + 1)!
(n + (m + 1) + 1)!
,
which proves this result for m + 1 and therefore by induction it is true for all m.
Problem 9 (watering the plant)
Let W be the event the neighbor waters the plant and let D be the event the plant dies. We
are told that P(D|Wc
) = 0.8, P(D|W) = 0.15, and P(W) = 0.90.

A a
A AA Aa
a aA aa
Table 12: The possible genotypes from two hybrid parents.
Part (a): We want to compute P(Dc
). We have
P(D) = P(D|W)P(W) + P(D|Wc
)P(Wc
)
= P(D|W)P(W) + P(D|Wc
)(1 − P(W))
= (0.15)(0.90) + (0.8)(1 − 0.90) = 0.135 + 0.080 = 0.215 .
Thus P(Dc
) = 1 − P(D) = 1 − 0.215 = 0.785.
Part (b): We want to compute P(Wc
|D). We find
P(Wc
|D) =
P(D|Wc
)P(Wc
)
P(D)
=
P(D|Wc
)(1 − P(W))
P(D)
=
(0.80)(1 − 0.90)
0.215
=
16
43
.
Problem 10 (black rat genotype)
We are told that in a certain species of rats, black dominates over brown. Let A be the
dominant black allele and let a be the recessive brown allele. Since the sibling rat is brown
and brown is recessive, the sibling rat must have a genotype aa. The only way for two black
parents to have a brown offspring is if both parents have the genotype Aa.
Part (a): As each parent contributes one allele to the genotype of the offspring, the possible
offspring of our two Black parents are given by the results in Table 12. There we see that any
offspring of these two parents will have genotypes of AA (with probability 1/4), Aa (with
probability 1/2), or aa (with probability 1/4). Since we know that this rat is black it must
be of genotypes AA or Aa. The probability it is of genotype Aa (or hybrid) is then
1
2
1
2
+ 1
4
=
2
3
.
The probability it is pure is then 1 − 2
3
= 1
3
.
Part (b): This black rat then mates with a brown rat. Since brown is recessive, the brown
rat must have a genotype of aa. As each parent contributes one allele to the genotype of
the offspring, if the black rat has genotype AA, then offspring will have genotype Aa. If
the black rat has genotype Aa, the offspring will have either the genotype Aa or aa. See
Table 13 for the possible offspring.
Let E be the event the rat has genotype AA or is “pure”, then Ec
is the event the rat has
genotype Aa or is “hybrid”. Let Ci be the event the ith offspring has genotype Aa. Then
from Table 13 we see that
P(Ci|E) = 1, and P(Ci|Ec
) =
1
2
.

a a
A Aa Aa
A Aa Aa
a a
A Aa Aa
a aa aa
Table 13: Possible offspring of a black rat mated with a brown rat.
Then we want to evaluate P(E | C1C2C3C4C5). From Bayes’ rule we have
P(E | C1C2C3C4C5) =
P(C1C2C3C4C5 | E)P(E)
P(C1C2C3C4C5)
.
Assume the 5 events the offspring have genotype Aa are conditionally independent, given
the rat has genotype AA or the rat has genotype Aa. We then have
P(C1C2C3C4C5 | E) =
5
Y
i=1
P(Ci|E) = (1)5
and
P(C1C2C3C4C5 | Ec
) =
5
Y
i=1
P(Ci|Ec
) =
1
2
5
.
Thus we have
P(C1C2C3C4C5) = P(C1C2C3C4C5|E)P(E) + P(C1C2C3C4C5|Ec
)P(Ec
)
=
1
3

(1)5
+
2
3
1
2
5
,
so our desired probability is given by
P(E | C1C2C3C4C5) =
P(C1C2C3C4C5 | E)P(E)
P(C1C2C3C4C5)
=
1
3

(1)5
1
3

(1)5 +
2
3
1
2
5
=
1
1 +
1
16
=
16
17
.
Problem 11 (circuit flow)
Let Ci be the event the ith relay is closed (so that current can flow through that connection)
and let E be the event current flows between A and B. If relay 1 is closed then the event
current flows is
E|C1 = C4 ∪ C3C5 ∪ C2C5 .
Note in the solution in the back of the book I think the expression there is missing the term
C2C5. If on the other hand relay 1 is open the event current flows is
E|Cc
1 = C2C3C4 ∪ C2C5 .
Thus the probability that current flows is
P(E) = P(E|C1)P(C1) + P(E|Cc
1)P(Cc
1) .

Since all relays are independent the condition that C1 (or Cc
1) hold true does not affect the
evaluation of the probabilities of the other sets of open/closed relays. Thus
P(C4 ∪ C2C5 ∪ C3C5|C1) = P(C4) + P(C2C5) + P(C3C5)
− P(C2C3C5) − P(C2C4C5) − P(C3C4C5) + P(C2C3C4C5) and
P(C2C3C4 ∪ C2C5|Cc
1) = P(C2C3C4) + P(C2C5) − P(C2C3C4C5) .
By independence each of term in the above can be expanded in terms of products of pi.
Using these two results we find
P(E) = p1(p4 + p2p5 + p3p5 − p2p3p5 − p2p4p5 − p3p4p5 + p2p3p4p5)
+ (1 − p1)(p2p3p4 + p2p5 − p2p3p4p5) .
Problem 12 (k-out-of-n-system)
Let Ci be the event the ith component is working and for each part let E be the event at
least k of n components are working.
Part (a): To have one of two components working means that E = C1 ∪ C2. Thus we have
P(E) = P(C1 ∪ C2) = P(C1) + P(C2) − P(C1C2)
= P(C1) + P(C2) − P(C1)P(C2)
= p1 + p2 − p1p2 .
The the probability of the event of interest or P(C1|E) can be computed as
P(C1|E) =
P(C1E)
P(E)
=
P(C1)
P(E)
=
p1
p1 + p2 − p1p2
,
since if C1 is true then E must also be true.
Part (b): We will evaluate P(C1|E) = P (C1E)
P (E)
. We thus need P(C1E) and P(E). To have
two of three components working means that
E = C1C2 ∪ C1C3 ∪ C2C3 .
Thus
P(E) = P(C1C2 ∪ C1C3 ∪ C2C3)
= P(C1C2) + P(C1C3) + P(C2C3)
− P(C1C2C1C3) + P(C1C2C2C3) + P(C1C3C2C3) + P(C1C2C1C3C2C3)
= P(C1C2) + P(C1C3) + P(C2C3) + 2P(C1C2C3)
= p1p2 + p1p3 + p2p3 − 2p1p2p3 .
From the above expression for E we have C1E given by
C1E = C1C2 ∪ C1C3 ∪ C1C2C3 = C1C2 ∪ C1C3 ,

since C1C2C3 ⊂ C1C2. Thus
P(C1E) = P(C1C2 ∪ C1C3) = P(C1C2) + P(C1C3) − P(C1C2C3)
= p1p2 + p1p3 − p1p2p3 ,
and we find
P(C1|E) =
P(C1E)
P(E)
=
p1p2 + p1p3 − p1p2p3
p1p2 + p1p3 + p2p3 − 2p1p2p3
.
Problem 13 (roulette)
This is a classic example of what is called “gamblers fallacy”. We can show that in fact
the probability of getting red has not changed given the 10 times that black has appeared
assuming independence of the sequential spins. Let Ri be the event the ball lands on red
on the ith spin. Let Bi be the event the ball lands on black on the ith spin. Then the
probability of the ball landing on red on spin i given the 10 other blacks is
P(Ri|Bi−1Bi−2 . . . Bi−10) =
P(RiBi−1Bi−2 . . . Bi−10)
P(Bi−1Bi−2 . . . Bi−10)
=
P(Ri)P(Bi−1)P(Bi−2) . . . P(Bi−10)
P(Bi−1)P(Bi−2) . . . P(Bi−10)
= P(Ri) ,
showing that there has been no change in his chance. It is interesting to note that even if
one believed in this strategy very strongly (which we argue above is not a sound idea) the
strategy itself would be onerous to implement since the event of 10 blacks in a row would
not happen very frequently, giving rise to long waiting times between bets.
Problem 14 (the odd man out)
On each coin toss the player A will be the odd man if he gets heads while the others get tails
or he gets tails while the others get heads. Each of these events happens with probability
p1(1 − p2)(1 − p3) and (1 − p1)p2p3 , (26)
respectively. The game continues if all players get heads or all players get tails. Each of
these events happen with probability
p1p2p3 and (1 − p1)(1 − p2)(1 − p3) , (27)
respectively. The game stops with A not the odd man out with probability one minus the
sum of the four probabilities above. Let E be the event A is the eventual odd man out and
the probability we want to compute. Then we can compute P(E) by conditioning on the
result of the first set of coin tosses. Let E1 be the event that A is the odd man out on one
coin toss and O1 the event that one of the players B or C is the odd man out on one coin
toss. We then have
P(E) = P(E|E1)P(E1) + P(E|O1)P(O1) + P(E|(E1 ∪ O1)c
)P((E1 ∪ O1)c
)
= P(E1) + P(E)P((E1 ∪ O1)c
) . (28)

Where we have used the facts that
P(E|E1) = 1 , P(E|O1) = 0 , and P(E|(E1 ∪ O1)c
) = P(E) ,
since in the last case no one was the odd man out and the game effectively starts over. From
Equation 28 we can solve for P(E). We find
P(E) =
P(E1)
1 − P((E1 ∪ O1)c)
.
We can evaluate P(E1) by summing the two terms in Equation 26 and we can evaluate
1 − P((E1 ∪ O1)c
) by recognizing that this is the probability no one is the odd man out on
the first toss and then use Equation 27. Thus we get
P(E) =
p1(1 − p2)(1 − p3) + (1 − p1)p2p3
1 − [p1p2p3 + (1 − p1)(1 − p2)(1 − p3)]
.
Problem 15 (the second trial is larger)
Let N and M be the outcome of the first and second experiment respectively. We want
P{M N}. We can do this by conditioning on the outcome of M. We have
P{M N} =
n
X
i=1
P{M N = i|N = i}P{N = i}
=
N
X
i=1
n
X
j=i+1
pj
!
pi =
n
X
i=1
n
X
j=i+1
pipj .
As another way to solve this problem let E be the event that the first experiment is smaller
than the second, let F the event that the two experiments have the same value and let
G be the event that the first experiment is larger than the second. Then by symmetry
P(E) = P(G) and we have
1 = P(E) + P(F) + P(G) = 2P(E) + P(F) .
Now we can explicitly evaluate P(F) since P(F) =
Pn
i=1 p2
i . Thus
P(E) =
1
2

(1 −
n
X
i=1
p2
i ) .
These two expressions can be shown to be equal by squaring the relationship
Pn
i=1 pi = 1.
Problem 16 (more heads)
Let A be the event A gets more heads than B after each has flipped n coins. Let B be the
event A gets fewer heads than B after each has flipped n coins. Let C be the event A and

B get the same number of heads after each has flipped n coins. Let E be the event A gets
more total heads than B after his n + 1st flip. The following the hint we have
P(E) = P(E|A)P(A) + P(E|B)P(B) + P(E|C)P(C)
= 1P(A) + 0p(B) +
1
2
P(C)
= P(A) +
1
2
P(C) .
But on the nth flip we have P(A) + P(B) + P(C) = 1, and by symmetry P(A) = P(B) thus
2P(A) + P(C) = 1 so P(C) = 1 − 2P(A) .
When we put this into the above expression for P(E) we find
P(E) = P(A) +
1
2
(1 − 2P(A)) =
1
2
.
Problem 17 (independence with E, F ∪ G, FG)
Part (a): This statement is False. Consider the following counter example. A die is rolled
and let the events E, F, and G be defined as the following outcomes from this roll E = {1, 6},
F = {1, 2, 3}, and G = {1, 4, 5}. These base events and their derivatives have the following
probabilities
P(E) =
2
6
, P(F) =
3
6
, P(G) =
3
6
P(EF) =
1
6
, P(E|F) =
1
3
= P(E)
P(EG) =
1
6
, P(E|G) =
1
3
= P(E) .
Note that since P(EF) = P(E)P(F) we have that E and F are independent. In the same
way E and G are independent. Now consider the events E(F ∪ G) and F ∪ G. We have
P(F ∪ G) =
5
6
, and P(E(F ∪ G)) =
1
6
.
Since P(E(F ∪ G)) = 1
6
6= P(E)P(F ∪ G) = 1
3
· 5
6
= 5
18
we have that E and F ∪ G are not
independent.
Part (b): This is true. Since E and F are independent, P(EF) = P(E)P(F) and since E
and G are independent we have P(EG) = P(E)P(G). Now consider
P(E(F ∪ G)) = P(EF ∪ EG) = P(EF) + P(EG) − P(EFEG) = P(EF) + P(EG) − P(EFG)
= P(E)P(F) + P(E)P(G) ,
using independence and the fact that since FG = ∅ we have EFG = ∅. In the same way
since FG = ∅ we have P(F ∪ G) = P(F) + P(G). Thus
P(E)P(F ∪ G) = P(E)P(F) + P(E)P(G) .

Since this is the same expression as P(E(F ∪G)) we have that P(E(F ∪G)) = P(E)P(F ∪G)
or the pair E and F ∪ G are independent.
Part (c): This is true. Since E and FG are independent, we have P(E(FG)) = P(E)P(FG).
Since F and G are independent we have P(FG) = P(F)P(G). Then
P(G(EF)) = P(E(FG)) = P(E)P(FG) = P(E)P(F)P(G) .
Since E and F are independent we have P(EF) = P(E)P(F) and thus
P(G)P(EF) = P(G)P(E)P(F) .
Since these are both the same expressions we have shown that, P(G(EF)) = P(G)P(EF),
and the pair G and EF are independent.
Problem 18 (∅ and independence)
Part (a): This is always false. The reason is that if AB = ∅, then P(AB) = P(∅) = 0,
but we are told that P(A) 0 and P(B) 0 thus the product the the two probabilities is
nonnegative P(A)P(B) 0. Thus it is not possible for P(AB) = P(A)P(B) and A and B
are not independent.
Part (b): This is always false. The reason is that if we assume A and B are independent,
then P(AB) = P(A)P(B). Since we assume that P(A) 0 and P(B) 0 we must
haveP(A)P(B) 0 and thus P(AB) 6= 0, which would be required if A and B were mutually
exclusive.
Part (c): This is always false. If we assume P(A) = P(B) = 0.6 and assume that A and B
could be mutually exclusive we would then conclude
P(A ∪ B) = P(A) + P(B) − P(AB) = 0.6 + 0.6 − 0 = 1.2 .
But we can’t have the probability of the event A ∪ B greater than 1. Thus A and B cannot
be mutually exclusive.
Part (d): This can possibly be true. Let an urn have 6 red balls and 4 white balls and draw
sequentially twice with replacement a ball. Let A and B be the event that we draw a red
ball. Then P(A) = P(B) = 0.6 and these two events are independent (since we are drawing
with replacement).
Problem 19 (ranking trials)
Let H be the event the coin toss is heads. Let Ei be the event the result of the ith trial is
success. The probabilities of each event in the order they are listed are

• P(H) =
1
2
= 0.5
• P(E1E2E3) = P(E1)P(E2)P(E3) = (0.8)3
= 0.512, when P(Ei) = 0.8.
• P(∩7
i=1Ei) =
Q7
i=1 P(Ei) = (0.9)7
= 0.4789269, when P(Ei) = 0.9.
Problem 20 (defective radios)
To start this problem we define several events, let A be the event the that the radios were
produced at factory A, B be the event they were produced at factory B and let Di be
the event the ith radio is defective. Then from the problem statement we are told that
P(A) = P(B) =
1
2
, P(Di|A) = 0.05, and P(Di|B) = 0.01. We observe that event D2 and
want to calculate P(D1|D2). To compute this probability we will condition on whether or
not the two radios came from factory A or B. We have
P(D2|D1) = P(D2|D1, A)P(A|D1) + P(D2|D1, B)P(B|D1)
= P(D2|A)P(A|D1) + P(D2|B)P(B|D1) ,
where we have assumed D1 and D2 are conditionally independent given A or B. Now to
evaluate P(A|D1) and P(B|D1) we will use Bayes’ rule. For example for either D1 or D2 we
have
P(A|D) =
P(D|A)P(A)
P(D)
=
P(D|A)P(A)
P(D|A)P(A) + P(D|B)P(B)
.
Using the numbers given for this problem we have
P(A|D) =
0.05(0.5)
0.05(0.5) + 0.01(0.5)
= 0.833
P(B|D) =
0.01(0.5)
0.05(0.5) + 0.01(0.5)
= 0.166 .
Thus we find
P(D2|D1) = 0.05(0.833) + 0.01(0.166) = 0.0433 .
Problem 21 (P(A|B) = 1 means P(Bc
|Ac
) = 1)
We are told that P(A|B) = 1, thus using the definition of conditional probability we have
P (A,B)
P (B)
= 1 or P(A, B) = P(B). The event AB is always a subset of the event B thus
P(AB) P(B) if AB 6= B. Thus AB = B and B ⊆ A. Complementing this relationship
gives Bc
⊇ Ac
or that P(BC
|Ac
) = 1 since if Ac
occurs then Bc
must have also occurred as
Ac
⊆ Bc
.

Problem 22 (i red balls)
For this problem let E(i, n) be the event there are exactly i red balls in the urn after n stages
and let Rn be the event a red ball is selected at stage n to generate the configuration of balls
in the next stage, where we take n ≥ 0. Then we want to show
P(E(i, n)) =
1
n + 1
for 1 ≤ i ≤ n + 1 . (29)
At stage n = 0 the urn initially contains 1 red and 1 blue ball
P(E(1, 0)) = 1 =
1
0 + 1
.
Thus we have the needed initial condition for the induction hypothesis. Now let n 0 and
assume that Equation 29 holds up to some stage n and we want to show that it then also
holds on stage n + 1.
To evaluate P(E(i, n + 1)) we consider how we could get i red balls on the n + 1st stage.
There are two ways this could happen, either we had i red balls during stage n and we drew
a blue ball, or we had i − 1 red balls during stage n and we drew a red ball. Since initially
there are 2 balls in the urn and one ball is added to the urn at each stage. Thus after stage
n, there are n + 2 balls in the urn. Thus on stage n the probability we draw a red or blue
ball is given by
P(Rn) =
i − 1
n + 2
P(Rc
n−1) = 1 − P(Rn−1) = 1 −
i − 1
n + 2
=
n + 2 − i
n + 2
.
Thus we have
P(E(i, n + 1)) = P(E(i − 1, n))P(Rn) + P(E(i, n))P(Rn
c
)
= P(E(i − 1, n))

i − 1
n + 2

+ P(E(i, n))

n + 2 − i
n + 2

=
1
n + 1

i − 1
n + 2

+
1
n + 1

n + 2 − i
n + 2

=
n + 1
(n + 1)(n + 2)
=
1
n + 2
,
where we have used the induction hypothesis to conclude P(E(i−1, n)) = P(E(i, n)) = 1
n+1
.
Since we have shown that Equation 29 is true at stage n + 1 by induction it is true for all n.
Problem 25 (a conditional inequality)
Now following the hint we have
P(E|E ∪ F) = P(E|E ∪ F, F)P(F) + P(E|E ∪ F, ¬F)P(¬F) .

Balls (W,W) (W,B) (W,0) (B,B) (B,0) (0,0)
X -2 +1 -1 4 2 0
Table 14: Possible values for our winnings X when two colored balls are selected from the
urn in Problem 1.
Chapter 4 (Random Variables)
Chapter 4: Problems
Problem 1 (winning by drawing balls from an urn)
The possibilities of the various X’s we can obtain are given in Table 14. We find that the
probabiltiies of the various X values are given by
P{X = −2} =

8
2

14
2
=
4
13
P{X = −1} =

8
1

2
1

14
2
=
16
91
P{X = 0} =

2
2

14
2
=
1
91
P{X = 1} =

8
1

4
1

14
2
=
32
91
P{X = 2} =

4
1

2
1

14
2
=
8
91
P{X = 3} = 0
P{X = 4} =

4
2

14
2
=
6
91
.

1 2 3 4 5 6
1 1 2 3 4 5 6
2 2 4 6 8 10 12
3 3 6 9 12 15 18
4 4 8 12 16 20 24
5 5 10 15 20 25 30
6 6 12 18 24 30 36
Table 15: The possible values for the product of two dice when two dice are rolled.
Problem 2 (the product of two dice)
We begin by constructing the sample space of possible outcomes. These numbers are com-
puted in table 15, where the row corresponds to the first die and the column corresponds to
the second die. In each square we have placed the product of the two dice. Each pair has
probability of 1/36, so by enumeration we find that
P{X = 1} =
1
36
, P{X = 2} =
2
36
P{X = 3} =
2
36
, P{X = 4} =
3
36
P{X = 5} =
2
36
, P{X = 6} =
4
36
P{X = 8} =
2
36
, P{X = 9} =
1
36
P{X = 10} =
2
36
, P{X = 12} =
4
36
P{X = 15} =
2
36
, P{X = 16} =
1
36
P{X = 18} =
2
36
, P{X = 20} =
2
36
P{X = 24} =
2
36
, P{X = 25} =
1
36
P{X = 30} =
2
36
, P{X = 36} =
1
36
,
with any other integer having zero probability.
Problem 4 (ranking five men and five women)
Note: In contrast to the explicitly stated instructions provided by the problem where X = 1
would correspond to the event that the hightest ranked woman is ranked first (best), I choose
to solve this problem with the backwards convention that the best ranking corresponds to
X = 10, effectivly the reverse of the standard convention. I’m sorry if this causes any
confusion.

As the variable X represents the ranking of the the highest female when we have five total
females and using the backwards ranking convention discussed above, the lowest the highest
ranking can be is five, so P{X = i} = 0 if 1 ≤ i ≤ 4. Now P{X = 5} is proportional to the
number of ways to get five women first in a line
P{X = 5} =
(5!)(5!)
10!
=
1
7 · 62
=
1
252
.
Since if the first five positions are taken up by women and the last five positions are taken
up by men we have 5! orderings of the women and 5! orderings of the men. Giving 5! · 5!
We now want to evaluate P{X = 6}. To do this we must place a women in the sixth place
which can be done in five possible ways (from among the five women). We then must place
four more women and one man in the positions 1, 2, 3, 4 and 5. We can pick the man in five
ways (from among all possible men) and his position in another five ways. We then have 4!
orderings of the remaining four women and 4! orderings of the remaining four men. Thus
our probability is then
P{X = 6} =
5 · 5 · 5 · 4! · 4!
10!
=
5
252
.
Now we need to evaluate P{X = 7} which has a numerator consisting of the following
product of terms.
5
1

·

5
2

· (2!) ·

6
2

· (4!) · (3!) .
The first term

5
1

, is the ways to pick the women in the seventh spot. The term

5
2

is the number of ways to pick the two men that will go to the left of this woman. The term
2! represents all possible permutations of these two men. The term

6
2

, is the number
of ways we can select the specific spots these two men go into. The term 4! is the ways to
pick the orderings of the remaining women. Finally the 3!, represents the number of ways
to pick the ordering of the three remaining men. We then need to divide the product by 10!
to convert this into a probability which gives
P{X = 7} =
5
84
.
We now need to evaluate P{X = 8}. To do this we reason as follows. We have

5
1

ways to pick the women to place in the eighth spot. Then

7
4

spots to the left of this
women to pick as the spots where the four remaining women will be placed. Then 4! different
placements of the women in these spots. Once all the women are placed we have 7 − 4 = 3
slots to place three men who will be to the left of the initial women at position eight. The
men to go in these spots can be picked in

5
3

ways and their ordering selected in 3!.

Finally, we have 2! arrangements of the remaining two men, giving a total count of the
number of instances where X = 8 of

5
1

·

7
4

· (4!) ·

5
3

· (3!) · (2!) = 504000 ,
which gives a probability
P{X = 8} =
5
36
.
To evaluate P{X = 9} we have

5
1

ways to pick the women at position nine. Then

8
4

ways to pick the remaining slots to the left of this women to place the remaining
other women into, and 4! ways to rearrange them. We then have 8 − 4 = 4 slots for men to
go into and

5
4

ways to pick four men to fill these spots and 4! ways to rearrange them.
So the number of instances when X = 9 is given by

5
1

·

8
4

· (4!) ·

5
4

· (4!) = 1008000 ,
which gives a probability of
P{X = 9} =
5
18
.
Finally, to evaluate P{X = 10} we have

5
1

ways to pick the women,

9
4

ways to
pick spots for the four remaining women and 4! ways to rearrange them. With the women
placed, we have five slots remaining for the men and 5! ways of arraignment them. This
gives
5
1

·

9
4

· (4!) · (5!) ,
Giving a probability of
P{X = 10} =
1
2
.
One can further check that if we add all of these probabilities up we obtain
1
252
+
5
252
+
5
84
+
5
36
+
5
18
+
1
2
= 1 ,
as we should.
We now present a simpler method that uses combinatorial counting to evaluate these prob-
abilities. As before we have P{X = i} = 0 for 1 ≤ i ≤ 4. Lets now assume 5 ≤ i ≤ 10 and
we can compute P{X = i} in the following way. We first select one of the five women to
occupy the leading position i. This can be done in 5 ways. Next we have a total of i − 1
positions behind the leading woman occupying position i in which we can to place the four
remaining women. We can select the spots in to place the women in

i − 1
4

ways and

their specific ordering in 4! ways. Next we can place the men in the remaining five spots in
5! ways. Using the multiplication principle we conclude that the probability for X = i is
thus given by
P{X = i} =
5

i − 1
4

(4!)(5!)
10!
.
Since the product of

i − 1
4

and 4! simplifies as

i − 1
4

(4!) =
(i − 1)!
(i − 5)!
,
the expression for P{X = i} above simplifies to
P{X = i} =
5
10 · 9 · 8 · 7 · 6

(i − 1)!
(i − 5)!

for 5 ≤ i ≤ 10 .
Evaluating the above for each value of i duplicated the results from above. One thing to
notice about the above formula is that the location of the men in this problem statement
becomes irrelevant. This can be seen if we write the final expression above as
5((i−1)!
(i−5)! )
10·9·8·7·6
. In
that expression the denominator of 10 · 9 · 8 · 7 · 6 represents the number of ways one can
place five women in ten spots, while the numerator in the above expression
5

(i − 1)!
(i − 5)!

= 5!
(i − 1)!
(i − 1 − 4)!4!
= 5!

i − 1
4

,
for 5 ≤ i ≤ 10 represents the number of ways to place five women where the top woman is in
the i-th spot. See the Matlab/Octave file chap 4 prob 4.m for the fractional simplifications
needed in this problem.
Problem 5 (the difference between heads and tails)
Define X = nH − nT with nH the number of heads and nT the number of tails. Then if
our sequence of n flips results in all heads (n of them) with no tails we have X = n. If we
have n − 1 heads (and thus one tail) the variable X is given by X = n − 1 − 1 = n − 2.
Continuiong, if we have n − 2 heads and therfore two tails our variable X then becomes,
X = n − 2 − 2 = n − 4. In general, we see by induction that
X ∈ {n, n − 2, n − 4, · · · , 4 − n, 2 − n, −n} ,
or as a formula X = n − 2i with i taken from 0, 1, 2, · · · , n. This result can be easiliy be
derived algebraically by recognizing the constrain that n = nH + nT , which implys when we
solve for nH that nH = n − nT , so that
X ≡ nH − nT = (n − nT ) − nT
= n − 2nT ,
where 0 ≤ nT ≤ n.

Problem 6 (the probabilities of heads minus tails)
From Problem 5 we see that the probability that X takes on a specific value is directly related
to the probability of obtaining some number nT of tails. The probability of obtaining nT tails
in n flips is a binomial random variable with parameters (n, p = 1/2) and thus probability

n
nT

pnT
(1 − p)n−nT
. Thus for a fair coin (where p = 1/2) we have
P{X = n} = P{nT = 0} =

n
0

2n
=
1
2n
P{X = n − 2} = P{nT = 1} =

n
1

2n
=
n
2n
P{X = n − 4} = P{nT = 2} =

n
2

2n
=
n(n − 1)
2n+1
,
etc. So in general we have
P{X = n − 2i} = P{nT = i} =
1
2n

n
i

.
So if n = 3 we have
P{X = 3} =
1
23
=
1
8
P{X = 1} =
1
23

3
1

=
3
8
P{X = −1} =
1
23

3
2

=
3
8
P{X = −3} =
1
23

3
3

=
1
8
Problem 7 (the functions of two dice)
In table 16 we construct a table of all possible outcomes associated with the two dice rolls.
In that table the row corresponds to the first die and the column corresponds to the second
die. Then for each part of the problem we find that
Part (a): X ∈ {1, 2, 3, 4, 5, 6}.
Part (b): X ∈ {1, 2, 3, 4, 5, 6}.
Part (c): X ∈ {2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12}.
Part (d): X ∈ {−5, −4, −3, −2, −1, 0, 1, 2, 3, 4, 5}.

1 2 3 4 5 6
1 (1,1,2,0) (2,1,3,-1) (3,1,4,-2) (4,1,5,-3) (5,1,6,-4) (6,1,7,-5)
2 (2,1,3,1) (2,2,4,0) (3,2,5,-1) (4,2,6,-2) (5,2,7,-3) (6,2,8,-4)
3 (3,1,4,2) (3,2,5,1) (3,3,6,0) (4,3,7,-1) (5,3,8,-2) (6,3,9,-3)
4 (4,1,5,3) (4,2,6,2) (4,3,7,1) (4,4,8,0) (5,4,9,-1) (6,4,10,-2)
5 (5,1,6,4) (5,2,7,3) (5,3,8,2) (5,4,9,1) (5,5,10,0) (6,5,11,-1)
6 (6,1,7,5) (6,2,8,4) (6,3,9,3) (6,4,10,2) (6,5,11,1) (6,6,12,0)
Table 16: The possible values for the maximum, minimum, sum, and first minus second die
observed when two dice are rolled.
Problem 8 (probabilities on dice)
The solution to this problem involves counting up the number of times that X equals the
given value and then dividing by 62
= 36. For each part we have the following
Part (a): From table 16, for this part we find that
P{X = 1} =
1
36
, P{X = 2} =
3
12
, P{X = 3} =
5
36
,
P{X = 4} =
7
36
, P{X = 5} =
1
4
, P{X = 6} =
11
36
Part (b): From table 16, for this part we find that
P{X = 1} =
11
36
, P{X = 2} =
1
4
, P{X = 3} =
7
36
,
P{X = 4} =
1
12
, P{X = 5} =
7
36
, P{X = 6} =
1
36
Part (c): From table 16, for this part we find that
P{X = 2} =
1
36
, P{X = 3} =
1
18
, P{X = 4} =
1
12
,
P{X = 5} =
1
9
, P{X = 6} =
5
36
, P{X = 7} =
1
6
,
P{X = 8} =
5
36
, P{X = 9} =
1
9
, P{X = 10} =
1
12
,
P{X = 11} =
1
18
P{X = 12} =
1
36
.
Part (d): From table 16, for this part we find that
P{X = −5} =
1
36
, P{X = −4} =
1
18
, P{X = −3} =
1
9
,
P{X = −2} =
1
9
, P{X = −1} =
5
36
, P{X = 0} =
1
6
,
P{X = 1} =
5
36
, P{X = 2} =
1
9
, P{X = 3} =
1
12
,
P{X = 4} =
1
18
, P{X = 5} =
1
36
.

Problem 9 (sampling balls from an urn with replacement)
For this problem balls selected with replacement from an urn and we define the random
variable X as X = max(x1, x2, x3) where x1, x2, and x3 are the numbers on the balls from
each three draws. We know that
P{X = 1} =
1
203
.
Now P{X = 2} can be computed as follows. To count the number of sets of three draws
that contain at least one two and so the max will be 2 (one such set is (1, 1, 2)) we consider
all sets of three we could build from the components 1 and 2. Then for each slot we have
two choices so we have 2 · 2 · 2 = 23
possible choices. But one of these (the one selected by
assembling an ordered set of three elements from only the element one) so the number of
sets with a two as the largest element is 23
− 13
= 23
− 1 = 7. To compute
P{X = 3}
we consider all ordered sets we can construct from the elements 1, 2, 3 since we have three
choices for the first spot, three for the second spot, and three for the third we have 33
= 27.
The number of sets that have a three in them are this number minus the number of sets that
have only two’s and one’s in them. Which is given by 23
thus we have 33
−23
= 27 −8 = 19.
The general pattern then is
P{X = i} =
i3
− (i − 1)3
203
. (30)
As a more general way to derive the result above consider that with replacement the sample
space for three draws is {1, 2, · · · , 20}3
. Then each possible draw of three numbers has the
same probability 1
203 . If we let Hi denote the event that the highest numbered ball drawn
(from the three) has a value of i then the event Hi can be broken down into several mutually
independent events. The first is the draw (i, i, i) where all three balls show the number i.
This can happen in only one way. The second type of draw under which event Hi can be
said to have occurred are draws where there are two balls that have the highest number i
and the third draw has a lower number. An example draw like this would be (i, i, X), with
X i. This can happen in

3
1

= 3 ways and each draw has a probability of

1
20

1
20

i − 1
20

,
of happening. The third type of draw we can get and have event Hi is one of the type
(i, X, Y ) where the two numbers X and Y are such that X i and Y i. This draw has a
probability of happening given by

1
20

i − 1
20

i − 1
20

,
and there are

3
1

= 3 ways that draws like this can happen. Thus to compute P{X = i}
we sum the three results above to get
P{X = i} =
1
203
+
3(i − 1)
203
+
3(i − 1)2
203
=
3i2
− 3i + 1
203
.

By expanding the numerator of Equation 30 we can show that these two expressions are
equivalent.
Using these results above to calculate P{X ≥ 17} or the probability that we win the bet we
find
173
− 163
+ 183
− 173
+ 193
− 183
+ 203
− 193
203
=
203
− 163
203
=
61
125
.
Problem 10 (if we win i dollars)
For this problem we desire to compute the conditional probability we win i dollars given
we win something. Let E be the event that we win something and we want to evaluate
P{X = i|E} using Bayes’ rule we find that
P{X = i|E} =
P{E|X = i}P{X = i}
P{E}
.
Now P{E} =
P
i=1,2,3 P{E|X = i}P{X = i} for these are the i that we make a profit on
and therefore have P{E|X = i} 6= 0 (the other i’s all have P{E|X = i} = 0). For these i’s
we have that
P{E|X = i} = 1 ,
so we get for P{E} given by
P{E} =
39
165
+
15
165
+
1
165
=
1
3
.
So that we have P{X = i|E} = 0, when i = 0, −1, −2, −3, and that
P{X = 1|E} =
P{X = 1}
P{E}
=
39
165
1
3
= 0.709
P{X = 2|E} =
P{X = 2}
P{E}
=
15
165
1
3
= 0.2727
P{X = 3|E} =
P{X = 3}
P{E}
=
1
165
1
3
= 0.01818 .
Problem 11 (the Riemann hypothesis)
Part (a): Note that there are ⌊103
3
⌋ = 333 multiples of three in the set {1, 2, 3, ., 103
}. The
multiples are specifically
3 · 1, 3 · 2, · · · , 3 · 333 .
Note that the last element equals 999. Since we are given that any number N is equally
likely to to be chosen from the 1000 numbers we see that
P(N is a multiple of 3) =
333
103
.

In the same way we would compute
⌊103
5
⌋
103
=
200
103
⌊103
7
⌋
103
=
142
103
⌊103
15
⌋
103
=
66
103
⌊103
105
⌋
103
=
9
103
.
In each of the above cases we see that as k gets larger and larger we expect
lim
k→∞
1
10k

10k
N

=
1
N
.
Player Opponent Random Variable
Guesses Shows Guesses Shows Amount Won: X
1 1 1 1 0
1 1 1 2 −3
1 1 2 1 2
1 1 2 2 0
1 2 1 1 3
1 2 1 2 0
1 2 2 1 0
1 2 2 2 −4
2 1 1 1 −2
2 1 1 2 0
2 1 2 1 0
2 1 2 2 3
2 2 1 1 0
2 2 1 2 4
2 2 2 1 −3
2 2 2 2 0
Table 17: Two-Finger Morra Outcomes
Problem 12 (two-finger Morra)
Part (a): All possible outcomes for a round of play of two-finger Morra are shown in
Table 17. Under the given assumptions, each row of the table is equally likely and can
therefore be assigned a probability of 1
16
. Using that table the associated probabilities for

the possible values of X are given by
P{X = 0} =
8
16
=
1
2
P{X = 2} = P{X = −2} =
1
16
P{X = 3} = P{X = −3} =
2
16
=
1
8
P{X = 4} = P{X = −4} =
1
16
.
Part (b): In this case the strategy corresponds to using only rows 1, 4, 13, and 16 of Table
17. We see that either both players guess correctly or both players guess incorrectly on every
play. Thus the only output is X = 0 and we have P{X = 0} = 1.
Problem 13 (selling encyclopedias)
There are 9 possible outcomes, as summarized in Table 18. Summing all possible ways to
get the various values of X we find
P{X = 0} = .28
P{X = 500} = .21 + .06 = .27
P{X = 1000} = .21 + .045 + .06 = .315
P{X = 1500} = .045 + .045 = .09
P{X = 2000} = .045 .
Sale from Customer 1 Sales from Customer 2 X Probability
0 0 0 (1 − .3)(1 − .6) = .28
0 500 500 (1 − .3)(.6)(.5) = .21
0 1000 1000 (1 − .3)(.6)(.5) = .21
500 0 500 (.3)(.5)(1 − .6) = .06
500 500 1000 (.3)(.5)(.6)(.5) = .045
500 1000 1500 (.3)(.5)(.6)(.5) = .045
1000 0 1000 (.3)(.5)(1 − .6) = .06
1000 500 1500 (.3)(.5)(.6)(.5) = .045
1000 1000 2000 (.3)(.5)(.6)(.5) = .045
Table 18: Encyclopedia Sales
Problem 14 (getting the highest number)
To begin we note that there are 5! equally likely possible possible orderings of the numbers
1 − 5 that could be dealt to the five players. Now player 1 will win 4 times if he has the

highest of the five numbers. Thus the first number must be a 5 followed by any of the 4!
possible orderings of the other numbers. This gives a probability
P{X = 4} =
1 · 4!
5!
=
1
5
.
Next player 1 will win 3 times if his number exceeds the numbers of players 2, 3, and 4, but
is less than the number of player 5. In other words, player 1 must have the second highest
number and player 5 the highest. This means that player 5 must have been given the number
5 and player 1 must have been given the number 4 and the other 3 numbers can be in any
order among the remaining players. This gives a probability of
P{X = 3} =
1 · 1 · 3!
5!
=
1
20
.
For player 1 to win twice he must have a number greater than the numbers of players 2 and
3 but less than that of player 4; i.e., of the first four players, player four has the highest
number and player 1 has the second highest. We select the four numbers to assign to the
first four player in 5
4

ways. This leaves a single number for player 5. We then select the
largest number from this group of four, for player four (in one way), and then select the
second largest number (in one way) for the first player. This gives two remaining numbers
which can be ordered in two ways. This gives a probability of
P{X = 2} =
5
4

· 1 · 1 · 2! · 1
5!
=
1
12
.
Player 1 wins exactly once if his number is higher than that of player 2 and lower than that
of player 3. Following the logic when we win twice we select the three numbers for players
1 − 3 in 5
3

ways. This gives two numbers for the players 4 and 5 which can be ordered in 2
ways. From the initial set of three, we assign the largest of this set to player 3 and the next
largest to player 1. The last number goes to player 2. Taken together this gives a probability
of
P{X = 1} =
5
3

· 1 · 1 · 1 · 2!
5!
=
1
6
.
Finally, player 1 never wins if his number is less than that of player 2. The same logic as
above gives for this probability the following
P{X = 0} =
5
2

· 1 · 1 · 3!
5!
=
1
2
Problem 15 (the NBA draft pick)
Notice first that once a ball belonging to a team has been drawn, any other balls belonging
to that team are subsequently ignored, so we may treat the problem as if all balls belonging
to a team are removed from the urn once any ball belonging to that team is drawn. Let us
adopt the following terminology in analyzing the problem:
Fi = “First pick goes to team with i-th worst record”
Si = “Second pick goes to team with i-th worst record”
Ti = “Third pick goes to team with i-th worst record”

In the above notation we have 1 ≤ i ≤ 11. Note that i = 1 is the worst team and has 11
balls in the urn initially, i = 2 is the second worst team and has 10 balls in the urn initially,
etc. In general, the ith worst team has 12 − i balls in the urn until it is selected. With this
shorthand, much of this problem can be solved by conditioning on what happens “first” i.e.
that the event Fi comes before the event Si and both come before Ti.
P {X = 1} = P {F1} =
11
66
= 0.1667 .
P {X = 2} = P {S1}
= P {F2S1 ∨ F3S1 ∨ · · · ∨ F11S1}
= P {F2} P {S1 | F2} + P {F3} P {S1 | F3} + · · · + P {F11} P {S1 | F11}
=
10
66
·
11
66 − 10
+
9
66
·
11
66 − 9
+ · · · +
1
66
·
11
66 − 1
=
11
X
k=2
12 − k
66
·
11
66 − (12 − k)
=
11
X
k=2
12 − k
66
·
11
54 + k
= 0.15563 .
P {X = 3} = P {T1}
=
11
X
k=2
P {SkT1} =
11
X
k=2




11
X
j=2
j6=k
P {FjSkT1}




=
11
X
k=2




11
X
j=2
j6=k
12 − j
66
·
12 − k
66 − (12 − j)
·
11
66 − (12 − j) − (12 − k)




=
11
X
k=2
11
X
j=2
j6=k
12 − j
66
·
12 − k
54 + j
·
11
42 + j + k
= 0.1435
P {X = 4} = 1 −
3
X
i=1
P {X = i} = 0.53423
P {X = i} = 0, i /
∈ {1, 2, 3, 4}
Note that P {X = i} = 0 for i ≥ 5 since if it is not drawn in the first 3 draws it will be given
the fourth draft pick according to the rules. These sums are computed in the MATLAB file
chap 4 prob 15.m.
Problem 16 (more draft picks)
Following the notation introduced in the previous problem, we have

Part (a):
P {Y1 = i} = P {Fi} =
12 − i
66
for 1 ≤ i ≤ 11 .
Part (b):
P {Y2 = i} = P {Si} =
11
X
j=1;j6=i
P {FjSi}
=
11
X
j=1;j6=i
12 − j
66

12 − i
66 − (12 − j)

=
11
X
j=1;j6=i
12 − j
66

12 − i
54 + j

.
Part (c):
P {Y3 = i} = P {Ti}
=
11
X
j=1;j6=i
P {SjTi} =
11
X
j=1;j6=i
11
X
k=1;k6=i ,j
P {FkSjTi}
!
=
11
X
j=1;j6=i
11
X
k=1;k6=i ,j
12 − k
66
·
12 − j
66 − (12 − k)
·
12 − i
66 − (12 − j) − (12 − k)
!
.
Problem 17 (probabilities from the distribution function)
Part (a): We find that
P{X = 1} = P{X ≤ 1} − P{X 1} =
1
2
−
1
4
=
1
4
P{X = 2} = P{X ≤ 2} − P{X 2} =
11
12
−

1
2
+
2 − 1
4

=
1
6
P{X = 3} = P{X ≤ 3} − P{X 3} = 1 −
11
12
=
1
12
.
Part (b): We find that
P{
1
2
X
3
2
} = lim
n→∞
P{X ≤
3
2
−
1
n
} − P{X ≤
1
2
}
=

1
2
+
3
2
− 1
4

−
1
2
4
=
1
2
.
Problem 18 (the probabilities mass from the number of heads)
The event Hk, that in n flips of a coin we get k heads is given by a binomial random variable
and so we have
P{Hk} =

n
k

pk
(1 − p)n−k
.

When n = 4 and p = 1
2
as for this problem we have P{Hk} = 4
k
1
2
4
. Thus
P{X = −2} = P{Hk = 0} =

1
2
4
P{X = −1} = P{Hk = 1} = 4

1
2
4
P{X = 0} = P{Hk = 2} = 6

1
2
4
P{X = 1} = P{Hk = 3} = 4

1
2
4
P{X = 2} = P{Hk = 4} =

1
2
4
,
are the values of the probability mass function.
Problem 19 (probabilities from the distribution function)
Since F(b) =
P
x≤b p(x) from the given expression for F(b) we see that
p(0) =
1
2
p(1) =
3
5
−
1
2
=
1
10
p(2) =
4
5
−
3
5
=
1
5
p(3) =
9
10
−
4
5
=
1
10
p(3.5) = 1 −
9
10
=
1
10
.
First Game Second Game Third Game X Probability
win N.A. N.A. +1 p
loss win win +1 p2
q
loss win loss -1 pq2
loss loss win -1 pq2
loss loss loss -3 q3
Table 19: Playing “no loose” roulette
Problem 20 (a winning roulette strategy)
This problem can more easily be worked by imagining a tree like structure representing the
possible outcomes and their probabilities. For example, from the problem in the text, if we

win on the first play (with probability p = 9
19
) we stop playing and have won +1. If we
loose on the first play we will play the game two more times. In these two games we can
win twice, win once and lose once or loose twice. Ignoring for the time begin the initial loss
these three outcomes occur with probabilities given by a binomial distribution with n = 3
and p = 9
19
or
p2
, 2pq , q2
.
The reward (payoff) for each of the outcomes is given by
+2 , 0 , −2 .
Since these second two games are only played if we loose the first game we must condition
them on the output from that event. Thus the total probabilities (and win amounts X) then
are given in Table 19 Using these we can answer the questions given.
Part (a): We find
P{X 0} = p + p2
q =
9
19
+
10
19

9
19
2
= 0.5917 .
Part (b): No, there are two paths where we win but 3 where we loose. One of these paths
has a loss of −3 which is relatively large given the problem.
Part (c): We find
E[X] = 1p + 1p2
q − 1pq2
− 1pq2
− 3q3
= −0.108 ,
when we evaluate.
Problem 21 (selecting students or buses)
Part (a): The probability of selecting a student on a bus is proportional to the number of
students on that bus while the probability we select a given bus driver is simply 1/4, since
there is no weighting based on the number of students in each bus. Thus E[X] should be
larger than E[Y ].
Part (b): We have that
E[X] =
4
X
i=1
xip(xi) = 40

40
148

+ 33

33
148

+ 25

25
148

+ 50

50
148

= 39.28 .
while
E[Y ] =
4
X
i=1
yip(yi) = 40

1
4

+ 33

1
4

+ 50

1
4

= 37 .
So we see that E[X] E[Y ] as expected.

Problem 22 (winning games)
Warning: Due to time constraints this problem has not been checked as thoroughly as
others and may not be entirely complete. If anyone finds anything wrong with these please
let me know.
We will consider the two specific cases where i = 2 and i = 3 before the general case. When
i = 2 to evaluate the expected number of games played we want to evaluate P{N = n}
where N is the random variable determining the expected number of games played before a
win (by either team A or B). Then P{N = 1} = 0 since we need two wins for A or B to
win overall. Now P{N = 2} = p2
+ q2
, since from the four possible outcomes from the two
experiments (A, A), (A, B), (B, A), and (B, B) only two result in a win. The first (A, A)
occurs with probability p2
and the last with probability q2
. Since they are mutually exclusive
events we have the desired probability above. Continuing, we have that
P{N = 3} = 2pqp + 2qpq = 2p2
q + 2q2
p ,
since to have A win on three games (and not win on two) we must place the last of A’s wins
as the third win. Thus only two sequences give wins for A in three flips i.e. (A, B, A) and
(B, A, A). Each occurs with probability p2
q. The second term in the above is equivalent but
with q replaced with p.
The expression P{N = 4} is not a valid probability since one player A or B would have won
before four games. We can also check that we have a complete formulation by computing
the probability A or B wins after any number of flips i.e. consider
p2
+ q2
+ 2p2
q + 2q2
p = p2
+ (1 − p)2
+ 2p2
(1 − p) + 2(1 − p)2
p
= p2
+ 1 − 2p + p2
+ 2p2
− 2p3
+ 2(1 − 2p + p2
)p = 1 .
Thus the expected number of games to play before one team wins is
E[N] = 2(p2
+ q2
) + 3(2p2
q + 2q2
p) = 2 + 2p − 2p2
In the general case it appears that
P{N = i} = pi
+ qi
P{N = i + 1} =

i
1

qpi
+

i
1

pqi
P{N = i + 2} =

i + 1
2

q2
pi
+

i + 1
2

p2
qi
P{N = i + 3} =

i + 2
3

q3
pi
+

i + 2
3

p3
qi
.
.
.
P{N = i + (i − 1)} =

2i − 2
i − 1

qi−1
pi
+

2i − 2
i − 1

pi−1
qi
.

In the case i = 3 the above procedure becomes
P{N = 3} = p3
+ q3
P{N = 4} = 3qp3
+ 3pq3
P{N = 5} = 6q2
p3
+ 6p2
q3
.
Checking that we have included every term we compute the sum of all of the above terms
to obtain
p3
+ q3
+ 3qp3
+ 3pq3
+ 6q2
p3
+ 6p2
q3
,
Which is simplified (to the required one) in the Mathematica file chap 4 prob 22.nb. To
compute the expectation we have
E[N] = 3(p3
+ q3
) + 4(3qp3
+ 3pq3
) + 5(6q2
p3
+ 6p2
q3
) .
We take the derivative of this expression and set it equal to zero in the Mathematical file
chap 4 prob 22.nb.
Problem 23 (trading commodities)
Part (a): Let assume one buys x of the commodity at the start of the week, then in cash
one has C = 1000 − 2x. Here we have x ounces of our commodity with 0 ≤ x ≤ 500. Then
at the end of the week our total value is given by
V = 1000 − 2x + Y x ,
where Y is the random variable representing the cost per ounce of the commodity. We desire
to maximize E[V ]. We have
E[V ] = 1000 − 2x + x
2
X
i=1
yip(yi)
= 1000 − 2x + x

1

1
2

+ 4

1
2

= 1000 +
x
2
.
Since this is an increasing linear function of x, to maximize our expected amount of money,
we should buy as much as possible. Thus let x = 500 i.e. buy all that one can.
Part (b): We desire to maximize the expected amount of the commodity that one posses.
Now by purchasing x at the beginning of the week, one is then left with 1000 − 2x cash to
buy more at the end of the week. The amount of the commodity A, that we have at the end
of the week is given by
A = x +
1000 − 2x
Y
,

where Y is the random variable denoting the cost per ounce of our commodity at the end of
the week. Then the expected value of A is then given by
E[A] = x +
2
X
i=1

1000 − 2x
yi

p(yi)
= x +

1000 − 2x
1

1
2

+

1000 − 2x
4

1
2

= 625 −
x
4
.
Which is linear and decreases with increasing x. Thus we should pick x = 0 i.e. buy none
of the commodities now and buy it all at the end of the week.
Problem 24
Part (a): Let XB be the gain of B when playing the game. Then if A has written down
one we have
E[XB] = p(1) + (1 − p)

−3
4

=
7p − 3
4
.
However if A has written down two, then our expectation becomes
E[XB] = p

−3
4

+ (1 − p)2 =
8 − 11p
4
.
To derive the value of p that will maximize player B’s return, we incorporate the fact that
the profit XB depends on what A does by conditioning on the possible choices. Thus we
have
E[XB] =
7p−3
4
A picks 1
8−11p
4
A picks 2
Plotting these two linear lines we have Figure ?? (left). From this graph we recognize that
we will guarantee the maximal possible expected return independent of what A does if we
select p such that
7p − 3
4
=
8 − 11p
4
.
which gives p = 11
18
. Thus the expected gain with this value of p is given by
7p − 3
4 p= 11
18
=
23
72
.
Now consider the expected loss of player A under his randomized rule. To do so, let YA be
the random variable specifying the loss received by player A. Then if B always picks number
one we have
E[YA] = q(−1) + (1 − q)

3
4

=
3
4
−
7
4
q ,

while if B always picks number two we have
E[YA] = q

3
4

+ (1 − q)(−2) =
11
4
q − 2 .
Plotting these expected losses as function of q we have Figure ?? (right). Then to find the
smallest expected loss for player A independent of what player B does we have to find q such
that
3
4
−
7
4
q =
11
4
q − 2 .
When we solve for q we find that q = 11
18
, which is the same as before. Now the optimal
expected loss is given by
3
4
−
7
4

11
18

= −
23
72
,
which is the negative of the expected gain for player B.
Problem 25 (expected winnings with slots)
To compute the expected winnings when playing one game from a slot machine, we first
need to compute the probabilities for each of the winning combinations. To begin with we
note that we have a total of 203
possible three dial combinations. Now lets compute a count
of each dial combination that results in a payoff. We find
N(Bar, Bar, Bar) = 3
N(Bell, Bell, Bell) = 2 · 2 · 3 = 12
N(Bell, Bell, Bar) = 2 · 2 · 1 = 4
N(Plum, Plum, Plum) = 4 · 1 · 6 = 24
N(Orange, Orange, Orange) = 3 · 7 · 6 = 126
N(Orange, Orange, Bar) = 3 · 7 · 1 = 21
N(Cherry, Cherry, Anything) = 7 · 7 · 20 = 980
N(Cherry, No Cherry, Anything) = 7 · (20 − 7) · 20 = 1820 .
So the number of non winning rolls is given by 203
−
P
Above = 203
− 2990 = 5101. Thus
the expected winnings are then given by
E[W] = 60

3
203

+ 20

12
203

+ 18

4
203

+ 14

24
203

+ 10

126
203

+ 8

21
203

+ 2

980
203

+ 0

1820
203

− 1

5010
203

= −0.09925 .

Problem 26 (guess my number)
Part (a): If at stage n by asking the question “is it i”, one is able to eliminate one possible
choice from further consideration (assuming that we have not guessed the correct number)
before stage n. Thus let En be the event at stage n we guess the number correctly, assuming
we have not guessed it correctly in the n − 1 earlier stages. Then
P(En) =
1
10 − (n − 1)
=
1
11 − n
.
so we have that
P(E1) =
1
10
P(E2) =
1
9
P(E3) =
1
8
.
.
.
P(E10) = 1 .
The expected number of guesses to make using this method is then given by
E[N] = 1

1
10

+ 2

1 −
1
10

1
9

+ 3

1 −
1
10

1 −
1
9

1
8
+ · · ·
= 1

1
10

+ 2

1
10

+ 3

1
10

+ · · ·
=
10
X
n=1
n

1
10

=
1
10

10(10 + 1)
2

= 5.5 .
Part (b): In this second case we will ask questions of the form: “is i less than the cur-
rent midpoint of the list”. For example, initially the number can be any of the numbers
1, 2, 3, · · · , 9, 10 so one could ask the question “is i less than five”. If the answer is yes,
then we repeat our search procedure on the list 1, 2, 3, 4. If the answer is no, we repeat our
search on the list 5, 6, 7, 8, 9, 10. Thus we never know the identity of the hidden number
until O(ceiling(lg(10))) steps have been taken. Since lg(10) = 3.32 we require O(4) steps.
To determine the expected number of steps, lets enumerate the number of guesses each spe-
cific integer would require using the above method Note, that it might be better to ask the
question is i less than or equal to x). Then since any given number is equally likely to be
selected the expected number of question to be asked is given by
E[N] =
1
10
(7 · 3 + 2 + 8) = 3.1 .

Questions (in order) Number of Questions
1 ( 5);( 3);( 2) 3
2 ( 5);( 3);( 2) 3
3 ( 5);( 3);( 2) 3
4 ( 5);( 3); 2
5 ( 5);( 7);( 6) 3
6 ( 5);( 7);( 6) 3
7 ( 5);( 7);( 6) 3
8 ( 5);( 7);( 8) 3
9 ( 5);( 7);( 8);( 9) 4
10 ( 5);( 7);( 8);( 9) 4
Table 20: The sequence of questions asked and the number for the situations where the
hidden number is somewhere between 1 and 10.
Problem 27
The company desires to make 0.1 A of a profit. Assuming the cost charged to each customer
is C, the expected profit of the company then given by
C + p(−A) + (1 − p)(0) = C − pA .
This can be seen as the fixed cost received from the paying customers minus what is lost if
a claim must be paid out. For this to be 0.1A we should have c − pA = 0.1A or solving for
C we have
C =

p +
1
10

A .

Problem 28
We can explicitly calculate the number of defective items obtained in the sample of twenty.
We find that
P0 =

16
3

4
0

20
3
= 0.491
P1 =

16
2

4
1

20
3
= 0.421
P2 =

16
1

4
2

20
3
= 0.084
P3 =

16
0

4
3

20
3
= 0.0035 ,
so the expected number of defective items is given by
3P3 + 2P2 + 1P1 + 0P0 = 0.6 =
3
5
.
Problem 29 (a machine that breaks down)
Under the first strategy we would check the first possibility and if needed check the second
possibility. This has an expected cost of
C1 + R1 ,
if the first possibility is true (which happens with probability p) and
C1 + C2 + R2 ,
if the second possibility is true (which happens with probability 1 − p). Here I am explicitly
assuming that if the first check is a failure we must then check the second possibility (at a
cost C2) before repair (at a cost of R2). Another assumption would be that if the first check
is a failure then we know that the second cause is the real one and we don’t have to check
for it. This results in a cost of C1 + R2 rather than C1 + C2 + R2. The first assumption
seems more consistent with the problem formulation and will be the one used. Thus under
the first strategy we have an expected cost of
p(C1 + R1) + (1 − p)(C1 + C2 + R2) ,

so our expected cost becomes
C1 + pR1 + (1 − p)(C2 + R2) = C1 + C2 + R2 + p(R1 − C2 − R2) .
Now under the second strategy we would first check the second possibility and if needed
check the first possibility. This first action has an expected cost of
C2 + R2 ,
if the second possibility is true cause (this happens with probability 1 − p) and
C2 + C1 + R1 ,
if the first possibility is true (which happens with probability p). This gives an expected
cost when using the second strategy of
(1 − p)(C2 + R2) + p(C2 + C1 + R1) = C2 + R2 + p(C1 + R1 − R2) .
The expected cost under strategy number one will be less than the expected cost under
strategy number if
C1 + C2 + R2 + p(R1 − C2 − R2) C2 + R2 + p(C1 + R1 − R2) .
When we solve for p the above simplifies to
p
C1
C1 + C2
.
As the threshold value to use for the different strategies. This result has the intuitive
understanding in that if p is “significantly” large (meaning the break is more likely to be
caused by the first possibility) we should check the first possibility first. While if p is not
significantly large we should check the second possibility first.
Problem 30 (the St. Petersburg paradox)
The probability that the first tail appears on the nth flip means that the n − 1 heads must
first appear and then a tail. This gives a probability of

1
2

1
2
n−1
=

1
2
n
.
Then the expected value of our winnings is given by
∞
X
n=1
2n

1
2
n
=
∞
X
n=1
1 = +∞ .
Part (a): If a person payed 106
to play this game he would only “win” if the first tail
appeared on toss greater than or equal to n∗
where n∗
≥ log2(106
) = 6 log2(10) = 6ln(10)
ln(2)
=
19.931, or n∗
= 20. In that case this event would occur with probability
∞
X
k=n∗

1
2
k
=

1
2
n∗ ∞
X
k=0

1
2
k
=

1
2
n∗−1
,

since
P∞
k=0
1
2
k
= 2. With n∗
= 20 we see that this probability is given by 9.5367 × 10−7
a
rather small number. Thus many would not be willing to play under these conditions.
Part (b): In this case, if we play k games then we will definitely “win” if the first tail
appears on a flip n∗
(or greater) where n∗
solves
−k 106
+ 2n∗
0 ,
or
n∗
6 log2(10) + log2(k) = 19.931 + log2(k) .
Since this target n∗
grows logarithmically with k one would expect that enough random
experiments were ran that eventually a very high paying result would appear. Thus many
would be willing to pay this game.
Problem 31 (scoring your guess)
Since the meterologist truly believes that it will rain with probability p∗
if he quotes a
probability p, then the expected score he will receive is given by
E[S; p] = p∗
(1 − (1 − p)2
) + (1 − p∗
)(1 − p2
) .
We want to pick a value of p such that we maximize this expression. To do so, consider the
derivative of this expression set equal to zero and solve for the value of p. We find that
dE[S; p]
dp
= p∗
(2(1 − p)) + (1 − p∗
)(−2p) = 0 .
solving for p we find that p = p∗
. Taking the second derivative of this expression we find
that
d2
E[S; p]
dp2
= −2p∗
− 2(1 − p∗
) = −2 0 ,
showing that p = p∗
is a maximum. This is a nice reason for using this metric, since it
behooves the meteriologist to quote the probability of rain that he truly believes is true.
Problem 32 (testing diseased people)
We have one hundred people which we break up into ten groups of ten for the purposes of
testing for a disease. For each group we will test the entire group of people with one test.
This test will be “positive” (meaning at least one person has the disease) with probability
1 − (0.9)10
. Since 0.910
is the probability that all people are normal and the complement of
this probability is the probability that at least one person has the disease. Then the expected
number of tests for each group of ten is then
1 + 0((0.9)10
) + 10(1 − (0.9)10
) = 11 − 10(0.9)10
= 7.51 .
Where the first 1 is because we will certainly test the pooled people and the remaining to
expressions represent the case where the entire pooled test result comes back negative (no
more tests needed) and the case where the entire pooled test result comes back positive
(meaning we have ten individual tests to then do).

0 5 10 15
−120
−100
−80
−60
−40
−20
0
20
b
expected
profit
Figure 1: The expected profit for the newspaper ordering problem when b papers are ordered.
Problem 33 (the number of papers to purchase)
Let b be the variable denoting the number of papers bought and N the random variable de-
noting the number of papers demanded. Finally, let the random variable P be the newsboys’
profits. Then with these definitions the newsboys’ profits is given by
P = −10b + 15 min(N, b) for b ≥ 1 ,
This is because if we only buy b papers we can only sell a maximum of b papers independent
of what the demand N is. Then to calculate the expected profit we have that
E[P] = −10b + 15E[min(N, b)]
= −10b + 15
10
X
n=0
min(n, b)

10
n

1
3
n
2
3
10−n
.
To evaluate the optimal number of papers to buy we can plot this as a function of b for
1 ≤ b ≤ 15. In the Matlab file chap 4 prob 33.m, where this function is computed and
plotted. See Figure 1, for a figure of the produced plot. There one can see that the maximum
expected profit occurs when we order b = 3 newspapers. The expected profit in that case is
given by 8.36.
Problem 35 (a game with marbles)
Part (a): Define W to be the random variable expression the winnings obtained when one
plays the proposed game. The expected value of W is then given by
E[W] = 1.1Psc − 1.0Pdc
where the notation “sc” means that the two drawn marbles are of the same color and the
notation “dc” means that the two drawn marbles are of different colors. Now to calculate
each of these probabilities we introduce the four possible events that can happen when we

draw to marbles: RR, BB, RB, and BR. As an example the notation RB denotes the event
that we first draw a red marble and then second draw a black marble. With this notation
we see that Psc is given by
Psc = P{RR} + P{BB}
=
5
10

4
9

+
5
10

4
9

=
4
9
.
while Pdc is given by
Pdc = P{RB} + P{BR}
=
5
10

5
9

+
5
10

5
9

=
5
9
.
With these two results the expected profit is then given by
1.1

4
9

− 1.0

5
9

= −
1
15
.
Part (b): The variance of the amount one wins can be computed by the standard expression
for variance in term of expectations. Specifically we have
Var(W) = E[W2
] − E[W]2
.
Now using the results from Part (a) above we see that
E[W2
] =
4
9
(1.1)2
+
5
9
(−1.0)2
=
82
75
.
so that
Var(W) =
82
75
−

1
15
2
=
49
45
≈ 1.08 .
Problem 36 (the variance of the number of games played)
From Problem 22 we have that (for i = 2)
E[N2
] = 4(p2
+ q2
) + 9(2p2
q + 2q2
p) = 4 + 10p − 10p2
.
Thus the variance is given by
Var(N) = E[N2
] − (E[N])2
= 4 + 10 − 10p2
− (2 + 2p − 2p2
)2
= 2p(1 − 3p + 4p2
− 2p3
) .
Which has an inflection point at p = 1/2.

Problem 38 (evaluating expectations and variances)
Part (a): We find, expanding the quadratic and using the linearity property of expectations
that
E[(2 + X)2
] = E[4 + 4X + X2
] = 4 + 4E[X] + E[X2
] .
In terms of the variance, E[X2
] is given by E[X2
] = Var(X) + E[X]2
, both terms of which
we know from the problem statement. Using this the above becomes
E[(2 + X)2
] = 4 + 4(1) + (5 + 12
) = 14 .
Part (b): We find, using properties of the variance that
Var(4 + 3X) = Var(3X) = 9Var(X) = 9 · 5 = 45 .
Exercise 39 (drawing two white balls in four draws)
The probability of drawing a white ball is 3/6 = 1/2. Thus if we consider event that we
draw a white ball a success, the probability requested is that in four trials, two are found
to be successes. This is equal to a binomial distribution with n = 4 and p = 1/2, thus our
desired probability is given by

4
2

1
2
2
1
2
4−2
=
6
4 · 4
=
3
8
.
Problem 40 (guessing on a multiple choice exam)
With three possible answers possible for each question we have a 1/3 chance of guessing any
specific question correctly. Then the probability that the student gets four or more correct
by guessing would be the required sum of a binomial distribution. Specifically we have

5
4

1
3
4
2
3
1
+

5
5

1
3
5
2
3
0
=
11
243
.
Where the first term is the probability the student guess four questions (from five) correctly
and the second term is the probability that the student guesses all five questions correctly.
Problem 41 (proof of extrasensory perception)
Randomly guessing the man would get seven correct answers (out of ten) with probability

10
7

1
2
7
1
2
3
= 0.11718 .

Since the book then requests the probability that he does at least this well we would need
to consider the probability that he gets eight, nine, or ten answers correct. This would be
the sum of three more numbers computed in exactly the same way as above. This sum is
0.17188.
Problem 42 (failing engines)
The number of of engines that fail (or function) is a Binomial random variable with prob-
ability 1 − p and p respectively. For a three engine plane the probability that it makes a
successful flight if the probability that three or two engines function (for in that case we will
have a majority). This probability is

3
3

p3
(1 − p)0
+

3
2

p2
(1 − p)1
= p2
(3 − 2p) .
For a five engine plane the probability that it makes a successful flight is the probability that
five, four, or three engines function. This probability is

5
5

p5
(1 − p)0
+

5
4

p4
(1 − p)1
+

5
3

p3
(1 − p)2
= p3
(6p2
− 15p + 10) .
Then a five engine plane will be preferred to a three engine plane if
p3
(6p2
− 15p + 10) ≥ p2
(3 − 2p) .
Putting all the p variable to one side of the inequality (and defining a function f(·)) the
above is equivalent to
f(p) ≡ 2p3
= 5p2
+ 4p − 1 ≥ 0 .
Plotting the function f(p) in Figure ?? we see that it is positive for 1
2
≤ p ≤ 1. Thus for ps
in this range we derive a benefit by using the five engine plane.
Problem 46 (convictions)
Let E be the event that a jury renders a correct decision and let G be the event that a
person is guilty. Then
P(E) = P(E|G)P(G) + P(E|Gc
)P(Gc
) .
From the problem statement we know that P(G) = 0.65 and P(Gc
) = 0.35 so P(E|G) is
the probability we get the correct decision given that the defendant is guilty. To reach the
correct decision we must have nine or more guilty votes so
P(E|G) =
12
X
i=9

12
i

(0.8)i
(0.2)12−i
,

which is the case where the person is guilty and at least nine members vote on this persons
guilt. This is because
P(Vote Guilty|G) = 1 − P(Vote Innocent|G) = 1 − 0.2 = 0.8 .
Thus we can compute P(E|G) using the above sum. We find P(E|G) = 0.58191. Now
P(E|Gc
) = 1 − P(Ec
|Gc
) or one minus the probability the jury makes a mistake and votes
an innocent man guilty. This is
P(E|Gc
) = 1 −
12
X
i=9

12
i

(0.1)i
(0.9)12−i
,
Since P(Vote Guilty|Gc
) = 0.1. The above can be computed and equals P(E|Gc
) = 0.7938,
so that
P(E) = 0.5819(0.65) + 0.793(0.35) = 0.656 .
Problem 47 (military convictions)
Part (a): With nine judges we have
P(G) =
9
X
i=5

9
i

(0.7)i
(0.3)9−i
= 0.901 .
With eight judges we have
P(G) =
8
X
i=5

8
i

(0.7)i
(0.3)8−i
= 0.805 .
With seven judges we have
P(G) =
7
X
i=4

7
i

(0.7)i
(0.3)7−i
= 0.8739 .
Part (b): For nine judges
P(Gc
) = 1 − P(G) = 1 −
9
X
i=5

9
i

(0.3)i
(0.7)9−i
= 0.901 .
For eight judges we have
P(Gc
) = 1 −
8
X
i=5

8
i

(0.3)i
(0.7)8−i
= 0.942 .
For seven judges we have
P(Gc
) = 1 −
7
X
i=4

7
i

(0.3)i
(0.7)7−i
= 0.873 .

Part (c): Assume the defense attorney would like to free his or her client. Let D be the
event that the client goes free then
P(D) = P(D|G)P(G) + P(D|Gc
)P(Gc
) ,
with P(D|G) indexed by the number of judges then
P(D|n = 9) = (1 − 0.9011)(0.6) + (0.901)(0.4) = 0.419
P(D|n = 8) = (1 − 0.805)(0.6) + (0.942)(0.4) = 0.493
P(D|n = 7) = (1 − 0.8739)(0.6) + (0.873)(0.4) = 0.429 .
Thus the defense attorney has the best chance of getting his client off if there are two judges
and so he should request that one be removed.
Problem 48 (defective disks)
For this problem lets take the guarantee that the company provides to mean that a package
will be considered “defective” if it has more than one defective disk. The probability that
more than one disk in a pack is defective (Pd) is given by
Pd = 1 −

10
0

(0.01)0
(0.99)10
−

10
1

(0.01)1
(0.99)9
≈ 0.0042 ,
since

10
0

(0.01)0
(0.99)10
is the probability that no disks are defective in the package of
ten disks, and

10
1

(0.01)1
(0.99)9
is the probability that one of the ten disks is defective.
If a customer buys three packs of disks the probability that he returns exactly one pack is the
probability that from his three packs one package is defective. This is given by a binomial
distribution with parameters n = 3 and p = 0.0042. We find this to be

3
1

(0.0042)1
(1 − 0.0042)2
= 0.0126 .
Problem 49 (flipping coins)
We are told in the problem statement that the event the first coin C1, lands heads happens
with probability 0.4, while the event that the second coin C2 lands heads happens with
probability 0.7.
Part (a): Let E be the event that exactly seven of the ten flips land on heads then condi-
tioning on the initially drawn coin (either C1 or C2) we have
P(E) = P(E|C1)P(C1) + P(E|C2)P(C2) .

Now we can evaluate each of these conditional probabilities as
P(E|C1) =

10
7

(0.4)7
(0.6)3
= 0.0424
P(E|C2) =

10
7

(0.7)7
(0.3)3
= 0.2668 .
So P(E) is given by (assuming uniform probabilities on the coin we initially select)
P(E) = 0.5 · 0.0424 + 0.5 · 0.2668 = 0.1546 .
Part (b): If we are told that the first three of the ten flips are heads then we desire to
compute what is the conditional probability that exactly seven of the ten flips land on
heads. To compute this let A be the event that the first three flips are heads. Then we want
to compute P(E|A), which we can do by conditioning on the initial coin selected, i.e.
P(E|A) = P(E|A, C1)P(C1) + P(E|A, C2)P(C2) .
Now as before we find that
P(E|A, C1) =

7
4

(0.4)4
(0.6)3
= 0.1935
P(E|A, C2) =

7
4

(0.7)4
(0.3)3
= 0.2268 .
So the above probability is given by
P(E|A) = 0.5 · 0.1935 + 0.5 · 0.2668 = 0.2102 .
Problem 55 (errors when typing)
Let A and B the event that the paper is typed by typist A or typist B respectively. Let E
be the event that our article has a least one error, then
P(E) = P(E|A)P(A) + P(E|B)P(B) ,
since both typist are equally likely P(A) = P(B) = 1
2
and
P(E|A) =
∞
X
i=1
P{E = i|A} =
∞
X
i=1
e−λA
λi
A
i!
= 1 − e−λA
= 1 − e−3
= 0.9502 .
and in the same way
P(E|B) = 1 − e−4.2
= 0.985 .
so that P(E) = 0.5(0.9502) + 0.5(0.985) = 0.967 so the probability of no errors is given by
1 − P(E) = 0.03239

Problem 56 (at least two birthdays)
The probability that at least one person will have the same birthday as myself is the com-
plement of the probability that no other person has a birthday equivalent to myself. An
individual person will not have the same birthday as myself with probability p = 364
365
= 0.997.
Thus the probability that n people all do not have my birthday is then pn
. So the probability
that at least one person does have my birthday is given by 1 − pn
. To have this greater than
1/2 requires that 1 − pn
≥ 1/2 or pn
≤ 1/2 or
n ≥
ln(1/2)
ln(p)
=
ln(2)
ln(365/364)
= 252.6 .
To make n a integer take n ≥ 253.
Problem 57 (accidents on a highway)
Part (a): P{X ≥ 3} = 1 − P{X = 0} − P{X = 1} − P{X = 2}, with
P{X = i} =
e−λ
λi
i!
.
Then with λ = 3 we have
P{X ≥ 3} = 1 − e−λ
− λe−λ
−
1
2
e−λ
λ2
= 0.576 .
Part (b): P{X ≥ 3|X ≥} = P {X≥3,X≥1}
P {X≥1}
= P {X≥3}
P {X≥1}
. Now P{X ≥ 1} = 1 − e−λ
= 0.95. So
P{X ≥ 3|X ≥ 1} = 0.576
0.95
= 0.607.
Problem 61 (a full house)
The probability of obtaining i full houses from n (n = 1000) is given by a binomial random
variable with p = 0.0014 and n = 1000. Thus the probability of obtaining at least two full
house is
n
X
i=2

n
i

pi
(1 − p)n−i
= 1 −
1
X
i=0

n
i

pi
(1 − p)n−i
= 1 −

1000
0

p0
(1 − p)1000
−

1000
1

p1
(1 − p)999
.
In this problem since p = 0.0014 and n = 1000 we have pn = 1.4 is rather small and we can
use the Poisson approximation to the binomial distribution as
P{X = i} ≈
e−λ
λi
i!
with λ = pn = 1.4 ,
so the above probability is approximately 1 − e−1.4
− e−1.4
(1.4) = 0.408.

Problem 62 (the probability that no wife sits next to her husband)
From Problem 66, the probability that couple i is selected next to each other is given by
2
2n−1
= 1
n−1/2
. Then we can approximate the probability that the total number of couples
sitting together is a Poisson distribution with parameter λ = n 1
n−1/2
= 2n
2n−1
. Thus the
probability that no wife sits next to her husband is given by evaluating a Poisson distribution
with count equal to zero and λ = 2n
2n−1
or
exp {−
2n
2n − 1
} .
When n = 10 this expression is exp {−20
19
} ≈ 0.349. The exact formula is computed in
example 5n from Chapter 2, where the exact probability is given as 0.3395 showing that our
approximation is rather close.
Problem 63 (entering the casino)
Part (a): Lets assume the number of people entering the casino follows a Poisson approxi-
mation with rate λ = 1 person in two minutes. Then in our five minutes interval from 12:00
to 12:05 we have a Poisson random variable with parameter of λt = 1 5
2

= 2.5, so the
probability that no one enters in that five minutes is given by
P{N = 0} = e−2.5
= 0.0821 .
Part (b): The probability that at least four people enter is
P{N ≥ 4} = 1 − P{N ≤ 3} = 1 − e−λ

1 + λ +
λ2
2
+
λ3
6

= 0.242 ,
when λ = 2.5.
Problem 64 (suicides)
Assume that the number of people who commit suicide is a Poisson random variable with
suicide rate one per every 105
inhabitants per month.
Part (a): Since we have 4 105
inhabitants the suicide rate for the population will be λ =
4 · 1 = 4. The desired probability is then
P{N ≥ 8} = 1 − P{N ≤ 7}
= 1 − e−4

1 +
4
1
+
42
2
+
43
6
+
44
24
+
45
120
+
46
720
+
47
5040

= 0.051 .

Part (b): If we now assume that P{N ≥ 8} computed above is independent from month
to month we are looking for the probability this event happens at least twice or
= 1 −

12
0

(1 − P{N ≥ 8})12
−

12
1

P{N ≥ 8}(1 − P{N ≥ 8})11
= 1 − (1 − 0.511)12
− 12(0.0511)(1 − 0.0511)11
= 0.122 .
Part (c): I would assume that month to month, the event eight or more suicides would
be independent and thus the probability that in the first month when eight suicides occurs
would be given by a geometric random variable with parameter p = P{N ≥ 8} = 0.0511.
A geometric random variable represents the probability of repeating an experiment until a
success occurs and is given by
P{X = n} = (1 − p)n−1
p for n = 1, 2, 3, · · ·
Problem 65 (the diseased)
Part (a): Since the probability that the number of soldiers with the given disease is a bino-
mial distribution with parameters (n, p) = (500, 1
103 ), we can approximate this distribution
with a Poisson distribution with rate λ = 500 1
103 = 0.5. Then the required probability is
given by
P{N ≥ 1} = 1 − P{N = 0} = 1 − e−0.5
≈ 0.3934 .
Part (b): We are now looking for
P{N ≥ 2|N 0} =
P{N ≥ 2 , N 0}
P{N 0}
=
1 − P{N 0}
P{N 0}
≈
1 − e−0.5
(1 + 0.5)
0.3934
= 0.2293 .
Part (c): If Jones knows that he has the disease then the news that the test result comes
back positive is not informative to him. Therefore he believes that the distribution of the
number of men with the disease is binomial with parameters (n, p) = (499, 1
103 ). As such,
it can be approximated with a Poisson distribution with parameter λ = np = 499
103 = 0.499.
Then to him the probability that more than one person has the disease is given by
P{N ≥ 2|N 0} = 1 − P{N 1} = 1 − e−0.499
≈ 0.3928 .
Part (d): We desire to compute the probability that any of the 500 − i remaining people
have the disease that is (with the number N the total number of people with the disease)

let E be the event that the people 1, 2, 3, · · ·i − 1 do not have the disease while i does the
probability we desire is then
P{N ≥ 2|E} =
P{N ≥ 2 , E}
P{E}
.
Now the probability P{E} = (1 − p)i
p, since E is a geometric random variable. Now
P{N ≥ 2 , E} is the probability that since person i has the disease that at least one more
person has the disease in the M − i additional people (here M = 500) and is given by
M−i
X
k=1

M − i
k

pk
(1 − p)M−i−k
so this probability (the entire conditional probability) is then
P{N ≥ 2|E} =
PM−i
k=1

M − i
k

pk
(1 − p)M−i−k
(1 − p)ip
,
which becomes (when we put the numbers for this problem in the expression above) the
following
P{N ≥ 2|E} =
P500−i
k=1

500 − i
k

1
103
k
1 − 1
103
500−i−k
1 − 1
103
i 1
103
.
Problem 66 (seating couples next to each other)
Part (a): There are (2n − 1)! different possible seating orders around a circular table when
each person is considered unique. For couple i to be seated next to each other, consider this
couple as one unit, then we have in total now
2n − 2 + 1 = 2n − 1 ,
unique “items” to place around our table. Here an item can be an individual person or
the ith couple considered as one unit. Specifically we have taken the total 2n people and
subtracted the specific ith couple (of two people) and put back the couple considered as one
unit (the plus one). Thus there are (2n − 1 − 1)! = (2n − 2)! rotational orderings of the
remaining 2n − 2 people and the “fused” couple. Since there are an additional ordering of
the individual people in the pair, we have a total of 2(2n − 2)! orderings where couple i is
together. Thus our probability is given by
P(Ci) =
2(2n − 2)!
(2n − 1)!
=
2
2n − 1
.
Part (b): To compute P(Cj|Ci) when j 6= i we note that it is equal to
P(Cj, Ci)
P(Ci)
.

Here P(Cj, Ci) is the joint probability where both couple i and couple j are together. Since
we have evaluated P(Ci) in Part a of this problem we will now evaluate P(Cj, Ci) in the same
way as earlier. With couple i and j considered as individual units, the number of “items”
we have to distribute around our table is given by
2n − 2 + 1 − 2 + 1 = 2n − 2 .
Here as before we subtract the individual people in the couple and then add back in a
“fused” couple considered as one unit. Thus the number of unique permutations of these
items around our table is given by 4(2n − 2 − 1)! = 4(2n − 3)!. The factor of four is for the
different orderings of the husband and wife in each fused pair. Thus our our joint probability
is then given by
P(Cj, Ci) =
4(2n − 3)!
(2n − 1)!
=
2
(2n − 1)(n − 1)
,
so that our conditional probability P(Cj|Ci) is given by
P(Cj|Ci) =
2/(2n − 1)(n − 1)
2/(2n − 1)
=
1
n − 1
.
Part (c): When n is large we want to approximate 1 −P(C1 ∪C2 ∪. . .∪Cn), which is given
by
1 − P(C1 ∪ C2 ∪ . . . ∪ Cn) = 1 −
n
X
i=1
P(Ci) −
X
ij
P(Ci, Cj) + · · ·
!
= 1 −
2n
2n − 1
−
X
ij
P(Cj|Ci)P(Ci) + · · ·
!
= 1 −

2n
2n − 1
−

n
2

2
(2n − 1)(n − 1)
+ · · ·

But since P(Cj|Ci) = 1
n−1
≈ 1
n−1/2
= P(Cj), when n is very large. Thus while the events
Ci and Cj are not independent, their dependence is weak for large n. Thus by the Poisson
paradigm we can expect the number of couples sitting together to have a Poisson approxi-
mation with rate λ = n 2
2n−1

≈ 1. Thus the probability that no married couple sits next
to each other is P{N = 0} = e−1
.
Problem 6 (the sum of cumulative probabilities)
Consider the claimed expression for E[N] that is
P∞
i=1 P{N ≥ i}. Now P{N ≥ i} =
P∞
k=i P{N = k} and inserting this into the above summation gives
∞
X
i=1
∞
X
k=i
P{N = k} .

We can graphically represent this summation in the (i, k) plane as where the summation is
done along the i axis first and then along the k axis i.e. we sum by columns upward first.
Now changing the order of the the summation to sum along rows first row i.e. k is the outer
index and i is the inner index we have that the above is equivalent to
∞
X
k=1
k
X
i=1
P{N = k} .
Which can be written (since P{N = k} does not depend on the index i) as
∞
X
k=1
kP{N = k} = E[N] ,
as expected.
Problem 7 (the first moments of cumulative probabilities)
Following the hint we have
∞
X
i=0
iP{N i} =
∞
X
i=0
i
∞
X
k=i+1
P{N = k} =
∞
X
i=0
∞
X
k=i+1
iP{N = k} .
Since P{N i} =
P∞
k=i+1 P{N = k} when N is an integral valued random variable. Now
to proceed we will change the order of summation. This can be explained by graphically
denoting the summation points in the (i, k) plane. Now in the formulation given above the
summation is done in columns moving from left to right in the triangle of points above.
Equivalently we will instead perform our summation over rows in the triangle of points.
Doing this we would have
∞
X
k=1
k−1
X
i=0
iP{N = k} .
Where the outer sum represents selecting the individual rows and the inner sum the sum-
mation across that row. This sum simplifies to
∞
X
k=1
P{N = k}
k−1
X
i=0
i =
∞
X
k=1
P{N = k}

k(k − 1)
2

=
1
2
∞
X
k=1
k2
P{N = k} −
∞
X
k=1
kP{N = k}
!
=
1
2
E[N2
] − E[N]

,
as requested.

Problem 8 (an exponential expectation)
If X is a random variable such that P{X = 1} = p = 1 − P{X = −1}, then
E[eX
] = P{X = 1}c + P{X = −1}c−1
= 1 .
With the above information we have that this becomes pc + (1 − p)c−1
= 1. On solving for
c (using the quadratic equation) we get that
c =
1 ± |1 − 2p|
2p
.
Thus we have (taking the two possible signs) that
c =
(
1+(1−2p)
2p
= 1−p
p
1−(1−2p)
2p
= 1
.
These are the two possible values for c. Since c 6= 1 then we must have c = 1−p
p
.
Problem 9 (the expected value of standardized variables)
Warning: I believe there is an error in the variance calculation for Y and I need to revisit
the algebra below.
Define our random variable Y by Y = X−µ
σ
then
E[Y ] =
X
i

xi − µ
σ

p(xi) =
1
σ
X
i
(xi − µ)p(xi)
=
1
σ
X
i
xip(xi) − µ
X
i
p(xi)
!
=
1
σ
(E[X] − µ) = 0 .
And also
E[Y 2
] =
X
i

xi − µ
σ
2
p(xi) =
X
i

x2
i − 2xiµ + µ2
σ2

p(xi)
=
1
σ2
X
i
x2
i p(xi) − 2µ
X
i
xip(xi) + µ2
X
i
p(xi)
!
.
Now since E[X2
] =
P
i x2
i p(xi), E[X] =
P
i xip(xi), and
P
i p(xi) = 1, we see that
E[Y 2
] =
1
σ2
E[X2
] − 2µE[X] + µ2

.
Recalling that Var(X) = E[X2
] − E[X], and using everything thus far we have
Var(Y ) =
1
σ2
(E[X2
] − 2µE[X] + µ2
) −
1
σ2
(E[X] − µ)2
=
1
σ2
(E[X2
] − 2µE[X] + µ2
− (E[X]2
− 2µE[X] + µ2
)) = 0 .

Problem 10 (an expectation with a binomial random variable)
If X is a binomial random variable with parameters (n, p) then
E

1
X + 1

=
n
X
k=0

1
k + 1

P{X = k}
=
n
X
k=0

1
k + 1

n
k

pk
(1 − p)n−k
.
Factoring out 1/(n + 1) we obtain
E

1
X + 1

=
1
n + 1
n
X
k=0

n + 1
k + 1

n
k

pk
(1 − p)n−k
.
This result is beneficial since if we now consider the fraction and the n choose k term we see
that
n + 1
k + 1

n
k

=

n + 1
k + 1

n!
k!(n − k)!
=
(n + 1)!
(k + 1)!(n − k)!
=

n + 1
k + 1

.
This substitution turns our summation into the following
E

1
X + 1

=
1
n + 1
n
X
k=0

n + 1
k + 1

pk
(1 − p)n−k
.
the following manipulations allow us to evaluate this summation. We have
E

1
X + 1

=
1
p(n + 1)
n
X
k=0

n + 1
k + 1

pk+1
(1 − p)n+1−(k+1)
=
1
p(n + 1)
n+1
X
k=1

n + 1
k

pk
(1 − p)n+1−k
=
1
p(n + 1)
n+1
X
k=0

n + 1
k

pk
(1 − p)n+1−k
− (1 − p)n+1
#
=
1
p(n + 1)
(1 − (1 − p)n+1
)
=
1 − (1 − p)n+1
p(n + 1)
,
as we were to show.
Problem 11 (each sequence of k successes is equally likely)
Each specific instance of k success and n − k failures has probability pk
(1 − p)n−k
. Since
each success occurs with probability p each failure occurs with probability 1 − p. As each
arraignment has the same number of p’s and 1 − p’s each has the same probability.

Problem 12
Warning: Here are some notes that I had lying around on this problem. I should state that
I’ve not had the time I would like to fully verify this solution. Caveat emptor.
n antennas with m defective, and n − m functioning. Then we have

n − m + 1
m

is the
number of orderings with no two defective antenna consecutive from a total of n antennas.
Therefore the probability of finding m defective antennas from n with no two defective
antennas consecutive is
P{M = m} =

n − m + 1
m

n − m + 1
.
If we don’t worry about consecutive defective antennas. The number of orderings of consec-
utive defective antennas is given by (n − m + 1)m
so that
P{M = m} =

n − m + 1
m

(n − m + 1)m
.
Thus the probability that no two neighboring components are non functional can be obtained
by conditioning on the number of defective components
P(A) =
n
X
m=0
P(A|M = m)P(M = m)
=
n
X
m=0

n − m + 1
m

(n − m + 1)m

n
m

(1 − p)m
pn−m
.
Problem 13 (maximum likelihood estimation with a binomial random variable)
Since X is a binomial random variable with parameters (n, p) we have that
P{X = k} =

n
k

pk
(1 − p)n−k
.
Then the p that maximizes this expression is given by taking the derivative of the above
(with respect to p) setting the resulting expression equal to zero and solving for p. We find
that this derivative is given by
d
dp
P{X = k} =

n
k

kpk−1
(1 − p)n−k
+

n
k

pk
(1 − p)n−k−1
(n − k)(−1) .
Which when set equal to zero and solve for p we find that p = k
n
, or the empirical counting
estimate of the probability of success.

Problem 14 (having children)
Now P{X = n} = αpn
so imposing
P∞
n=0 p(n) = 1 requires that
α
X
n≥0
pn
= 1 ⇒ α
1
1 − p
= 1 ⇒ α = 1 − p .
so that P{X = n} = (1 − p)pn
.
Part (a): The proportions of families with no children is P{X = 0} = 1 − p
P{B = k} =
X
i≥k
P{B = k|X = i}P{X = i} ,
where in computing P{B = k} we have conditioned on the number of children a given
family has. Now we know P{X = i} = (1 − p)pi
. In addition, P{B = k|X = i} =

i
k

1
2
k
1 − 1
2
i−k
=

i
k

1
2
i
. This later probability is because the probability that
we have k boys (given that we have i children) is a binomial random variable with probability
of success 1/2. Combining these two results we find
P{B = k} =
X
i≥k

i
k

1
2
i
(1 − p)pi
= (1 − p)
X
i≥k

i
k
p
2
i
= (1 − p)
p
2
k
+

k + 1
k
p
2
k+1
+

k + 2
k
p
2
k+2
+ · · ·

.
Problem 15
Warning: Here are some notes that I had lying around on this problem. I should state that
I’ve not had the time I would like to fully verify this solution. Caveat emptor.
Let Pn be the probability that we we obtain an even number heads in n flips. Now condi-
tioning on the results of the first flip we find that
Pn = p(1 − Pn−1) + (1 − p)Pn−1 .
To explain this, the first term p(1 − Pn−1) is the probability we get a head (p) times the
probability that we have an odd number of heads in n−1 flips. The second term (1−p)Pn−1
is the probability we have a tail times the probability of an even number of heads in n − 1
tosses. The above simplifies to
p + (1 − 2p)Pn−1 .

We can check that the suggested expression satisfies this recurrence relationship. That is we
ask if
1
2
(1 + (1 − 2p)n
) = p + (1 − 2p)(
1
2
(1 + (1 − 2p)n−1
)
= p +
1 − 2p
p
+
1
2
(1 − 2p)n
=
1
2
(1 + (1 − 2p)n
) ,
giving a true identity. This result should also be able to be shown by explicitly enumerated
all n tosses with an even number if heads as done in the book.
Problem 16 (the location of the maximum of the Poisson distribution)
Since X is a Poisson random variable the probability mass function for X is given by
P{X = i} =
e−λ
λi
i!
.
Following the hint we compute the requested fraction. We find that
P{X = i}
P{X = i − 1}
=

e−λ
λi
i!

(i − 1)!
e−λλi−1

=
λ
i
.
Now from the above expression if i λ then the “lambda” fraction λ
i
1, meaning that the
probabilities satisfy P{X = i} P{X = i−1} which implies that P{X = i} is increasing for
these values of i. On the other hand if i λ then λ
i
1 we P{X = i} P{X = i − 1} and
P{X = i} is decreasing for these values of i. Thus when i λ, our probability P{X = i} is
increasing, while when i λ, our probability P{X = i} is decreasing. From this we see that
the maximum of P{X = i} is then when i is the largest integer still less than or equal to λ.
Problem 17 (the probability of an even Poisson sample)
Since X is a Poisson random variable the probability mass function for X is given by
P{X = i} =
e−λ
λi
i!
.
To help solve this problem it is helpful to recall that a binomial random variable with
parameters (n, p) can be approximated by a Poisson random variable with λ = np, and that
this approximation improves as n → ∞. To begin then, let E denote the event that X is
even. Then to evaluate the expression P{E} we will use the fact that a binomial random
variable can be approximated by a Poisson random variable. When we consider X to be a
binomial random variable we have from theoretical Exercise 15 in this chapter that
P{E} =
1
2
(1 + (q − p)n
) .

Using the Poisson approximation to the binomial we will have that p = λ/n and q = 1−p =
1 − λ/n, so the above expression becomes
P{E} =
1
2

1 +

1 −
2λ
n
n
.
Taking n to infinity (as required to make the binomial approximation by the Poisson distri-
bution exact) and remembering that
lim
n→∞

1 +
x
n
n
= ex
,
the probability P{E} above goes to
P{E} =
1
2
1 + e−2λ

,
as we were to show.
Part (b): To directly evaluate this probability consider the summation representation of
the requested probability, i.e.
P{E} =
∞
X
i=0,2,4,···
e−λ
λi
i!
= e−λ
∞
X
i=0
λ2i
(2i)!
.
When we look at this it looks like the Taylor expansion of cos(λ) but without the required
alternating (−1)i
factor. This observation might trigger the recollection that the above series
is in fact the Taylor expansion of the cosh(λ) function. This can be seen from the definition
of the cosh function which is
cosh(λ) =
eλ
+ e−λ
2
,
when one Taylor expands the exponentials on the right hand side of the above expression.
Thus the above probability for P{E} is given by
e−λ

1
2
(eλ
+ e−λ
)

=
1
2
(1 + e−2λ
) ,
as claimed.
Problem 18 (maximizing λ in a Poisson distribution)
If X is a Poisson random variable then P{X = k} = e−λλk
k!
. Now to determine the value of
λ that maximizes this expression we differentiate P{X = k} with respect to λ and set the
resulting expression equal to zero. We have the derivative (equated equal to zero) given by
−
e−λ
λk
k!
+
e−λ
kλk−1
k!
= 0 .

or
−λk
+ kλk−1
= 0 .
Since λ 6= 0, we have λ = k. We should check this value is indeed a maximum by computing
the second derivative of P{X = k} and showing that when λ = k it is negative.
Problem 19 (the Poisson expectation of powers)
If X is a Poisson random variable then from the definition of the expectation we have that
E[Xn
] =
∞
X
i=0
in
e−λ λi
i!
= e−λ
∞
X
i=0
in
λn
i!
e−λ
=
∞
X
i=1
in
λi
i!
,
since (assuming n 6= 0) when i = 0 the first term vanishes. Continuing our calculation we
can cancel a factor of i and find that
E[Xn
] = e−λ
∞
X
i=1
in−1
λi
(i − 1)!
= e−λ
∞
X
i=0
(i + 1)n−1
λi+1
i!
= λ
∞
X
i=0
(i + 1)n−1
e−λ
λi
i!
.
Now this sum can be recognized as the expectation of the variable (X + 1)n−1
so we see that
E[Xn
] = λE[(X + 1)n−1
] .
From the result we have
E[X3
] = λE[(X + 1)2
] = λ2
E[(X + 2)] = λ2
(λ + 2) = λ3
+ 2λ .
Problem 20 (flipping many coins)
Part (a): The total event of tossing all n coins is equivalent to to performing n, Bernoulli
trials with a probability of success equal on each trial of p. Thus the the total number
of successes is a Binomial random variable which we know can be approximated well as a
Poisson random variable i.e.
P{X = i} ≈
e−λ
λi
i!
,
so that P{X = 1} ≈ e−λ
λ, thus the reasoning is correct.
Part (b): This is false since P{X = 1} is the probability that only one head appears, when
we have no other information about the number of heads. The expression P{Y = 1|Y 0}
is the probability we have one head given that we know at least one head appears. Since
before the experiment begins we don’t know that we will have at least one head we can’t
condition on that fact.

Part (c): This is not true since P{X = 1} is the probability that any one but one of the n
trials result in a one, while P{Y = 0} is the probability that a fixed set of n − 1 flips results
in no heads. That is we don’t allow the set of n−1 of flips chosen to change. If we did, then
we have n choices for which flip lands heads giving ne−λ
and a probability p that the chosen
position does indeed give a head, giving a combined probability pne−λ
= λe−λ
which is the
correct answers.
Problem 21 (the birthdays of i and j)
Part (a): The events E3,4 and E1,2 would be independent since they consider different
people so
P(E3,4|E1,2) = P(E3,4) =
1
365
.
Part (b): Now E1,3 and E1,2 are still independent since if persons one and two have the same
birthday, this information tells us nothing about the coincidence of the birthdays of persons
one and three. Now since E1,2 means that person one and two share the same birthday (one
of the 365 days) then since person three must have this exact same day as his birthday we
see that P(E1,3|E1,2) = 1
365
.
Part (c): Now E2,3 and E1,2 ∩ E1,3 are not independent since the sets depend on all the
same people. Then since person one and two have the same birthday as persons one and
three, it follows that two and three have the same birthday. This means
P(E2,3|E1,2 ∩ E1,3) = 1 .
Since the probabilities Eij given other pairings can jump from 1
365
to 1 we can conclude that
these events are not independent. To be independent would require
P(E2,3|E1,2 ∩ E1,3) =
P(E2,3 ∩ E1,2 ∩ E1,3)
P(E1,2 ∩ E1,3)
=
P(E2,3)P(E1,2)P(E1,3)
P(E1,2)P(E1,3)
= P(E2,3) .
But the left hand side of the above expression is equal to 1 while the right hand side is equal
to 1
365
. As these two are not equal the events Eij are not independent.
Problem 25 (events registered with probability p)
We can solve this problem by conditioning on the number of true events (from the original
Poisson random variable N) that occur. We begin by letting M be the number of events
counted by our “filtered” Poisson random variable. Then we want to show that M is another
Poisson random variable with parameter λp. To do so consider the probability that M has
counted j “filtered events”, by conditioning on the number of observed events from the
original Poisson random variable. We find
P{M = j} =
∞
X
n=0
P{M = j|N = n}

e−λ
λn
n!

The conditional probability in this sum can be computed using the acceptance rule defined
above. For if we have n original events the number of derived events is a binomial random
variable with parameters (n, p). Specifically then we have
P{M = j|N = n} =




n
j

pj
(1 − p)n−j
j ≤ n
0 j n .
Putting this result into the original expression for P{M = j} we find that
P{M = j} =
∞
X
n=j

n
j

pj
(1 − p)n−j

e−λ
λn
n!

To evaluate this we note that

n
j

1
n!
= 1
j!(n−j)!
, so that the above simplifies as following
P{M = j} =
e−λ
pj
j!
∞
X
n=j
1
(n − j)!
(1 − p)n−j
λn
=
e−λ
pj
j!
∞
X
n=j
1
(n − j)!
(1 − p)n−j
(λ)j
λn−j
=
e−λ
(pλ)j
j!
∞
X
n=j
((1 − p)λ)n−j
(n − j)!
=
e−λ
(pλ)j
j!
∞
X
n=0
((1 − p)λ)n
n!
=
e−λ
(pλ)j
j!
e(1−p)λ
= e−pλ (pλ)j
j!
,
from which we can see M is a Poisson random variable with parameter λp as claimed.
Problem 26 (an integral expression for the CDF of a Poisson random variable)
We will begin by evaluating
R ∞
λ
e−x
xn
dx. To perform repeated integration by parts we
remember the integration by parts “formula” udv = uv − vdu, and in the following we will
let u be the polynomial in x and dv the exponential. To start this translates into letting
u = xn
and dv = e−x
, and we have
Z ∞
λ
e−x
xn
dx = −xn
e−x ∞
λ
+
Z ∞
λ
nxn−1
e−x
dx
= λn
e−λ
+ n
Z ∞
λ
xn−1
e−x
dx
= λn
e−λ
+ n

−xn−1
e−x ∞
λ
+
Z ∞
λ
(n − 1)xn−2
e−x
dx

= λn
e−λ
+ nλn−1
e−λ
+ n(n − 1)
Z ∞
λ
xn−2
e−x
dx .

Continuing to perform one more integration by parts (so that we can fully see the pattern)
we have that this last integral given by
Z ∞
λ
xn−2
e−x
dx = −xn−2
e−x ∞
λ
+
Z ∞
λ
(n − 2)xn−3
e−x
dx
= λn−2
e−λ
+ (n − 2)
Z ∞
λ
xn−3
e−x
dx .
Then we have for our total integral the following
Z ∞
λ
e−x
xn
dx = λn
e−λ
+ nλn−1
e−λ
+ n(n − 1)λn−2
e−λ
+ n(n − 1)(n − 2)
Z ∞
λ
xn−3
e−x
dx .
Using mathematical induction the total pattern can be seen as
Z ∞
λ
e−x
xn
dx = λn
e−λ
+ nλn−1
e−λ
+ n(n − 1)λn−2
e−λ
+ · · ·
+ n(n − 1)(n − 2) · · ·(n − k)
Z ∞
λ
xn−k
e−x
dx
= λn
e−λ
+ nλn−1
e−λ
+ n(n − 1)λn−2
e−λ
+ · · · + n!
Z ∞
λ
e−x
dx
= λn
e−λ
+ nλn−1
e−λ
+ n(n − 1)λn−2
e−λ
+ · · · + n!e−x
.
When we divide this sum by n! we find it is given by
λn
n!
e−λ
+
λn−1
(n − 1)!
e−λ
+
λn−2
(n − 2)!
e−λ
+ · · · + λe−λ
+ e−λ
or the left hand side of the expression given in the problem statement i.e.
n
X
i=0
e−λ
λi
i!
,
as we were to show.
Problem 29 (ratios of hypergeometric probabilities)
For a Hypergeometric random variable we have
P(i) =

m
i

N − m
n − i

N
n
for i = 0, 1, · · · , m .

So that the requested ratio is given by
P(k + 1)
P(k)
=

m
k + 1

N − m
n − k − 1

N
n
·

N
n

m
k

N − m
n − k

=

m
k + 1

m
k

N − m
n − k − 1

N − m
n − k

=
m!
(k+1)!(m−k−1)!
· (N−m)!
(n−k−1)!(N−m−n+k+1)!
m!
k!(m−k)!
· (N−m)!
(n−k)!(N−m−n+k)!
=
k!(m − k)!
(k + 1)!(m − k − 1)!
·
(n − k)!(N − m − n + k)!
(n − k − 1)!(N − m − n + k + 1)!
=
(m − k)(n − k)
(k + 1)(N − m − n + k + 1)
.
Problem 32 (repeated draws from a jar)
At each draw, the boy has a memory bank of numbers he has seen and X is the random
variable that determines the number of draws before he sees a number twice (once to fill his
memory and then viewed a second time). Then F(x) = P{X ≤ x}, now
F(1) = P{X ≤ 1} = 0
F(2) = P{X ≤ 2} =
1
n
,
since he has seen only one chip and therefore has a 1/n chance of redrawing this chip. Now
F(3) = P{X ≤ 2} + P{X = 3|X 2} ,
and P{X = 3|X 2} is the probability that the boy draws two chips and then his third
chip is a duplicate of one of the first two draws. We are assuming that X is not less than or
equal to two i.e. the first two draws result in unseen numbers. Thus P{X = 3|X 2} = 2
n
.
In the same way
P{X = i|X i − 1} =
i − 1
n
for 1 ≤ i ≤ n + 1 .
Therefore for 1 ≤ i ≤ n + 1 we have
F(i) =
i
X
k=1
k − 1
n
=
1
n
i
X
k=1
(k − 1) =
1
n
i
X
k=1
k − i
#
=
1
n

i(i + 1)
2
− i

=
i(i − 1)
2n
.

Problem 35 (double replacement of balls in our urn)
Let X be the selection number for the first selection of a blue ball. Recall that P{X i} =
1 − P{X ≤ i}. Now we can compute P{X = i} by induction. First we see that
P{X = 1} =
1
2
.
Next we have
P{X = 2} = P{X = 2|B1 = R}P{B1 = R} =

1
3

1
2

=
1
2 · 3
,
where we have conditioned on the fact that the first ball must be red. Continuing we see
that
P{X = 3} = P{X = 3|B1, B2 = R, R}P{RR} =

1
4

2
3

1
2

=
1
3 · 4
P{X = 4} = P{X = 4|B1, B2, B3 = R, R, R}P{RRR} =

1
5

1
2

2
3

3
4

=
1
4 · 5
.
By induction we conclude that
P{X = i} =
1
i(i + 1)
.
So that
P{X ≤ i} =
i
X
k=1
P{X = k} =
i
X
k=1
1
k(k + 1)
.
Now using partial fractions we see that the fraction we are summing is given by
1
k(k + 1)
=
1
k
−
1
k + 1
.
So that we see that our sum above is of the “telescoping” type and simplifies as
i
X
k=1
1
k(k + 1)
=
i
X
k=1

1
k
−
1
k + 1

=

1 −
1
2

+

1
2
−
1
3

+

1
3
−
1
4

+ · · · +

1
i − 1
−
1
i

+

1
i
−
1
i + 1

= 1 −
1
i + 1
.
Thus P{X i} = 1 − P{X ≤ i} = 1
i+1
for i ≥ 1.
Part (b): From our expression P{X ≤ i} = 1 − 1
i+1
, we have that P{X ≤ ∞} = 1 − 0 = 1,
thus with probability one X will be finite i.e. the blue ball is eventually chosen.
Part (c): Using the definition of expectation we have
E[X] =
∞
X
i=1
iP{X = i} =
∞
X
i=1
1
i + 1
= +∞ ,
thus the blue ball is eventually chosen but on average it is not chosen until very late.

Chapter 5 (Continuous Random Variables)
Chapter 5: Problems
Problem 1 (normalizing a continuous random variable)
Part (a): The integral of the f must evaluate to one, which requires
Z 1
−1
c(1 − x2
)dx = 2c
Z 1
0
(1 − x2
)dx
= 2c

x −
x3
3
1
0
= 2c

1 −
1
3

=
4c
3
.
For this to equal one, we must have c = 3
4
.
Part (b): The cumulative distribution is given by
F(x) =
Z x
−1
3
4
(1 − ξ2
)dξ
=
3
4

ξ −
ξ3
3
x
−1
=
3
4

x −
x3
3

+
1
2
for − 1 ≤ x ≤ 1 .
Problem 2 (how long can our system function?)
We must first evaluate the constant in our distribution function. Specifically to be a proba-
bility density we must have Z ∞
0
cxe−x/2
dx = 1 .
Integrating by parts we find that
Z ∞
0
cxe−x/2
dx = c

xe−x/2
(−1/2)
∞
0
−
1
(−1/2)
Z ∞
0
e−x/2
dx

= c

2
Z ∞
0
e−x/2
dx

= 2c
e−x/2
(−1/2)
∞
0
= −4c(0 − 1) = 4c .

So for this to equal one we must have c = 1/4. Then the probability that our system last at
least five months is given by
Z ∞
5
1
4
xe−x/2
dx =
1
4

xe−x/2
(−1/2)
∞
5
−
Z ∞
5
e−x/2
(−1/2)
dx

=
1
4

0 + 10e−5/2
+ 2
Z ∞
5
e−x/2
dx

= · · · =
7
2
e−5/2
.
Problem 3 (possible density functions)
Even with a value of C specified a problem with this function f is that it is negative for some
values of x. Specifically f will be zero when x(2 − x2
) = 0, which happens when x = 0 or
x = ±
√
2 = ±1.4142. With these zeros found we see that if x is less than
√
2 then x(2 − x2
)
is positive, however if x is greater than
√
2 (but still less than 5/2) the expression x(2−x2
) is
negative. Thus whatever the sign of c, f(x) will be negative for some region of the interval.
Since f cannot be negative this functional form cannot be a probability density function.
For the second function this f is zero when x(2 − x) = 0, which happens when x = 0 and
x = 2. Since 2 5/2 = 2.5. This f will also change signs regardless of the constant C as x
crosses the value 2. Since f takes on both positive and negative signed values it can’t be a
distribution function.
Problem 4 (the lifetime of electronics)
Part (a): The requested probability is given by
P{X 20} =
Z ∞
20
10
x2
dx =
1
2
.
Part (b): The requested cumulative distribution function is given by
F(x) =
Z ∞
10
10
ξ2
dξ =
10ξ−1
(−1)
x
10
= 1 −
10
x
for 10 ≤ x .
Part (c): To function for at least fifteen hours will happen with probability 1 − F(15) =
1 − (1 − 10
15
) = 2
3
. To have three of six such devices function for at least fifteen hours is given
by sums of binomial probability density functions. Specifically we have this probability given
by
6
X
k=3

6
k

2
3
k
1
3
6−k
,

which we recognized as the “complement” of the binomial cumulative distribution function.
To evaluate this we can use the Matlab command binocdf(2,6,2/3). See the Matlab
file chap 5 prob 4.m for these calculations and we find that the above equals 0.8999. In
performing this analysis we are assuming independence of the devices.
Problem 11 (picking a point on a line)
An interpretation of this statement is that a point is picked randomly on a line segment
of length L would be that the point “X” is selected from a uniform distribution over the
interval [0, L]. Then the question asks us to find
P

min(X, L − X)
max(X, L − X)

1
4

.
This probability can be evaluated by integrating over the appropriate region. Formally we
have the above equal to Z
E
p(x)dx
where p(x) is the uniform probability density for our problem, i.e. 1
L
and the set “E” is
x ∈ [0, L] and satisfying the inequality above, i.e.
min(x, L − x) ≤
1
4
max(x, L − x) .
Plotting the functions max(x, L−x), and min(x, L−x) in Figure 2, we see that the regions of
X where we should compute the integral above are restricted to the two ends of the segment.
Specifically, the integral above becomes,
Z l1
0
p(x)dx +
Z L
l2
p(x)dx .
since the region min(x, L − x) 1
4
max(x, L − x) in satisfied in the region [0, l1] and [l2, L]
only. Here l1 is the solution to
min(x, L − x) =
1
4
max(x, L − x) when x L − x ,
i.e. we need to solve
x =
1
4
(L − x)
which has as its solution x = L
5
. For l2 we must solve
min(x, L − x) =
1
4
max(x, L − x) when L − x x ,
i.e. we need to solve
L − x =
1
4
x ,
which has as its solution x = 4
5
L. With these two limits we have for our probability
Z L
5
0
1
L
dx +
Z L
4
5
L
1
L
dx =
1
5
+
1
5
=
2
5
.

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
0
0.05
0.1
0.15
0.2
0.25
0.3
0.35
0.4
0.45
0.5
x (unitless)
min(x,L−x)
max(x,L−x)/4
Figure 2: A graphical view of the region of x’s over which the integral for this problem
should be computed.
Problem 15 (some normal probabilities)
In general to solve this problem we will convert each probability to a corresponding one
involving unit normal random variables and then compute this second probability using the
cumulative distribution function Φ(·).
Part (a): We find
P{X 5} = P{
X − 10
6

5 − 10
6
}
= P{Z −0.833} = 1 − P{Z −0.833} = Φ(−0.833) .
Part (b): P{4 X 16} = P{4−10
6
X−10
6
16−10
6
} = P{−1 Z +1} = Φ(1) −
Φ(−1).
Part (c): P{X 8} = P{X−10
6
8−10
6
} = P{Z −0.333} = Φ(−0.333).
Part (d): P{X 20} = Φ(20−10
6
) = Φ(1.66), following the same steps as in Part (a).
Part (e): P{X 16} = P{X−10
6
16−10
6
} = P{Z 1} = 1 − P{Z 1} = 1 − Φ(1).
Problem 16 (annual rainfall)
The probability that we have over 50 inches of rain in one year is
P{X 50} = P{
X − 40
4

50 − 40
4
} = P{Z 2.5} = 1 − P{Z 2.5} = 1 − Φ(2.5) .
If we assume that the probabilities each year are uncorrelated and also assuming that the
probability this event happens at year n is a geometric random variable, the probability that

it takes over ten years before this event happens is one minus the probability that it takes
1, 2, 3, · · · , 10 years for this event to occur. That is if E is the event that over ten years pass
before this event happens, than
P(E) = 1 −
10
X
i=1
p(1 − p)i
.
Since
PN
i=1 ai
= a−aN+1
1−a
we have that
P(E) = 1 − p

(1 − p) − (1 − p)11
1 − (1 − p)

= 1 − ((1 − p) − (1 − p)11
) = p + (1 − p)11
.
When we put p = 1 − Φ(2.5) we get the desired probability.
Problem 17 (the expected number of points scored)
We desire to calculate E[P(D)], where P(D) is the points scored when the distance to the
target is D. This becomes
E[P(D)] =
Z 10
0
P(D)f(D)dD
=
1
10
Z 10
0
P(D)dD
=
1
10
Z 1
0
10dD +
Z 3
1
5dD +
Z 5
3
3dD +
Z 10
5
0dD

=
1
10
(10 + 5(2) + 3(2)) =
26
10
= 2.6 .
Problem 18 (variable limits on a normal random variable)
Since X is a normal random variable we can evaluate the given probability P{X 9} as
P{X 9} = P{
X − 5
σ

9 − 5
σ
}
= P

Z
4
σ

= 1 − P

Z
4
σ

= 1 − Φ(
4
σ
) = 0.2 ,
so solving for Φ(4/σ) we have that Φ(4/σ) = 0.8, which can be inverted by using the Matlab
command norminv and we calculate that
4
σ
= Φ−1
(0.8) = 0.8416 .
which then implies that σ = 4.7527, so Var(X) = σ2
≈ 22.58.

Problem 19 (more limits on normal random variables)
Since X is a normal random variable we can evaluate the given probability P{X c} as
P{X c} = P{
X − 12
2

c − 12
2
}
= P

Z
c − 12
2

= 1 − P

Z
c − 12
2

= 1 − Φ(
c − 12
2
) = 0.1 ,
so solving for Φ(c−12
2
we have that Φ(c−12
2
) = 0.9, which can be inverted by using the Matlab
command norminv and we calculate that
c − 12
2
= Φ−1
(0.9) = 1.28 ,
which then implies that c = 14.56.
Problem 20 (the expected number of people in favor of a proposition)
Now the number of people who favor the proposed rise in taxes is a binomial random variable
with parameters (n, p) = (100, 0.65). Using the normal approximation to the binomial, we
have a normal with a mean of np = 100(0.65) = 65, and a variance of σ2
= np(1 − p) =
100(0.65)(0.35) = 22.75, so the probabilities desired are given as
Part (a):
P{N ≥ 50} = P{N 49.5}
= P

N − 65
√
22.75

49.5 − 65
4.76

= P {Z −3.249}
= 1 − Φ(−3.249) = 1 − (1 − Φ(3.249)) = Φ(3.249) .
Where in the first equality we have used the “continuity approximation”. Using the Matlab
command normcdf(x) to evaluate the function Φ(x) we have the above equal to ≈ 0.9994.
Part (b):
P{60 ≤ N ≤ 70} = P{59.5 N 70.5}
= P

59.5 − 65
√
22.75
Z
70.5 − 65
√
22.75

= P {−1.155 Z 1.155}
= Φ(1.155) − Φ(−1.155) ≈ 0.7519 .

Part (c):
P{N 75} = P{N 74.5}
= P

Z
74.5 − 65
4.76

= P {Z 1.99}
= Φ(1.99) ≈ 0.9767 .
Problem 21 (percentage of men with height greater than six feet two inches)
We desire to compute P{X 6 · 12 + 2}, where X is the random variable expressing height
(measured in inches) of a 25-year old man. This probability can be computed by converting
to the standard normal in the usual way. We have
P{X 6 · 12 + 2} = P

X − 71
√
6.25

3
√
6.25

= P

Z
3
√
6.25

= 1 − P

Z
3
√
6.25

= 1 − Φ(1.2) ≈ 0.1151 .
For the second part of this problem we are looking for
P{X 6 · 12 + 5|X 6 · 12} .
Again this can be computed by converting to a standard normal, after first considering the
joint density. We have
P{X 6 · 12 + 5|X 6 · 12} =
P {X 77, X 72}
P{X 72}
=
P {X 77}
P{X 72}
=
1 − P
n
Z 6
√
6.25
o
1 − P
n
Z 1
√
6.25
o
=
1 − Φ( 6
√
6.25
)
1 − Φ( 1
√
6.25
)
≈ 0.0238 .
Some of the calculations for this problem can be found in the file chap 5 prob 21.m.

Problem 22 (number of defective products)
Part (a): Lets calculate the percentage that are acceptable if we let the variable X be the
width of our normally distributed slot this percentage is given by
P{0.895 X 0.905} = P{X 0.905} − P{X 0.895} .
Each of these individual cumulative probabilities can be calculated by transforming to the
standard normal, in the usual way. We have that the above is equal to (since the population
mean is 0.9 and the population standard deviation is 0.003)
P

X − 0.9
0.003

0.905 − 0.9
0.003

− P

X − 0.9
0.003

0.895 − 0.9
0.003

= Φ (1.667) − Φ (−1.667) = 0.904 .
So that the probability (or percentage) of defective forgings is one minus this number (times
100 to convert to percentages). This is 0.095 × 100 = 9.5.
Part (b): This question is asking to find the value of σ such that
P{0.895 X 0.905} =
99
100
.
Since these limits on X are symmetric about X = 0.9 we can simplify this probability by
using
P{0.895 X 0.905} = 1 − 2P{X 0.895} = 1 − 2P

X − 0.9
σ

0.905 − 0.9
σ

We thus have to solve for σ in
1 − 2Φ(
−0.005
σ
) = 0.99
or inverting the Φ function and solving for σ we have
σ =
−0.005
Φ−1(0.005)
.
Using the Matlab command norminv to evaluate the above we have σ = 0.0019. See the
Matlab file chap 5 prob 22.m for these calculations.
Problem 23 (probabilities on the number of 5’s to appear)
The probability that one six occurs is p = 1/6 so the total number of sixes rolled is a
binomial random variable. We can approximate this density as a Gaussian with a mean
given by np = 1000
6
≈ 166.6 and a variance of σ2
= np(1 − p) = 138.8. Then the desired
probabilities are
P{150 ≤ N ≤ 200} = P{149.5 N 200.5}
= P {−1.45 Z 2.877}
= Φ(2.87) − Φ(−1.45) ≈ 0.9253 .

If we are told that a six appears two hundred times then the probability that a five will appear
on the other rolls is 1/5 and it must appear on one of the 1000−200 = 800 other rolls. Thus
we can approximate the binomial random variable (with parameters (n, p) = (800, 1/5)) with
a normal with mean np = 800
5
= 160 and variance σ2
= np(1 − p) = 128. So the requested
probability is
P{N 500} = P{N 149.5}
= P

Z
149.5 − 160
√
128

= P {Z −0.928}
= Φ(−0.928) ≈ 0.1767 .
Problem 24 (probability of enough long living batteries)
If each chips lifetime is denoted by the random variable X (assumed Gaussian with the
given mean and variance), then each chip will have a lifetime less than 1.8 106
hours with
probability given by
P{X 1.8 106
} = P

X − 1.4 106
3 105

(1.8 − 1.4) 106
3 105

= P

Z
4
3

= Φ(4/3) ≈ 0.9088 .
With this probability, the number N, in a batch of 100 that will have a lifetime less than
1.8 106
is a binomial random variable with parameters (n, p) = (100, 0.9088). Therefore, the
probability that a batch will contain at least 20 is given by
P{N ≥ 20} =
100
X
n=20

100
n

(0.908)n
(1 − 0.908)100−n
.
Rather than evaluate this exactly we can approximate this binomial random variable N
with a Gaussian random variable with a mean given by µ = np = 100(0.908) = 90.87, and a
variance given by σ2
= np(1 − p) = 8.28 (equivalently σ = 2.87). Then the probability that
a given batch of 100 has at least 20 that have lifetime less than 1.8 106
hours is given by
P{N ≥ 20} = P{N ≥ 19.5}
= P

N − 90.87
2.87
≥
19.5 − 90.87
2.87

≈ P {Z ≥ −24.9}
= 1 − P {Z ≤ −24.9}
= 1 − Φ(−24.9) ≈ 1 .
Where in the first line above we have used the continuity correction required when we
approximate a discrete density by a continuous one, and in the third line above we use our
Gaussian approximation to the binomial distribution.

Problem 25 (the probability of enough acceptable items)
The number N of acceptable items is a binomial random variable so we can approximate it
with a Gaussian with mean µ = pn = 150(0.95) = 142.5, and a variance of σ2
= np(1 − p) =
7.125. From the variance we have a standard deviation of σ ≈ 2.669. Thus the desired
probability is given by
P{N ≥ 140} = P{N ≥ 139.5}
= P

N − 142.5
2.669
≥
139.5 − 142.5
2.669

≈ P {Z ≥ −1.127}
= 1 − P {Z ≤ −1.127}
= 1 − Φ(−1.127) ≈ 0.8701 .
Gaussian approximation to the binomial distribution. We note that we solved this problem in
terms of the number of items that are acceptable. An equivalent formulation could easily be
done in terms of the number that are unacceptable by using the complementary probability
q ≡ 1 − p = 1 − 0.95 = 0.05.
Problem 26 (calculating the probability of error)
Let N be the random variable that represents the number of heads that result when we flip
our coin 1000 times. Then N is distributed as binomial random variable with a probability
of success p that depends on whether we are considering the biased or unbiased (fair) coin.
It the coin is actually fair we will make an error in our assessment of its type if N is greater
than 525 according to the statement of this problem. Thus the probability that we reach a
false conclusion is given by
P{N ≥ 525} .
To compute this probability we will use the normal approximation to the binomial distribu-
tion. In this case the normal to use to approximate this binomial distribution has a mean
given by µ = np = 1000(0.5) = 500 and a variance given by σ2
= np(1 − p) = 250 since we
know we are looking at the fair coin where p = 0.5. To evaluate this probability we have
P{N ≥ 525} = P{N ≥ 524.5}
= P

N − 500
√
250
≥
524.5 − 500
√
250

≈ P {Z ≥ 1.54}
= 1 − P {Z ≤ 1.54}
= 1 − Φ(1.54) ≈ 0.0606 .
approximate a discrete density by a continuous one, and in the third line above we use

our Gaussian approximation to the binomial distribution. In the case where the coin is
actually biased our probability of obtaining a head becomes p = 0.55 and we will reach a
false conclusion in this case when
P{N 525} .
To compute this probability we will use the normal approximation to the binomial distribu-
tion. In this case the normal to use to approximate this binomial distribution has a mean
given by µ = np = 1000(0.55) = 550 and a variance given by σ2
= np(1 − p) = 247.5. To
evaluate this probability we have
P{N 525} = P{N 524.5}
= P

N − 550
√
247.5

524.5 − 550
√
247.5

≈ P {Z −1.62}
= Φ(−1.62) ≈ 0.0525 .
Gaussian approximation to the binomial distribution.
Problem 27 (fair coins)
Now P{N = 5800} = P{5799.5 ≤ N ≤ 5800.5}, by the continuity approximation. The
second probability can be approximated by a normal with mean np = 104
(0.5) = 5000 and
variance np(1 − p) = 2500, so that the above becomes
P{
5799.5 − 5000
√
2500
≤ Z ≤
5800.5 − 5000
√
2500
} = Φ

5800.5 − 5000
√
2500

− Φ

5799.5 − 5000
√
2500

= 1 − 1 = 0 ,
so there is effectively no probability this could have happened by chance.
Problem 28 (the number of left handed students)
The number of students that are left handed (denoted by N) is a Binomial random variable
with parameters (n, p) = (200, 0.12). From the normal approximation to the binomial we
can approximate this distribution with a Gaussian with mean µ = pn = 200(0.12) = 24, and
a variance of σ2
= np(1 − p) = 21.120. From the variance we have a standard deviation of
σ ≈ 4.59. Thus the desired probability is given by
P{N 20} = P{N 19.5}
= P

N − 24
4.59

19.5 − 24
4.59

≈ P {Z −0.9792}
= 1 − P {Z ≤ −0.9792}
= 1 − Φ(−0.9792) ≈ 0.8363 .

Where in the second line above we have used the continuity correction that improves our
accuracy when we approximate a discrete density by a continuous one, and in the third line
above we use our Gaussian approximation to the binomial distribution. These calculations
can be find in the file chap 5 prob 28.m.
Problem 29 (a simple model of stock movement)
If we count each time the stock rises in value as a “success”, we see that the movement of the
stock for one timestep is a Bernoulli random variable with parameter p. So after n timesteps
the number of rises is a binomial random variable with parameters (n, p). The price of the
security after n timesteps where we have k “successes” will then be given by suk
dn−k
. The
probability we are looking for then is given by
P{suk
dn−k
≥ 1.3 s} = P{uk
dn−k
≥ 1.3}
= P
u
d
k
≥
1.3
dn

= P

k ≥
ln(1.3
dn )
ln(u
d
)

= P

k ≥
ln(1.3) − n ln(d)
ln(u) − ln(d)

.
Using the numbers given in the problem i.e. d = 0.99 u = 1.012, and n = 1000, we have that
ln(1.3) − n ln(d)
ln(u) − ln(d)
≈ 469.2 .
To approximate the above probability we can use the Gaussian approximation to the binomial
distribution, which would have a mean given by np = 0.52(1000) = 520 and a variance given
by np(1 − p) = 249.6, so using this approximation the above probability then becomes
P{k ≥ 469.2} = P{k ≥ 470}
= P{k 469.5}
= P{Z
469.5 − 520
15.7
}
= P{Z −3.21}
= 1 − P{Z −3.21}
= 1 − Φ(−3.23) ≈ 0.9994 .
Problem 30 (priors on the type of region)
Let E be the event that we make an error in our classification of the given pixel. Then we
can make an error in two symmetric ways. The first is that we classify the pixel as black
when it should be classified as white. The second is where we classify the pixel as white

when it should be black. Thus we can compute P(E) by conditioning on the true type of
the pixel i.e. whether it is B (for black) or W (for white). We have
P(E) = P(E|B)P(B) + P(E|W)P(W) .
Since we are told that the prior probability that the pixel is black is given by α, the prior
probability that the pixel is W is then given by 1 − α and the above becomes
P(E) = P(E|B)α + P(E|W)(1 − α) .
The problem then asks for the value of α such that the error in making each type of error is
the same, we desire to pick α such that
P(E|B)α = P(E|W)(1 − α) ,
or upon solving for α we find that
α =
P(E|W)
P(E|W) + P(E|B)
.
We now need to evaluate P(E|W) and P(E|B). Now P(E|W) is the probability that we
classify this pixel as black given that it is white. If we classify the pixel with a value of 5 as
black, then all points with pixel value greater than 5 would also be classified as black and
P(E|W) is then given by
P(E|W) =
Z ∞
5
N (x; 4, 4)dx =
Z ∞
(5−4)/2
N (z; 0, 1)dz = 1 − Φ(1/2) = 0.3085 .
Where N (x; µ, σ2
) is an expression for the normal probability density function with mean µ
and variance σ2
. In the same way we have that
P(E|B) =
Z 5
−∞
N (x; 6, 9)dx =
Z (5−6)/3
−∞
N (z; 0, 1)dz = Φ(−1/3) = 0.3694 .
Thus with these two expressions α becomes
α =
1 − Φ(1/2)
(1 − Φ(1/2)) + Φ(−1/3)
= 0.4551 .
Problem 31 (the optimal location of a fire station)
Part (a): If x (the location of the fire) is uniformly distributed in [0, A) then we would like
to select a (the location of the fire station) such that
F(a) ≡ E[ |X − a| ] ,

is a minimum. We will compute this by breaking the integral involved in the definition of
the expectation into regions where x − a is negative and positive. We find that
E[ |X − a| ] =
Z A
0
|x − a|
1
A
dx
= −
1
A
Z a
0
(x − a)dx +
1
A
Z A
a
(x − a)dx
= −
1
A
(x − a)2
2
a
0
+
1
A
(x − a)2
2
A
a
= −
1
A

0 −
a2
2

+
1
A

(A − a)2
2
− 0

=
a2
2A
+
(A − a)2
2A
.
To find the a that minimizes this we compute F′
(a) and set this equal to zero. Taking the
derivative and setting this equal to zero we find that
F′
(a) =
a
A
+
2(A − a)(−1)
2A
= 0 .
Which gives a solution a∗
given by a∗
= A
2
. A second derivative of our function F shows
that F′′
(a) = 2
A
0 showing that the point a∗
= A/2 is indeed a minimum.
Part (b): The problem formulation is the same as in part (a) but since the distribution of
the location of fires is now an exponential we now want to minimize
F(a) ≡ E[ |X − a| ] =
Z ∞
0
|x − a|λe−λx
dx .
We will compute this by breaking the integral involved in the definition of the expectation
into regions where x − a is negative and positive. We find that
E[ |X − a| ] =
Z ∞
0
|x − a|λe−λx
dx
= −
Z a
0
(x − a)λe−λx
dx +
Z ∞
a
(x − a)λe−λx
dx
= −λ

(x − a)
−λ
e−λx
a
0
+
1
λ
Z a
0
e−λx
dx

+ λ

(x − a)
−λ
e−λx
∞
a
+
1
λ
Z ∞
a
e−λx
dx

= −λ

−a
λ
−
1
λ2
e−λx
a
0

+ λ

0 −
1
λ2
e−λx
∞
a

= a +
1
λ
(e−λa
− 1) −
1
λ
(−e−λa
)
= a +
1 + 2e−λa
λ
.
To find the a that minimizes this we compute F′
(a) and set this equal to zero. Taking the
derivative we find that
F′
(a) = 1 − 2e−λa
= 0 .

Which gives a solution a∗
given by a∗
= ln(2)
λ
. A second derivative of our function F shows
that F′′
(a) = 2λe−λa
0 showing that the point a∗
= ln(2)
λ
is indeed a minimum.
Problem 32 (probability of repair times)
Part (a): We desire to compute P{T 2} which is given by
P{T 2} =
Z ∞
2
1
2
e−1/2 t
dt .
To evaluate this let v = t
2
, giving dv = dt
2
, from which the above becomes
P{T 2} =
Z ∞
1
e−v
dv = −e−v ∞
1
= e−1
.
Part (b): The probability we are after is given by P{T 10|T 9} which equals P{T
10 − 9} = P{T 1} by the memoryless property of the exponential distribution. This is
given by
P{T 1} = 1 − P{T 1} = 1 − (1 − e−1/2
) = e−1/2
.
Problem 33 (a functioning radio?)
Because of the memoryless property of the exponential distribution the fact that the ratio
is used is irrelevant. The probability requested is then
P{T 8 + t|T t} = P{T 8} = 1 − P{T 8} = 1 − (1 − e− 1
8
(8)
) = e−1
.
Problem 34 (getting additional miles from a car)
Since the exponential random variable has no memory, the fact that the car has been driven
100,000 miles makes no difference. The probability we are looking for is
P{T 20000} = 1 − P{T 20000} = 1 − (1 − e− 1
20
(20)
) = e−1
.
If the lifetime distribution is not exponential but is uniform over (0, 40) then the desired
probability is given by
P{Tthous 30|Tthous 20} =
P{T 30}
P{T 10}
=
(1/4)
(3/4)
=
1
3
.
Here Tthous is the distance driven in thousands of miles.

Problem 35 (lung cancer hazard rates)
Given a hazard rate of λ(t) then from Example 5f we see that if E is the event that an “A”
year old reaches age B is given by
P(E) = exp{−
Z B
A
λ(t)dt} ,
so for this problem since our person is age forty we want
exp{−
Z 50
40
λ(t)dt} .
First lets calculate
Z B
40
λ(t)dt =
Z B
40
(0.027 + 0.00025(t − 40)2
)dt = 0.027(B − 40) + 0.00025
(B − 40)3
3
.
When B = 50 this number is 0.353, so our survival probability is exp(−0.353) = 0.702.
While if B = 60 this number is 1.2 so our survival probability is 0.299.
Problem 36 (surviving with a given hazard rate)
From Example 5f the probability that our object survives to age B (from zero) is given by
exp{−
Z B
0
λ(t)dt} = exp{−
Z B
0
t3
dt} = exp{−
t4
4
B
0
} = exp{−
B4
4
} .
Part (a): The desired probability is when B = 2, which when put in the above gives 0.0183.
Part (b): The required probability is
exp{−
Z 1.4
0.4
λ(t)dt} = exp{−
1
4
(1.4)4
− (0.4)4

} = 0.3851 .
Part (c): This probability is
exp{−
Z 2
1
λ(t)dt} = exp{−
1
4
24
− 1

} = 0.0235 .
Problem 37 (uniform probabilities)
Part (a): The desired probability is
P{|x|
1
2
} =
Z
|x| 1
2
f(x)dx =
1
2
Z
|x| 1
2
dx =
2
2
Z 1
1/2
dx = 1 −
1
2
=
1
2
.

Part (b): Define Y = |x| and consider the distribution for Y i.e.
FY (a) = P{Y ≤ a} = P{|x| ≤ a} = 2P{0 ≤ x ≤ a}
= 2
Z a
0
f(x)dx = 2
Z a
0
1
2
dx = a ,
for 0 ≤ a ≤ 1 and is zero elsewhere. Then fY (y) = dFY (a)
da
= 1 and Y (over a smaller range
than X) is also a uniform distribution.
Problem 38 (the probability of roots)
The roots of the equation 4x2
+ 4Y x + Y + 2 = 0 are given by
x =
−4Y ±
p
16Y 2 − 4(4)(y + 2)
2(4)
=
−Y ±
√
Y 2 − Y − 2
2
,
which will be real if and only if Y 2
− Y − 2 0. Noticing that this expression factors into
(Y − 2)(Y + 1) 0, we see that this expression will be positive when Y 2. Thus the
probability we seek is given (if E is the event that x is real) by
P(E) = P{Y 2} =
Z 5
2
f(y)dy =
Z 5
2
1
5
dy =
1
5
(3) =
3
5
.
Problem 39 (the variable Y = log(X))
We begin by computing the cumulative distribution function for Y i.e. we have FY (a)
FY (a) = P{Y ≤ a} = P{log(X) ≤ a} = P{X ≤ ea
} = FX(ea
) .
Now since X is an exponential random variable it has a cumulative distribution function
given by FX (a) = 1 − e−λa
= 1 − e−a
, so that the above then becomes
FY (a) = 1 − e−ea
.
The probability density function for Y is then the derivative of this expression with respect
to a or
fY (a) =
d
da
FY (a) =
d
da
(1 − e−ea
) = −e−ea
(−ea
) = ea
eea
= ea+ea
.
Problem 40 (the variable Y = eX
)
We begin by computing the cumulative distribution function for Y i.e.
FY (a) = P{Y ≤ a} = P{eX
≤ a} = P{X ≤ log(a)} = FX(log(a)) .

Now Since X is uniform distributed its cumulative distribution function is linear i.e. FX(a) =
a, so FY (a) = log(a) and the density function for Y is given by
fY (a) =
d
da
FY (a) =
d
da
(log(a)) =
1
a
for 1 ≤ a ≤ e .
Problem 41 (the variable R = A sin(θ))
We begin by computing the distribution function of the random variable R given by
FR(a) = P{R ≤ a} = P{A sin(θ) ≤ a} = P{sin(θ) ≤
a
A
} .
To compute this we can plotting the function sin(θ) for −π
2
≤ θ ≤ π
2
and see that the
line a
A
crosses this function sin(θ) at various points. These points determine the integration
region of θ required to determined the probability above. We have that the probability above
becomes Z sin−1( a
A
)
− π
2
fθ(θ)dθ =
Z sin−1( a
A
)
− π
2

1
π

dθ =
1
π

sin−1
(
a
A
) +
π
2

.
From which we have a density function given by
fR(a) =
d
da
FR(a) =
d
da

1
π
sin−1
a
A

+
1
2

=
1
π
1
q
1 − a
A
2

1
A

=
A
π
√
A2 − a2
.
for |a| ≤ |A|.
Problem 1
Since f(x) is a probability density if must integrate to one
R ∞
−∞
f(x)dx = 1. In the case here
using integration by parts this becomes
Z ∞
0
ax2
e−bx2
dx = a
xe−bx2
(−b)
∞
0
− a
Z ∞
0
e−bx2
(−b)
!
dx
= 0 − 0 +
a
b
Z ∞
0
e−bx2
dx .
To evaluate this integral let v = bx2
so that dv = 2bxdx, x = ±
pv
b
, dv = 2b
pv
b
dx, which
gives
dx =

b1/2
2b

v−1/2
dv =
v−1/2
2
√
b
dv ,

and our integral above becomes
1 =
a
b
1
2
√
b
Z ∞
0
v−1/2
e−v
dv .
Now the integral remaining can be seen to be
Z ∞
0
v−1/2
e−v
dv =
Z ∞
0
v1/2−1
e−v
dv ≡ Γ(1/2) =
√
π .
Using this we have
1 =
a
2b3/2
√
π .
Thus a = 2b3/2
√
π
is the relationship between a and b.
Problem 2
Consider the first integral Z ∞
0
P{Y −y}dy .
Using integration by parts with the substitutions u(y) = P{Y −y} and dv(y) = dy.
Then the standard integration by parts differential formula becomes u(y)dv(y) = u(y)v(y)−
v(y)du(y), and the above becomes
Z ∞
0
P{Y −y}dy = P{Y −y}y|∞
0 −
Z ∞
0
y
d
dy
P{Y −y}dy
= 0 −
Z ∞
0
yfY (−y)(−1)dy =
Z ∞
0
yfY (−y)dy .
To evaluate this let v = −y and the above integral becomes
Z −∞
0
−vfY (v)(−dv) = −
Z 0
−∞
vfY (v)dv .
In the same way, for the second integral we have
Z ∞
0
P{Y y}dy = yP{Y y}|∞
0 −
Z ∞
0
y
d
dy
P{Y y}dy
= −
Z ∞
0
y
d
dy
P{Y y}dy = −
Z ∞
0
y
d
dy
(1 − P{Y y})dy .
Where we have used the fact that P{Y y} = 1 − P{Y y}. Taking the above derivative
we get the above integral equal to
Z ∞
0
yfY (y)dy .
Now then
E[Y ] =
Z ∞
−∞
yfY (y)dy =
Z 0
−∞
yfY (y)dy +
Z ∞
0
yfY (y)dy .
Using the above two integral derived above we get this equal to
−
Z ∞
0
P{Y −y}dy +
Z ∞
0
P{Y y}dy .

Problem 3
From Problem 2 above we have that
E[g(X)] =
Z ∞
0
P{g(X) y}dy −
Z ∞
0
P{g(X) −y}dy
=
Z ∞
0
Z
x:g(x)y
f(x)dxdy −
Z ∞
0
Z
x:g(x)−y
f(x)dxdy .
Now as in the proof of proposition 2.1 we can change the order of each integration with the
following manipulations
E[g(X)] =
Z
x:g(x)0
Z g(x)
0
dyf(x)dx −
Z
x:g(x)0
Z −g(x)
0
dyf(x)dx
=
Z
x:g(x)0
f(x)g(x)dx +
Z
x:g(x)0
f(x)g(x)dx =
Z ∞
−∞
g(x)f(x)dx .
Problem 4
Corollary 2.1 is E[ax+b] = aE[x]+b and can be proven using the definition of the expectation
E[ax + b] =
Z ∞
−∞
(ax + b)f(x)dx = a
Z ∞
−∞
xf(x)dx + b
Z ∞
−∞
f(x)dx = aE[X] + b .
Problem 5
Compute E[Xn
] by using the given identity i.e.
E[Xn
] =
Z ∞
0
P{Xn
t}dt .
To evaluate this let t = xn
then dt = nxn−1
and we have
E[Xn
] =
Z ∞
0
P{Xn
xn
}nxn−1
dx =
Z ∞
0
nxn−1
P{X x}dx ,
Using the fact that P{Xn
xn
} = P{X x} when X is a non-negative random variable.
Problem 6
We want a collection of events Ea with 0 a 1 such that P(Ea) = 1 but P(∩aEa) = 0.
Let X be a uniform random variable over (0, 1) and let the event Ea be the event that X 6= a.
Since X is a continuous random variable P(Ea) = 1, since the probability X = a (exactly)
must be zero. Now the event ∩aEa is the event that X is not any of the elements from (0, 1).
Since X must be at least one of these elements the probability of this intersection must be
zero i.e. P(∩aEa) = 0.

Problem 7
We have
Std(aX + b) =
p
Var(aX + b) =
p
Var(aX)
=
p
a2Var(X) = |a|σ .
Problem 8
We know that P{0 ≤ X ≤ c} = 1 and we want to show Var(X) ≤ c2
4
. Following the hint we
have
E[X2
] =
Z c
0
x2
fX(x)dx ≤ c
Z c
0
xfX (x)dx = cE[X] ,
since x ≤ c for all x ∈ [0, c]. Now
Var(X) = E[X2
] − E[X]2
≤ cE[X] − E[X]2
= c2

E[X]
c
−
E[X]2
c2

.
Define α = E[X]
c
and we have
Var(X) ≤ c2
(α − α2
) = c2
α(1 − α) .
Now to select the value of α that maximizes the expression α(1 − α) for α in the range
0 ≤ α ≤ 1 we take the derivative with respect to α, set this expression equal to zero, and
solve for α. The derivative gives
c2
(1 − 2α) = 0 ,
which gives α = 1
2
. A second derivative gives −2c2
which is negative, showing that α = 1
2
is
a maximum. The value at this maximum is
1
2

1 −
1
2

=
1
4
,
and so we have that Var(X) ≤ c2
4
.
Problem 9
Part (a): P{Z x} =
R ∞
x
1
√
2π
e−z2/2
dz. Let v = −z so that dv = −dz and we have the
above given by
Z −∞
−x
1
√
2π
e−v2/2
(−dv) =
Z −x
−∞
1
√
2π
e−z2/2
dz = P{Z −x} .

Part (b): We find
P{|Z| x} =
Z −x
−∞
1
√
2π
e−z2/2
dz +
Z ∞
x
1
√
2π
e−z2/2
dz
=
Z x
∞
1
√
2π
e−z2/2
(−dz) +
Z ∞
x
1
√
2π
e−z2/2
dz
= 2
Z ∞
x
1
√
2π
e−z2/2
dz = 2P{Z x} .
Part (c): We find
P{|Z| ≤ x} =
Z x
−x
1
√
2π
e−z2/2
dz
=
Z x
−∞
1
√
2π
e−z2/2
dz −
Z −x
−∞
1
√
2π
e−z2/2
dz
= P{Z x} +
Z x
∞
1
√
2π
e−z2/2
dz
= P{Z x} −
Z ∞
x
1
√
2π
e−z2/2
dz
= P{Z x} −
Z ∞
x
1
√
2π
e−z2/2
dz −
Z ∞
−∞
1
√
2π
e−z2/2
dz +
Z ∞
−∞
1
√
2π
e−z2/2
dz
= P{Z x} +
Z x
−∞
1
√
2π
e−z2/2
dz − 1
= 2P{Z x} − 1 .
Problem 10 (points of inflection of the Gaussian
We are told that f(x) = 1
√
2πσ
exp
n
−1
2
(x−µ)2
σ2
o
. And points of inflection are given by f′′
(x) =
0. To calculate f′′
(x) we need f′
(x). We find
f′
(x) ≈ −

x − µ
σ2

exp

−
1
2
(x − µ)2
σ2

.
So that the second derivative is given by
f′′
(x) ≈ −
1
σ2
exp

−
1
2
(x − µ)2
σ2

+

(x − µ)2
σ2

exp

−
1
2
(x − µ)2
σ2

.
Setting f′′
(x) equal to zero we find that this requires x satisfy
exp{−
1
2
(x − µ)2
σ2
}

−1 +
(x − µ)2
σ2

= 0, .
or (x − µ)2
= σ2
. Which has as solutions x = µ ± σ.

Problem 11 (E[X2
] of an exponential random variable)
Theoretical Exercise number 5 states that
E[Xn
] =
Z ∞
0
nxn−1
P{X x}dx .
For an exponential random variable we have our cumulative distribution function given by
P{X ≤ x} = 1 − e−λx
.
so that P{X x} = e−λx
, and thus our expectation becomes
E[Xn
] =
Z ∞
0
nxn−1
e−λx
dx
Now if n = 2 we find that this expression becomes in this case
E[X2
] =
Z ∞
0
2xe−λx
dx
= 2
Z ∞
0
xe−λx
dx
= 2

xe−λx
−λ
∞
0
+
1
λ
Z ∞
0
e−λx
dx

=
2
λ

e−λx
−λ
∞
0

=
2
λ2
,
as expected.
Problem 12 (the median of a continuous random variable)
Part (a): When X is uniformly distributed over (a, b) the median is the value m that solves
Z m
a
dx
b − a
=
Z b
m
dx
b − a
.
Integrating both sides gives that m − a = b − m, which has a solution of m = a+b
2
.
Part (b): When X is a normal random variable with parameters (µ, σ2
) we find that m
must satisfy
Z m
−∞
1
√
2πσ
exp{−
1
2
(x − µ)2
σ2
}dx =
Z ∞
m
1
√
2πσ
exp{−
1
2
(x − µ)2
σ2
}dx .
To evaluate the integral on both sides of this expression we let v = x−µ
σ
, so that dv = dx
σ
and
each integral becomes
Z m−µ
σ
−∞
1
√
2π
exp{−
v2
2
}dv =
Z ∞
m−µ
σ
1
√
2π
exp{−
v2
2
}dv
= 1 −
Z m−µ
σ
−∞
1
√
2π
exp{−
v2
2
}dv .

Remembering the definition of the cumulative distribution function Φ(·) as
Φ(x) =
1
√
2π
Z x
−∞
e−y2/2
dy ,
we see that the above can be written as Φ(m−µ
σ
) = 1 − Φ(m−µ
σ
), so that
2Φ(
m − µ
σ
) = 1 or Φ(
m − µ
σ
) =
1
2
Thus we have m = µ + σΦ−1
(1/2), since we can compute Φ−1
using the Matlab function
norminv, we find that Φ−1
(1/2) = 0, which intern implies that m = µ.
Part (c): If X is an exponential random variable with rate λ then m must satisfy
Z m
0
λe−λx
dx =
Z ∞
m
λe−λx
dx = 1 −
Z m
0
λe−λx
dx .
Introducing the cumulative distribution function for the exponential distribution (given by
F(x) =
R m
0
λe−λx
dx) the above equation can be seen to be F(m) = 1 − F(m) or F(m) = 1
2
.
So in general the median m is given by m = F−1
(1/2) where F is the cumulative distribution
function. For the exponential random variable this expression gives
1 − e−λm
=
1
2
or m =
ln(2)
λ
.
Problem 14 (if X is an exponential random variable then cX is)
If X is an exponential random variable with parameter λ, then defining Y = cX the distri-
bution function for Y is given by
FY (a) = P{Y ≤ a}
= P{cX ≤ a}
= P
n
X ≤
a
c
o
= FX(
a
c
) .
So, taking the derivative of the above expression, to obtain the density function for Y we
see that
fY (a) =
dFY
da
=
d
da
FX (
a
c
)
= F′
X (
a
c
)
1
c
=
1
c
fX(
a
c
)

But since X is an exponential random variable with parameters λ we have that
fX (x) =

λe−λx
x ≥ 0
0 x 0
so that we have for fY (y) the following
fY (y) =
1
c

λe−λy
c
y
c
≥ 0
0 y
c
0
or
fY (y) =
λ
c
e− λ
c
y
y ≥ 0
0 y 0
showing that Y is another exponential random variable with parameter λ
c
.
Problem 15
The hazard rate function λ(t) is defined by λ(t) = f(t)
F̄ (t)
= f(t)
1−F (t)
. For a uniform random
variable distributed between (0, a) we have
f(t) =
1
a
0 ≤ t ≤ a
0 otherwise
and
F(t) =
Z t
0
f(t′
)dt′
=
Z t
0
dt′
a
=
t
a
,
so the hazard rate function then is
λ(t) =
(1/a)
1 − t
a
=
1
a − t
,
for 0 ≤ t ≤ a.
Problem 16
For this problem if we are told that X has a hazard rate function λX(t) we desire to compute
the hazard rate function for Y = aX, with a 0. When Y = aX the probability density
function of Y is given by fY (y) = fX (y/a) 1
a

and its distribution function is given by
Fy(c) = P{Y ≤ c} = P{aX ≤ c} = P{X ≤
c
a
} = FX (
c
a
) ,
so the hazard rate for Y is given by
λY (t) =
fY (t)
1 − FY (t)
=
fX( t
a
) 1
a

1 − FX (t/a)
=

1
a

fX(t/a)
1 − FX (t/a)

=

1
a

λX(t/a) .

Problem 17
The Gamma density function is given by
f(x) =
λe−λx
(λx)α−1
Γ(α)
x ≥ 0 .
To check that it integrates to one we have
Z ∞
0
f(x)dx =
Z ∞
0
λe−λx
(λx)α−1
Γ(α)
dx .
To evaluate this let y = λx so that dy = λdx to get
Z ∞
0
e−y
yα−1
Γ(α)
dy =
1
Γ(a)
Z ∞
0
e−y
yα−1
dy .
We note that the integral on the right hand side
R ∞
0
e−y
yα−1
dy is the definition of the
gamma function, Γ(a), and the above becomes equal to one showing that the Gamma density
integrates to one.
Problem 18 (the expectation of Xk
when X is exponential)
If X is exponential with mean 1/λ then f(x) = λe−λx
so that
E[Xk
] =
Z ∞
0
λxk
e−λx
dx = λ
Z ∞
0
xk
e−λx
dx .
To transform to the gamma integral, let v = λx, so that dv = λdx and the above integral
becomes
λ
Z ∞
0
vk
λk
e−v dv
λ
= λ−k
Z ∞
0
vk
e−v
dv .
Remembering the definition of the Γ function as
R ∞
0
vk
e−v
dv ≡ Γ(k + 1) and that when k is
an integer Γ(k + 1) = k!, we see that the above integral is equal to k! and we have that
E[Xk
] =
k!
λk
,
as required.
Problem 19 (the variance of a gamma random variable)
If X is a gamma random variable then
f(x) =
λe−λx
(λx)α−1
Γ(α)
,

when x ≥ 0 and is zero otherwise. To compute the variance we require E[X2
] which is given
by
E[X2
] =
Z ∞
0
x2
f(x)dx
=
Z ∞
0
x2 λe−λx
(λx)α−1
Γ(α)
dx
=
λα
Γ(α)
Z ∞
0
xα+1
e−λx
dx .
To evaluate the above integral, let v = λx so that dv = λdx then the above becomes
λα
Γ(α)
Z ∞
0
vα+1
λα+1
e−v dv
λ
=
λα
λα+2Γ(α)
Z ∞
0
vα+1
e−v
dv =
Γ(α + 2)
λ2Γ(α)
.
Where we have used the definition of the gamma function in the above. If we “factor” the
gamma function as
Γ(α + 2) = (α + 1)Γ(α + 1) = (α + 1)αΓ(α) ,
we see that
E[X2
] =
α(α + 1)
λ2
,
when X is a gamma random variable with parameters (α, λ). Since E[X] = α
λ
we can
compute Var(X) = E[X2
] − E[X]2
as
Var(X) =
α(α + 1)
λ2
−
α2
λ2
=
α
λ2
,
as claimed.
Problem 20 (the gamma function at 1/2)
We want to consider Γ(1/2) which is defined as
Γ(1/2) =
Z ∞
0
x−1/2
e−x
dx .
Since the argument of the exponential is the square of the term x1/2
this observation might
motivate the substitution y =
√
x. Following the hint let y =
√
2x, so that
dy =
1
√
2x
dx .
So that with this substitution Γ(1/2) becomes
Γ(1/2) =
Z ∞
0
√
2 dy e−y2/2
=
√
2
Z ∞
0
e−y2/2
dy .

Now from the normalization of the standard Gaussian we know that
Z ∞
−∞
1
√
2π
exp{−
y2
2
}dy = 1 ,
which easily transforms (by integrating only over the positive real numbers) into
2
Z ∞
0
1
√
2π
exp{−
y2
2
}dy = 1 .
Finally manipulating this into the specific integral required to evaluate Γ(1/2) we find that
√
2
Z ∞
0
exp{−
y2
2
}dy =
√
π ,
which shows that Γ(1/2) =
√
π as requested.
Problem 21 (the hazard rate function for the gamma random variable)
The hazard rate function for a random variable T that has a density function f(t) and
distribution function F(t) is given by
λ(t) =
f(t)
1 − F(t)
.
For a gamma distribution with parameters (α, λ) we know our f(t) is given by
f(t) =
(
λe−λt(λt)α−1
Γ(α)
t ≥ 0
0 t 0 .
Lets begin by calculating the cumulative density function for a gamma random variable with
parameters (α, λ). We find that
F(t) =
Z t
0
f(ξ)dξ =
Z t
0
λe−λξ
(λξ)α−1
Γ(α)
dξ ,
which cannot be simplified further. We then have that
1 − F(t) =
Z ∞
0
f(ξ)dξ −
Z t
0
f(ξ)dξ
=
Z ∞
t
f(ξ)dξ
=
Z ∞
t
λe−λξ
(λξ)α−1
Γ(α)
dξ ,
which also cannot be simplified further. Thus our hazard rate is given by
λ(t) =
λe−λt(λt)α−1
Γ(α)
R ∞
t
λe−λξ(λξ)α−1
Γ(α)
dξ
=
tα−1
e−λt
R ∞
t
ξα−1e−λξdξ
=
1
R ∞
t
ξ
t
α−1
e−λ(ξ−t)dξ .

To try and simplify this further let v = ξ
t
so that dv = dξ
t
, and the above becomes
λ(t) =
1
R ∞
1
vα−1e−λt(v−1)tdv
=
1
teλt
R ∞
1
vα−1e−λtvdv
.
Which is one expression for the hazard rate for a gamma random variable. We can try and
reduce the integral in the bottom of the above fraction to that of the “upper incomplete
gamma function” by making the substitution y = λtv so that dy = λtdv and obtaining
λ(t) =
1
teλt
R ∞
λt
yα−1
(λt)α−1 e−y dy
λt
=
(λt)α
teλt
1
R ∞
λt
yα−1e−ydy
=
(λt)α
teλt
1
Γ(α, λt)
.
Where we have introduced the upper incomplete gamma function whos definition is
given by
Γ(a, x) =
Z ∞
x
ta−1
e−t
dt .
Problem 27 (modality of the beta distribution)
The beta distribution with parameters (a, b) has a probability density function given by
f(x) =
1
B(a, b)
xa−1
(1 − x)b−1
for 0 ≤ x ≤ 1 .
Part (a): Our mode of this distribution will equal either the endpoints of our interval i.e.
x = 0 or x = 1 or the location where the first derivative of f(x) vanishes. Computing this
derivative the expression df
dx
= 0 implies
df
dx
(x) = (a − 1)xa−2
(1 − x)b−1
+ (b − 1)xa−1
(1 − x)b−2
(−1) = 0
⇒ xa−2
(1 − x)b−2
[(2 − a − b)x + (a − 1)] = 0 ,
which can be solved for the x∗
that makes this an equality and gives
x∗
=
a − 1
a + b − 2
assuming a + b − 2 6= 0 .
In this case to guarantee that this is a maximum we should check that the second derivative
of f at the value of a−1
a+b−2
is indeed negative. This second derivative is computed in the
Mathematica file chap 5 te 27.nb where it is shown to be negative for the given domains of
a and b. To guarantee that this value is interior to the interval (0, 1) we should verify that
0
a − 1
a + b − 2
1

which since a + b − 2 0 is equivalent to
0 a − 1 a + b − 2 .
or from the first inequality we have that a 1 and from the second inequality (a−1 a+b−2)
we have that b 1 verifying that our point x∗
is in the interior of this interval and our
distribution is unimodal as was asked.
Part (b): Now the case when a = b = 1 is covered below, so lets consider a = 1. From the
requirement a + b 2 we must have b 1 and our density function in this case is given by
f(x) =
(1 − x)b−1
B(1, b)
.
This has a derivative given by
f′
(x) =
(1 − b)(1 − x)b−2
B(1, b)
,
and is positive over the entire interval since b 1. Because the derivative is positive over
the entire domain the distribution is unimodal and the single mode will occur at the right
most limit i.e. x = 1. Now if b = 1 in the same way we have a 1 and our density function
is given by
f(x) =
xa−1
B(a, 1)
.
Which has a derivative given by
f′
(x) =
(a − 1)xa−2
B(a, 1)
,
and is negative because a 1. Because the derivative is negative over the entire domain the
distribution is unimodal and the unique mode will occur at the left most limit of our domain
i.e. x = 0. Finally, we consider the case where a 1, b 1 and neither equal to one. In
this case from the derivative above our minimum or maximum is given by a−1
a+b−2
which for
the domain of a and b given here is positive implying that the point x∗
is a minimum. Thus
we have two local maximums at the endpoints x = 0 and x = 1. One can also show (in the
same way as above) that for this domain of a and b the point x∗
is in the interior of the
interval.
Part (c): If a = b = 1, then the density function for the beta distribution becomes (since
Beta(1, 1) ≡ B(1, 1) = 1) is
f(x) = 1 ,
and we have the density of the uniform distribution, which is “flat” and has all points modes.

Problem 28 (Y = F(X) is a uniform random variable)
If Y = F(X) then the distribution function of Y is given by
FY (a) = P{Y ≤ a}
= P{F(X) ≤ a}
= P{X ≤ F−1
(a)}
= F(F−1
(a)) = a .
Thus fY (a) = dFY
da
= 1, showing that Y is a uniform random variable.
Problem 29 (the probability density function for Y = aX + b)
We begin by computing the cumulative distribution function of the random variable Y as
FY (y) = P{Y ≤ y}
= P{aX + b ≤ y}
= P{X ≤
y − b
a
}
= FX(
y − b
a
) .
Taking the derivative to obtain the distribution function for Y we find that
fY (y) =
dFY
dy
= F′
X(
y − b
a
)
1
a
=
1
a
fX(
y − b
a
) .
Problem 30 (the probability density function for the lognormal distribution)
We begin by computing the cumulative distribution function of the random variable Y as
FY (a) = P{Y ≤ a}
= P{eX
≤ a}
= P{X ≤ log(a)}
= FX (log(a)) .
Since X is a normal random variable with mean µ and variance σ2
it has a cumulative
distribution function given by
FX(a) = Φ

a − µ
σ

so that the cumulative distribution function for Y becomes
FY (a) = Φ

log(a) − µ
σ

.

The density function for the random variable Y is given by the derivative of the cumulative
distribution function thus we have
fY (a) =
FY (a)
da
= Φ′

log(a) − µ
σ

1
σ

1
a

.
Since Φ′
(x) = 1
√
2π
e−x2/2
we have for the probability density function for a lognormal random
variable given by
fY (a) =
1
√
2πσa
exp

−
1
2
(log(a) − µ)2
σ

.
Problem 31 (Legendre’s theorem on relatively primeness)
Part (a): If k is the greatest common divisor of both X and Y then k must divide the random
variable X and the random variable Y . In addition, X/k and Y/k must be relatively prime
i.e. have no common factors. Now to show the given probability we first argue that that
k will divide X with probability 1/k (approximately) and divide Y with probability 1/k
(approximately). This can be reasoned heuristically by considering the case where X and
Y are drawn from say 1, 2, . . . , 10. Then if k = 2 the numbers five numbers 2, 4, 6, 8, 10
are all divisible by 2 and so the probability 2 will divide a random number from this set is
5/10 = 1/2. If k = 3 then the three numbers 3, 6, 9 are all divisible by 3 and so the probability
3 will divide a random number from this set is 3/10 ≈ 1/3. In the same way when k = 4
the probability that 4 will divide one of the numbers in our set is 2/10 = 1/5 ≈ 1/4. These
approximations become exact as N goes to infinity. Finally, X/k and Y/k will be relatively
prime with probability Q1. Letting EX,k to be event that X is divisible by k, EY,k the event
that Y is divisible by k, and EX/k,Y/k the event that X/k and Y/k are relatively prime we
have that
Qk = P{D = k}
= P{EX,k}P{EY,k}P{EX/k,Y/k}
=

1
k

1
k

Q1 .
which is the desired results.
Part (b): From above we have that Qk = Q1/(k2
), so summing both sides for k = 1, 2, 3, · · ·
gives (since
P
k Qk = 1, i.e. the greatest common divisor must be one of the numbers 1, 2, 3,
.
.
.)
1 = Q1
∞
X
k=1
1
k2
,
which gives the desired result of
Q1 =
1
P∞
k=1
1
k2
.
Since
P∞
k=1 1/k2
= π2
/6 the above expression for Q1 becomes
Q1 =
1
π2
6
=
6
π2
.

Part (c): Now Q1 is the probability that X and Y are relatively prime will be true if P1 = 2
is not a divisor of X and Y . The probability that P1 is not a divisor of X is 1/P1 and
the same for Y . So the probability that P1 is a divisor for both X and Y is (1/P1)2
. The
probability that P1 is not a divisor of both will happen with probability 1 − (1/P1)2
. The
same logic applies for P2 giving that the probability that X and Y don’t have P2 as a factor
is 1 − (1/P2)2
. Since for X and Y be be relatively prime they cannot have any Pi as a joint
factor, and thus we are looking for the conjunction of each of the individual probabilities.
This is that P1 is not a divisor, that P2 is not a divisor, etc. This requires the product of all
of these terms giving for Q1 that
Q1 =
∞
Y
i=1

1 −
1
P2
i

=
∞
Y
i=1

P2
i − 1
P2
i

.
Problem 32 (the P.D.F. for Y = g(X), when g is decreasing)
Theorem 7.1 expresses how to obtain the probability density function for Y when Y = g(X)
and the probability density function for X is known. To prove this result in the case when
g(·) is decreasing lets compute the cumulative distribution function for Y i.e.
FY (y) = P{Y ≤ y}
= P{g(X) ≤ y}
By plotting a typical decreasing function g(x) we see that the set above is given by the set
of x values such that x ≥ g−1
(y) and the above expression becomes
FY (y) =
Z ∞
g−1(y)
f(x)dx .
Talking the derivative of this expression with respect to y we obtain
F′
Y (y) = f(g−1
(y))(−1)
dg−1
(y)
dy
.
Since dg−1(y)
dy
is negative
(−1)
dg−1
(y)
dy
=
dg−1
(y)
dy
,
and using this in the above the theorem in this case is proven.
Problem 1 (playing times for basketball)
Part (a): The probability that the players plays over fifteen minute is given by
Z 40
15
f(x)dx =
Z 20
15
0.025dx +
Z 30
20
0.05dx +
Z 40
30
0.025dx
= 0.025 · (5) + 0.05 · (10) + 0.025 · (10) = 0.875 .

Part (b): The probability that the players plays between 20 and 35 minute is given by
Z 35
20
f(x)dx =
Z 30
20
0.05dx +
Z 35
30
0.025dx
= 0.05 · (10) + 0.025 · (5) = 0.625 .
Part (c): The probability that the players plays less than 30 minutes is given by
Z 30
10
f(x)dx =
Z 20
10
0.025dx +
Z 30
20
0.05dx
= 0.025 · (10) + 0.05 · (10) = 0.75 .
Part (d): The probability that the players plays more than 36 minutes is given by
Z 40
36
f(x)dx = 0.025 · (4) = 0.1 .
Problem 2 (a power law probability density)
Part (a): Our random variable must normalize so that
R
f(x)dx = 1, or
Z 1
0
cxn
dx = c
xn+1
n + 1
1
0
=
c
n + 1
.
so that from the above we see that c = n + 1. Our probability density function is then given
by
f(x) =

(n + 1)xn
0 x 1
0 otherwise
Part (b): This expression is then given by
P{X x} =
Z 1
x
(n + 1)ξn
dξ = ξn+1 1
x
= 1 − xn+1
for 0 x 1 .
Thus we have
P{X x} =



1 x 0
1 − xn+1
0 x 1
0 x 1
Problem 3 (computing E[X] and Var(X))
Given this f(x) we compute the constant c by requiring
R
f(x)dx = 1, which in this case is
Z 2
0
cx4
dx =
cx5
5
2
0
=
c
5
(25
) = 1 ,

so c = 5
32
. Thus we can compute the expectation of X as
E[X] =
Z 2
0
x

5
32

x4
dx =
5
32
Z 2
0
x5
dx =
5
32
x6
6
2
0
=
5
3
.
And Var(X) = E[X2
] − E[X]2
, which means that we need
E[X2
] =
Z 2
0
x2

5
32

x4
dx =
5
32
Z 2
0
x6
dx =
5
25
x7
7
2
0
=
5 · 27
25 · 7
=
20
7
.
Thus Var(X) = 20
7
− 25
9
= 5
63
.
Problem 4 (a continuous density)
Our f(x) must integrate to one
Z 1
0
(ax + bx2
)dx =
ax2
2
+
bx3
3
1
0
=
a
2
+
b
3
= 1 .
We also told that E[X] = 0.6 so
E[X] =
Z 1
0
(ax2
+ bx3
)dx =
ax3
3
+
bx4
4
1
0
=
a
3
+
b
4
= 0.6 .
Solving for the a and b we have the following system

1/2 1/3
1/3 1/4

a
b

=

1
0.6

.
By Cramer’s rule we have
a =
1 1/3
0.6 1/4
1/2 1/3
1/3 1/4
=

1
20

72
1

= 3.6 ,
and
b =
1/2 1
1/3 0.6
1
8
− 1
9
= −2.4 .
Now for Part (a)
P{X 1/2} =
Z 1/2
0
(3.6x − 2.4x2
)dx =
3.6x2
2
1/2
0
−
2.4x3
3
1/2
0
= 0.45 − 0.1 = 0.35 .
For Part (b) we have Var(X) = E[X2
] − E[X]2
so
E[X2
] =
Z 1
0
x2
(3.6x − 2.4x2
)dx =
Z 1
0
(3.6x3
− 2.4x4
)dx
=
3.6x4
4
−
2.4x5
5
1
0
=
3.6
4
−
2.4
5
= 0.42 .
So that Var(X) = 0.42 − (0.6)2
= 0.06.

Problem 5 (a discrete uniform random variable)
We want to prove that X = int(nU) + 1 is a uniform random variable. To prove this first
fix n, then X = i is true if and only if
Int(nU) + 1 = i for i = 1, 2, 3, · · · , n .
or
Int(nU) = i − 1 .
or
i − 1
n
≤ U
i
n
for i = 1, 2, 3, · · · , n
Thus the probability that X = i is equal to
P{X = i} =
Z i
n
i−1
n
1dξ =
i
n
−
i − 1
n
=
1
n
for i = 1, 2, 3, · · · , n .
Problem 6 (bidding on a contract)
Assume we select a bid price b. Then our profit will be b − 100 if get the contract and zero
if we don’t get the contract. Thus our profit is a random variable that depends on the bid
received by the competing company u. Our profit is then given by (here P is for profit)
P(b) =

0 b u
b − 100 b u
Lets compute the expected profit
E[P(b)] =
Z b
70
0 ·
1
140 − 70
dξ +
Z 140
b
(b − 100) ·
1
140 − 70
dξ
=
(b − 100)(140 − b)
70
=
240b − b2
− 14000
70
.
Then to find the maximum of the expected profit we take the derivative of the above expres-
sion with respect to b, setting that expression equal to zero and solve for b. The derivative
set equal to zero is given by
dE[P(b)]
db
=
1
70
(240 − 2b) = 0 .
Which has b = 120 as a solution. Since d2E[P (b)]
db2 = − 2
70
0, this value of b is indeed
a maximum of the function P(b). Using this value of b our expected profit is given by
400
70
= 40
7
.

Problem 7
Part (a): We want to compute P{U ≥ 0.1} =
R 1
0.1
dξ = 0.9.
Part (b): We want to compute
P{U ≥ 0.2|U ≥ 0.1} =
P{U ≥ 0.2, U ≥ 0.1}
P{U ≥ 0.1}
=
P{U ≥ 0.2}
P{U ≥ 0.1}
=
1 − 0.2
1 − 0.1
=
0.8
0.9
=
8
9
.
Part (c): We want to compute
P{U ≥ 0.3|U ≥ 0.1, U ≥ 0.2} =
P{U ≥ 0.1, U ≥ 0.2, U ≥ 0.3}
P{U ≥ 0.1, U ≥ 0.2}
=
P{U ≥ 0.3}
P{U ≥ 0.2}
=
0.7
0.8
=
7
8
.
Part (d): We have P(winner) = P{U ≥ 0.3} = 0.7.
Problem 8 (IQ scores)
We can transform all questions to those involving a standard normal. We have with S the
random variable denoting the score of our IQ test taker. Then we have
P{S ≥ 125} = 1 − P{S ≥ 125}
= 1 − P{
S − 100
15
≤
150 − 100
15
}
= 1 − P{Z ≤ 1.66} = 1 − Φ(1.66) .
Part (b): We desire to compute
P{90 ≤ S ≤ 110} = P

90 − 100
15
≤ Z ≤
110 − 100
15

= P{−0.66 ≤ Z ≤ 0.66} = Φ(0.66) − Φ(−0.66)
Problem 9
Let 1:00−T be the time we leave from the house. Then if X is the random variable denoting
how long it takes us to go to work, we arrive at work at the time 1:00−T + X. To guarantee
with 95% probability that we arrive on time we must require that 1:00−T + X ≤1:00 or
−T + X ≤ 0 or X ≤ T ,

with 95% probability. Thus we require T such that P{X ≤ T} = 0.95 or
P{
X − 40
7
≤
T − 40
7
} = 0.95 ⇒ Φ(
T − 40
7
) = 0.95 ,
T − 40
7
= Φ−1
(0.95) ⇒ T = 40 + 7Φ−1
(0.95) .
Problem 10 (the lifetime of automobile tires)
Part (a): We want to compute P{X ≥ 40000}, which we do by converting to a standard
normal. We find
P{X ≥ 40000} = P

X − 34000
4000
≥ 1.5

= 1 − P{Z 1.5} = 1 − Φ(1.5) = 0.0668 .
Part (b): We want to compute P{30000 ≤ X ≤ 35000}, which we do by converting to a
standard normal. We find
P{30000 ≤ X ≤ 35000} = P

30000 − 34000
4000
≤ Z ≤
35000 − 34000
4000

= P{−1 ≤ Z ≤ 0.25} ≈ 0.4401 .
Part (c): We want to compute
P{X ≥ 40000|X ≥ 30000} =
P{X ≥ 40000, X ≥ 30000}
P{X ≥ 30000}
=
P{X ≥ 40000}
P{X ≥ 30000}
.
We again do this by converting to a standard normal. We find
P{X ≥ 40000}
P{X ≥ 30000}
=
P

Z ≥ 40000−34000
4000
P

Z ≥ 30000−34000
4000
=
1 − Φ(1.5)
1 − Φ(−1.0)
= 0.0794 .
All of these calculations can be found in the Matlab file chap 5 st 10.m.
Problem 11 (the annual rainfall in Cleveland)
Part (a): Let X be the random variable denting the annual rainfall in Cleveland. Then we
want to evaluate P{X ≥ 44}. Which we can do by converting to a standard normal. We
find that
P{X ≥ 44} = P

X − 40.2
8.4
≥
44 − 40.2
8.4

= 1 − Φ(0.452) = 0.3255 .

Part (b): Following the assumptions stated for this problem lets begin by calculating P(Ai)
for i = 1, 2, . . . , 7. Assuming independence each is equal to the value calculated in part (a) of
this problem. Lets denote that common value by p. Then the random variable representing
the number of years where the rainfall exceeds 44 inches (in a seven year time frame) is a
Binomial random variable with parameters (n, p) = (7, 0.3255). Thus the probability that
three of the next seven years will have more than 44 inches of rain is given by

7
3

p3
(1 − p)4
= 0.2498 .
These calculations are performed in the Matlab file chap 5 st 11.m.
Problem 14 (hazard rates)
Part (a): We have
P{X 2} = 1 − P{X ≤ 2} = 1 − (1 − e−22
) = e−22
= e−4
.
Part (b): We find
P{1 X 3} = P{X ≤ 3} − P{X 1}
= (1 − e−9
) − (1 − e−1
) = e−1
− e−9
.
Part (c): The hazard rate function is defined as
λ(x) =
f(x)
1 − F(x)
.
Where f is the density function and F is the distribution function. We find for this problem
that
f(t) =
dF
dx
=
d
dx
(1 − e−x2
) = 2xe−x2
.
so λ(x) is given by
λ(x) =
2xe−x2
1 − (1 − e−x2
)
= 2x .
Part (d): The expectation is given by (using integration by parts to evaluate the first
integral)
E[X] =
Z ∞
0
xf(x)dx =
Z ∞
0
x(2xe−x2
)dx
= 2
xe−x2
−2
∞
0
+
1
2
Z ∞
0
e−x2
dx
!
=
Z ∞
0
e−x2
dx .

From the unit normalization of the standard Gaussian 1
√
2π
R ∞
−∞
e−s2/2
ds = 1 we can compute
the value of the above integral. Using this expression we find that
R ∞
0
e−x2
dx =
√
π/2 thus
E[X] =
√
π
2
.
Part (d): The variance is given by Var(X) = E[X2
] − E[X]2
so first computing the expec-
tation of X2
we have that
E[X2
] =
Z ∞
0
x2
f(x)dx =
Z ∞
0
x2
(2xe−x2
)dx
= 2
x2
e−x2
−2
∞
0
+
1
2
Z ∞
0
2xe−x2
dx
!
= 2
Z ∞
0
xe−x2
dx = 2
e−x2
−2
∞
0
!
= 1 .
Thus
Var(X) = 1 −
π
4
=
4 − π
4
.

X1, X2 P{X1, X2}
0, 0 8
13
7
12

0, 1 8
13
5
12

1, 0 5
13
8
12

1, 1 5
13
4
12

Table 21: The joint probability distribution for Problem 2 in Chapter 6
X1, X2 X3 P{X1, X2, X3}
0, 0, 0 8
13
7
12
6
11

0, 0, 1 8
13
7
12
5
11

0, 1, 0 8
13
5
12
7
11

0, 1, 1 8
13
5
12
4
11

1, 0, 0 5
13
8
12
7
11

1, 0, 1 5
13
8
12
4
11

1, 1, 0 5
13
4
12
8
11

1, 1, 1 5
13
4
12
3
11

Table 22: The joint probability distribution for Problem 3 in Chapter 6
Chapter 6 (Jointly Distributed Random Variables)
Chapter 6: Problems
Problem 2
We have five white and eight balls. Let Xi = 1 if the i ball selected is white, and equals zero
otherwise.
Part (a): We want to compute P{X1, X2}. Producing the Table 21 we have We can check
that the given numbers do sum to one i.e. that
P
X1,X2
P{X1, X2} = 1. We have
1
12 · 13
(56 + 40 + 40 + 20) =
156
12 · 13
= 1 .
Part (b): We want compute P{X1, X2, X3}. Enoumerating these probabilities we compute
our results in Table 21
Problem 3
We begin by defining Yi = 1 if the i white ball (from five) is selected at step i and zero
otherwise.

Part (a): Computing the joint probability by conditioning on the first ball drawn, we have
P{Y1 = 0, Y2 = 0} = P{Y2 = 0|B1 is W2}P{B1 is W2}
+ P{Y2 = 0|B1 is not W2}P{B1 is not W2}
= 1

1
13

+

11
12

12
13

.
Now the other probabilties are computed in the same way. We find
P{Y1 = 0, Y2 = 1} =

12
13

1
12

.
P{Y1 = 1, Y2 = 0} =

1
13

P{Y2 = 0} =

1
13

11
12
.
and
P{Y1 = 1, Y2 = 1} =

1
13

1
12

.
Problem 10
Part (a): We find (performing several manipulations on one line to save space) that
P{X Y } =
Z Z
Ω
f(x, y)dxdy =
Z ∞
x=0
Z ∞
y=x
f(x, y)dydx
=
Z ∞
x=0
Z ∞
y=x
e−(x+y)
dydx =
Z ∞
x=0
e−x e−y
(−1)
∞
x
dx
=
Z ∞
x=0
e−x
(e−x
)dx =
Z ∞
x=0
e−2x
dx =
e−2x
(−2)
∞
0
=
1
2
.
Part (b): We compute that
P{X a} =
Z Z
Ω
f(x, y)dxdy =
Z a
x=0
Z ∞
y=0
e−(x+y)
dydx
=
Z a
x=0
e−x e−y
(−1)
∞
0
dx =
Z a
x=0
e−x
dx =
e−x
(−1)
a
0
= 1 − e−a
.
Problem 11 (shopping for a television)
We have pTV = 0.45, pP T = 0.15, pB = 0.4, so this problem looks like a multinomial
distribution. If we let NTV , NP T , and NB be the number of ordinary televisions, plasma

televisions, and people browsing. Then we desire to compute
P{NTV = 2, NP T = 1, NB = 2} =

N
NTV , NP T , NB

pNT V
TV pNP T
P T pNB
B
=

5!
2!1!2!

(0.45)2
(0.15)1
(0.4)2
= 0.1458 .
Problem 12
We begin by recalling the example 2b from this from this chapter. There the book shows
that if the total number of people entering a store is given by a Poisson random variable with
rate λ and each person is of one type (male) with probability p and another type (female)
with probability q = 1 − p, then the number of males and females entering are given by
Poisson random variables with rates λp and λ(1 − p) respectively. In addition, and more
surprisingly, the random variables X and Y are independent. Thus the desired probability
for this problem is
P{X ≤ 3|Y = 10} = P{X ≤ 3} =
3
X
i=0
e−λp
(λp)i
i!
= e−λp
3
X
i=0
(λp)i
i!
.
If we assume that men and women are equally likely to enter so that p = 1/2 then the above
becomes with λ = 10
e−5
3
X
i=0
5i
i!
= e−5

1 + 5 +
52
2
+
53
3!

= e−5
(39.33) = 0.265 .
Problem 13
Let X be the random variable denoting the arrival time of the men and Y be the random
variable denoting the arrival time of the women. Then X is a uniform random variable with
X ∈ [12:15, 12:45], and Y is a uniform random variable with Y ∈ [12:00, 1:00]. To simplify
our calculations, we will let the time 12:30 be denote zero and measure time in minutes.
Under this convention X is a uniform random variable taking values in [−15, +15] and Y
is a uniform random variable taking values in [−30, +30]. Then the question asks us to
compute P{|X − Y | ≤ 5}. To compute this, condition on the case when the men arrives
first or second. This expression then becomes
P{|X − Y | ≤ 5} = P{Y − X ≤ 5|X Y }P{X Y }
+ P{X − Y ≤ 5|X Y }P{X Y } .
We will first compute P{X Y } which can easily be computed from the joint density
fX,Y (x, y) = fX(x)fY (y) since the arrival of X and Y are independent. From our chosen
coordinate system we have a valid domain for X and Y that looks like

WWX draw figure ...
In that figure form the line X = Y we see clearly the domain where X Y . Thus
P{X Y } =
Z Z
ΩXY
fX,Y (x, y)dxdy .
Since if we integrate over all of Ω we must obtain one, the joint density is given by
fX,Y (x, y) =
1
30 · 60
=
1
1800
.
To evaluate this integral we regonnize it as the area of the suggested trapezoid. Using the
result from elementary geometry that area of a trapezoid equals the height times the average
of its two bases. We have
P{X Y } =
1
1800
Z Z
ΩXY
dxdy
=
1
1800

30 ·

45 + 15
2

=
1
2
.
Thus the other conditioning probability P{X Y } is easy to compute
P{X Y } = 1 − P{X Y } =
1
2
.
Now lets compute P{Y −X ≤ 5|X Y } which from the definition of conditional probability
is given by
P{Y − X ≤ 5, X Y }
P{X Y }
.
The top probability is then given by the following integration region.
WWX: draw region
P{Y − X ≤ 5, X Y } =
Z 15
x=−15
Z y=x+5
y=−x
fX,Y (x, y)dydx
=
Z 15
x=−15
1
1800
(x + 5 + x)dx
=
1
1800
Z 15
x=−15
(2x + 5)dx
=
1
1800
x2
+ 5x
15
−15
=
1
12
.
Thus P{Y − X ≤ 5|X Y } = 1/12
1/2
= 1
6
. Now additional probabilities can be computed in
using these same methods but a direct solution to the problem is to compute P{|X −Y | ≤ 5}
directly. For example, to compute P{|X − Y | ≤ 5} we integrate over the region
WWX put plot here!!!

and is represented by the following integral
Z Z
Ω{|X−Y |≤5}
fX,Y (x, y)dxdy =
1
1800
Z +15
x=−15
Z x+5
y=x−5
dydx
=
1
1800
Z 15
x=−15
(x + 5 − (x − 5))dx =
1
6
.
Now the probability that the man arrives first is given by P{X ≤ Y } and was computed
before to be 1/2.
Problem 14
Let X denote the random variable providing the location of the accident along the road. The
according to the problem specification let X be uniformly distributed between [0, L]. Let Y
denote the random variable the location of the ambulance. Then we define D = |X − Y |
the random variable representing the distance between the accident and the ambulance. We
want to compute
P{D ≤ d} =
Z Z
X,Y ∈Ω
f(x, y)dxdy ,
with Ω the set of points where |X − Y | ≤ d. In the X, Y plane the set |x − y| ≤ d can be
graphically represented by
WWX put drawing!!!
Using the diagram we see how to analytically integrate over the domain Ω. Specifically we
have
P{D ≤ d} =
Z d
x=0
Z x+d
y=0
f(x, y)dydx +
Z L−d
x=d
Z x+d
y=x−d
f(x, y)dydx
+
Z L
x=L−d
Z L
x−d
f(x, y)dydx ,
Since X and Y are independent fX,Y (x, y) = 1
L2 so the expression above becomes
P{D ≤ d} =
1
L2
Z d
0
(x + d)dx +
1
L2
Z L−d
d
(x + d − (x − d))dx
+
1
L2
Z L
L−d
(L − x + d)dx
=
1
L2

x2
2
+ dx
d
0
+
1
L2
(2d(L − 2d)) +
1
L2

(L + d)x −
x2
2
L
L−d
=
(2L − d)d
L2
.
Then fD(d) = dFD(d)
d(d)
(meaning we take the derivative with resepct to the variable d) and we
find fD(d) = 2(L−d)
L2 .

Problem 32
Part (a): Since each week the gross sales is a draw from an independent normal random
variable the sum of n of these normal random variables is another normal with mean the
sum of the n means and variance the sum of the n variances. For the case given in this
problem we have two normals over the two weeks and the mean gross sales is
µ = 2200 + 2200 = 4400 .
and variance is
σ2
= 2302
+ 2302
= 105800 .
The we desire to compute
P{Sales ≥ 5000} = 1 − P{Sales ≤ 5500}
= 1 − P

S − 4400
325.2
≤
5000 − 4400
325.2

= 1 − Φ(1.8446) = 0.0325 .
Part (b): In this part of the problem I’ll compute the probability that the weekly sales
will exceed 2000 by treating this probability as the probability of success under a Binomial
random variable model. I’ll use the Binomial mass function to compute the probability that
our weekly sales exceeds 2000 in two of the next three weeks. Defining p to to be
p = P{Sales ≥ 2000}
= 1 − P

Sales − 2200
230
≤
2000 − 2200
230

.
Then with this probability we are after is
Ps =

3
2

p2
(1 − p)1
+

3
3

p3
= 0.9033 .
Problem 44
This is a problem in so called Bayesian inference. By this what I mean is that given informa-
tion about the number of accidents that have occurred, we want to find the density (given
this number of accidents) of the accident rate λ. Before observing the number of accidents
in a year, the unknown λ is governed by a gamma distribution with parameters s and α.
Specifically,
f(λ) =
(
se−sλ(sλ)α−1
Γ(α)
λ ≥ 0
0 λ 0
.
We wish to evaluate p(λ|N = n) where p(λ|N = n) is the probability that λ takes a given
value after the observation that n accident occurred last year. From Bayes’ rule we have
that
P(λ|N = n) =
p(N = n|λ)f(λ)
R
Λ
p(N = n|λ)f(λ)dλ
∝ p(N = n|λ)f(λ) .

Now p(N = n|λ) is a Poisson random variable with mean λ and therefore has a density
function by
p(N = n|λ) =
e−λ
λn
n!
=
e−λ
λn
Γ(n + 1)
n = 0, 1, 2, · · ·
so the above expression above becomes
p(λ|N = n) ∝

e−λ
λn
Γ(n + 1)

se−sλ
(sλ)α−1
Γ(α)

=
e−λ(1+s)
sα
λn+α+1
Γ(α)Γ(n + 1)
,
which is proportional to e−(1+s)λ
λn+α+1
. The density that is proportional to an expression
like this is another gamma distribution with parameters s + 1 and n + α. This is seen from
the functional form of the gamma distribution (presented above) and from this we can easily
compute the required normalizing factor. Specifically we have
p(λ|N = n) =
(
(s+1)e−(s+1)λ((s+1)λ)n+α−1
Γ(n+α)
λ ≥ 0
0 λ 0
.
This is the conditional density of the accident parameter λ.
To determine the expected number of accidents the following year (denoted N2) we will first
compute the probability that we observe m accidents in that year i.e. P{N2 = m}. This is
given by conditioning on λ as follows
P{N2 = m} =
Z
Λ
P{N2 = m|λ}p(λ)dλ ,
where in this expression everything is implicitly conditioned on N1 = n i.e. that in the first
year we observed n accidents. The above integral becomes
P{N2 = m} =
Z ∞
λ=0

e−λ
λm
m!

(s + 1)n+α
Γ(n + α)
e−(s+1)λ
λn+α−1
dλ
=
(s + 1)n+α
m!Γ(n + α)
Z ∞
0
e−(s+2)λ
λn+m+α−1
dλ .
To evaluate this integral let v = (s + 2)λ then dv = (s + 2)dλ and the above becomes
P{N2 = m} =
(s + 1)n+α
m!Γ(n + α)
Z ∞
0
e−v vn+m+α−1
(s + 2)n+m+α−1+1
dv
=
(s + 1)n+α
m!Γ(n + α)(s + 2)n+m+α
Γ(n + m + α)
=
Γ(n + m + α)
Γ(m + 1)Γ(n + α)
(s + 1)n+α
(s + 2)n+m+α
for m = 0, 1, 2, 3, · · ·
Now a generalization expression for the binomial coefficient

n
k

is given by

n
k

=
n!
k!(n − k)!
=
Γ(n + 1)
Γ(k + 1)Γ(n − k + 1)
,

which from the definition in terms of the gamma function is valid for non integer n and k.
Thus
Γ(n + m + α)
Γ(m + 1)Γ(n + α)
=

n + m + α − 1
m

=

n + m + α − 1
n + α − 1

,
so the distribution for P{N2 = m} becomes (using the second expression for the ratio of
gamma functions)
P{N2 = m} =

n + m + α − 1
n + α − 1

(s + 1)n+α
(s + 2)n+m+α
for m = 0, 1, 2, · · ·
We can shift this index N2 to be “offset” by n+ α to make this look like a negative binomial
random variable. To do this define Ñ2 = N2 + (n + α), then since the range of N2 is from
0, 1, 2 · · · the range of Ñ2 is from n + α, n + α + 1, · · ·. Thus
P{Ñ2 = m̃} =

m̃ − 1
r − 1

(s + 1)r
(s + 2)m̃
=

m̃ − 1
r − 1

s + 1
s + 2
r
1
(s + 2)m̃−r
,
Where we have defined r = n + α and m̃ take values in the range r, r + 1, r + 2, · · ·. Then
further defining p = s+1
s+2
so that
1 − p =
s + 2
s + 2
−
s + 1
s + 2
=
1
s + 2
,
we have that the above is given by
P{Ñ2 = m̃} =

m̃ − 1
r − 1

pr
(1 − p)m̃−r
for m̃ = r, r + 1, r + 2, · · ·
From this expression and Example 8f from the book we know that
E[Ñ2] =
r
p
=

s + 2
s + 1

(n + α) .
The expectation of N2, the expected number of accidents is then given by
E[N2] = E[Ñ2] − (n + α)
=

s + 2
s + 1

(n + α) − (n + α)
=

s + 2
s + 1
− 1

(n + α) =
n + α
s + 1
,
which is requested expression.
Problem 53
From the theorem on the transformation of coordinates for joint probability density functions
we have that
fX,Y (x, y) = fZ,U (z, u)|J(z, u)|−1
.

Now in this case the Jacobian expression becomes
J(z, u) =
∂x
∂z
∂x
∂u
∂y
∂z
∂y
∂u
=
√
2z−1/2
(1/2) cos(u) −
√
2z1/2
sin(u)
√
2z−1/2
(1/2) sin(u) +
√
2z1/2
cos(u)
=
2
2
cos2
(u) +
2
2
sin2
(u) = 1 .
We also have that fX,Y (x, y) = fZ(z)fU (u) by independence of the random variable Z and
U. Since Z is uniform [0, 2π] and U is exponential with a rate one we get for fX,Y (x, y) the
following
fX,Y (x, y) =

1
2π

1e−z
.
Now since
X
√
2Z
= cos(U) and
Y
√
2Z
= sin(U) ,
squaring both sides and adding we obtain
X2
2Z
+
Y 2
2Z
= 1 ,
or Z = 1
2
(X2
+ Y 2
), thus we have
fX,Y (x, y) =
1
2π
e− 1
2
(x2+y2)
=
1
√
2π
e− x2
2 ·
1
√
2π
e− y2
2 ,
which is the product of two probability density functions for standard normal random vari-
ables showing that X and Y are independent and normal.
Problem 19
We are asked to compute p(W|N = n), using Bayes’ rule this can be expressed as
P{N = n|W}P(W)
R
W
P{N = n|W}P(W)dw
.
Now ignoring the normalizing term the above is proportional to
e−w
wn
Γ(n + 1)
βe−βw
(βw)t−1
Γ(t)
,
which by factoring only the terms that depend on w is proportional to
e−(β+1)w
wn+t−1
,
which is the functional form (in w) for a gamma distribution with parameters n + t and
β + 1. Thus
p(W|N = n) =
(β + 1)e−(β+1)w
((β + 1)w)n+t−1
Γ(n + t)
.

Problem 22
We desire to compute P{[X] = n, X − [X] ≤ x} for n = 0, 1, 2, · · · and 0 ≤ x ≤ 1. This is
given by the following integral
Z n+x
ξ=n
λe−λξ
dξ = λ
e−λξ
−λ
n+x
n
= e−λn
− e−λ(n+x)
.
To see if [x] and X − [X] are independent, consider if the joint distribution function is the
product of the two marginalized distributions. The joint probability density function of this
random variable is given by the derivative of this with respect to x i.e.
fN,X(n, x) =
d
dx
P{[X] = n, X − [X] ≤ x} = λe−λ(n+x)
,
computing the marginal distribution we first compute
P{[X] = n} =
Z 1
x=0
fN,X(n, x)dx = λ
Z 1
0
e−λ(n+x)
dx
=
−λ
λ
e−λ(n+x)
1
0
= −(e−λ(n+1)
− e−λn
) = e−λn
− e−λ(n+1)
.
Also
p(X − [X] = x) =
∞
X
n=0
λe−λ(n+x)
= λe−λx
∞
X
n=0
e−λn
= λe−λx
∞
X
n=0
(e−λ
)n
= λe−λx 1
1 − e−λ
=
λe−λx
1 − e−λ
,
so if fN,X(n, x) = P{[X] = n}p(X − [X] = x), then our elements are independent. Comput-
ing the right hand side of this expression we have
(e−λn
− e−λ(n+1)
)
λe−λx
1 − e−λ
= λe−λ(x+n)
,
which does equal fN,X(n, x) so these two random variables are independent.
Problem 23
Part (a): Given Xi for i = 1, 2, · · · , n with common distribution function F(x) define
Y = max(Xi) for i = 1, 2, · · · , n. Then Y is called the nth order statistic and is often

written X(n). To compute the cumulative distribution function for X(n) i.e. P{X(n) ≤ x} we
can either use the result in the book
P{X(n) ≤ x} =
n
X
k=j

n
k

F(y)k
(1 − F(y))n−k
,
for j = n which gives FX(n)
= F(y)n
, or we can reason more simply as follows. For the
random variable X(n) to be less than x, each draw Xi must be less than x. Each draw is less
than x with probability F(x) and as this must happen n times, the probability that X(n) is
less than x is F(x)n
, verifying the above.
Part (b): In this case define Z = min(Xi), then Z is another order statistic and this time
corresponds to X(1). Using the distribution function from before we have
FX(1)
=
n
X
k=1

n
k

F(y)k
(1 − F(y))n−k
=
n
X
k=0

n
k

F(y)k
(1 − F(y))n−k
−

n
0

F(y)0
(1 − F(y))n
= 1 − (1 − F(y))n
.
Or we may reason as follows. The distribution function
FZ(x) = P{Z ≤ x} = P{min
i
(Xi) ≤ x} = 1 − P{min(Xi) ≥ x} .
This last probability is the intersection of n events i.e. each Xi that is drawn must be drawn
greater than the value of x. Each event of this type happens with probability 1 − F(x).
Giving that
P{min
i
(Xi) ≤ x} = 1 − (1 − F(x))n
.
Problem 27 (the sum of a uniform and an exponential)
Part (a): Since X and Y are independent, the density of the random variable Z = X + Y
is given by the convolution of the density function for X and Y . For example
fZ(a) =
Z ∞
−∞
fX(a − y)fY (y)dy =
Z ∞
−∞
fX(y)fY (a − y)dy .
Since X is uniform we have that
fX(x) =

1 0 ≤ x ≤ 1
0 else
,
and since Y is exponential we have that
fY (y) =

λe−λy
y ≥ 0
0 else

−3 −2 −1 0 1 2 3
−0.5
0
0.5
1
1.5
initial step function
−3 −2 −1 0 1 2 3
−0.5
0
0.5
1
1.5
flipped step function
Figure 3: Left: The initial probability density function for X or fX(x) (a step function).
Right: This function flipped or fX (−x).
−3 −2 −1 0 1 2 3
−0.5
0
0.5
1
1.5
flipped step function shifted right by 0.75
−3 −2 −1 0 1 2 3 4
0
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6
1.8
Figure 4: Left: The initial probability density function for X, flipped and shifted by a = 3/4
to the right or fX(−(x−a)). Right: The flipped and shifted function plotted together with
fY (y) allowing visualizations of function overlap as a is varied.

To evaluate the above convolution we might as well select a formulation that is simple to
evaluate. I’ll pick the first formulation since it is easy to shift to the uniform distribution
to produce fX(a − y). Now since fX(x) looks like the plot given in Figure 3 (left) we
see that fX(−x) then looks like Figure 3 (right). Inserting a right shift by a we have
fX(−(x − a)) = fX (a − x), and this function looks like that shown in Figure 4 (left). Thus
we can now evaluate the distribution function for fZ(a), we find that
fZ(a) =
Z a
−1+a
1fY (y)dy .
Now since fY (y) is zero when y is negative, to further evaluate this we evaluate it for some
specific a’s. One easy case is if a 0 then fZ(a) = 0. If a 0 but the lower limit of
integration is negative, that is −1 + a 0, i.e. 0 a 1, then we have
fZ(a) =
Z a
−1+a
fY (y)dy =
Z a
0
λe−λy
dy = λ
Z a
0
e−λy
dy
=
λe−λy
−λ
a
0
= −(e−λa
− 1) = 1 − e−λa
.
If a 1 then the integral for fZ(a) then is
fZ(a) =
Z a
−1+a
fY (y)dy =
Z a
−1+a
λe−λy
dy
=
λe−λy
−λ
a
−1+a
=
−λ
λ
(e−λa
− e−λ(−1+a)
)
= e−λ(a−1)
− e−λa
.
In summary then
fZ(a) =



0 a 0
1 − e−λa
0 a 1
e−λ(a−1)
− e−λa
a 1
In the MATLAB function chap 6 prob 27.m we have code to duplicate the above figures.
Part (b): If Z = X
Y
then to compute the distribution function for Z is
FZ(a) = P{Z ≤ a}
= P{
X
Y
≤ a}
= P{X ≤ aY }
=
Z Z
X≤aY
f(x, y)dxdy
=
Z Z
X≤aY
f(x)f(y)dxdy
where we get the last equality from the independence of X and Y . The above integral can
be evaluated by letting x range over 0 to 1, while y ranges over 1
a
x to +∞. With these limits

x
a
≤ y ≤ ∞ or x ≤ ay ≤ ∞. So the integral above becomes
Z 1
x=0
Z ∞
y=x/a
f(x)f(y)dydx =
Z 1
0
Z ∞
x/a
λe−λy
dydx
=
Z 1
x=0
−λ
λ
e−λy
∞
x/a
=
Z 1
0
−(0 − e−λx/a
)dx
=
Z 1
0
e−λx/a
dx =
e−λx/a
−λ
a

1
0
= −
a
λ
(e−λ/a
− 1) =
a
λ
(1 − e−λ/a
) .
Problem 33 (the P.D.F. of the ratio of normals is a Cauchy distribution)
As stated in the problem, let X1 and X2 be distributed as standard normal random variables
(i.e. they have mean 0 and variance 1). Then we want the distribution of the variable
X1/X2. To this end define the random variables U and V as U = X1/X2 and V = X2. The
distribution function of U is then what we are after. From the definition of U and V in terms
of X1 and X2 we see that X1 = UV and X2 = V . To solve this problem we will derive the
joint distribution function for U and V and then marginalize out V giving the distribution
function for U, alone. Now from Theorem 2 − 4 on page 45 of Schaums probability and
statistics outline the distribution of the joint random variable (U, V ), in term of the joint
random variable (X1, X2) is given by
g(u, v) = f(x1, x2)
∂(x1, x2)
∂(u, v)
.
Now
∂(x1, x2)
∂(u, v)
=
v u
0 1
= |v| ,
so that
g(u, v) = f(x1, x2)|v| = p(x1) p(x2)|x2| ,
as f(x1, x2) = p(x1)p(x2) since X1 and X2 are assumed independent. Now using the fact
that the distribution of X1 and X2 are standard normals we get
g(u, v) =
1
2π
exp(−
1
2
(u v)2
) exp(−
1
2
v2
) |v| .
Marginalizing out the variable V we get
g(u) =
Z ∞
−∞
g(u, v)dv =
1
π
Z ∞
0
v e− 1
2
(1+u2)v2
dv
To evaluate this integral let η =
q
1+u2
2
v, and after performing the integration we then find
that
g(u) =
1
π
1
1 + u2
.
Which is the distribution function for a Cauchy random variable.

Chapter 7 (Properties of Expectations)
Chapter 7: Problems
Problem 1 (expected winnings with coins and dice)
If we roll a heads then we win twice the digits on the die roll. If we roll a tail then we win
1/2 the digit on the die. Now we have a 1/2 chance of getting a head (or a tail) and a 1/6
chance of getting any individual number on the die. Thus the expected winnings are given
by
1
2
·
1
6

1
2
· 1

+
1
2
·
1
6

1
2
· 2

+ · · · +
1
2
·
1
6
(2 · 1) +
1
2
·
1
6
(2 · 2) + · · ·
or factoring out the 1/2 and the 1/6 we obtain
1
2
·
1
6

1
2
+
2
2
+
3
2
+
4
2
+
5
2
+
6
2
+ 2 + 2 · 2 + 2 · 3 + 2 · 4 + 2 · 5 + 2 · 6

which equals
105
24
.
Problem 2
Part (a): We have six choices for a suspect, six choices for a weapon and nine choices for a
room giving in total 6 · 6 · 9 = 324 possible combinations.
Part (b): Now let the random variables S, W, and R be the number of suspects, weapons,
and rooms that the player receives and let X be the number of solutions possible after
observing S, W, and R. Then X is given by
X = (6 − S)(6 − W)(9 − R) .
Part (c): Now we must have
S + W + R = 3 with 0 ≤ S ≤ 3 , 0 ≤ W ≤ 3 , 0 ≤ R ≤ 3
Each specification of these three numbers (S, W, R) occurs with a uniform probability given
by
1

3 + 3 − 1
3 − 1
=
1

5
2
=
1
10
,

using the results given in Chapter 1. Thus the expectation of X is given by
E[X] =
1
10
X
S
X
W
X
R
(6 − S)(6 − W)(9 − R)
=
1
10
3
X
s=0
(6 − s)
3
X
w=0
(6 − w)
3
X
r=0
(9 − r)
=
1
10

6
X
W +R=3
(6 − W)(9 − R) + 5
X
W +R=2
(6 − W)(9 − R)
+ 4
X
W +R=1
(6 − W)(9 − R) + 3
X
W +R=0
(6 − W)(9 − R)
#
= 190.4 .
Problem 3
We have by definition
E[|X − Y |α
] =
Z Z
|x − y|α
fX,Y (x, y)dxdy
=
Z Z
|x − y|α
dxdy .
Since the area of region of the x−y plane where y x is equal to the area of the x−y plane
where y x, we can compute the above integral by doubling the integration domain y x
to give
2
Z 1
x=0
Z x
y=0
(x − y)α
dydx = 2
Z 1
x=0
(−1)
(x − y)α+1
α + 1
|x
0dx
=
2
α + 1
Z 1
x=0
xα+1
dx
=
2
α + 1
xα+2
α + 2
|1
0
=
2
(α + 1)(α + 2)
.
Problem 4
If X and Y are independent and uniform, then
P{X = x, Y = y} = P{X = x}P{Y = y} =

1
m
2
.
Then
E[|X − Y |] =
X
X
X
Y
|x − y|

1
m
2
.

We can sum over the set y x and double the summation range to evaluate this expectation.
Doing this we find that the above is equal to
2
m2
m
X
x=1
x−1
X
y=1
(x − y) .
Note that the second sum above goes to x − 1 since when y = x the term in the above
vanishes. Now evaluating the inner summation we have that
x−1
X
y=1
(x − y) = (x − 1) + (x − 2) + (x − 3) + · · · + 2 + 1 =
x−1
X
y=1
y =
x(x − 1)
2
.
So the above double sum then becomes
2
m2
m
X
x=1
x(x − 1)
2
=
1
m2
m
X
x=1
x2
−
m
X
x=1
x
!
.
Now remembering or looking up in tables we have that
m
X
x=1
x =
m(m + 1)
2
m
X
x=1
x2
=
m(m + 1)(2m + 1)
6
,
so that our expectation then becomes
E[|X − Y |] =
1
m2

m(m + 1)(2m + 1)
6
−
m(m + 1)
2

=
(m + 1)(m − 1)
3m
,
as requested.

Problem 5
Then the distance traveled by the ambulance is given by D = |X| + |Y | so our expected
value is given by
E[D] =
Z 1.5
−1.5
Z 1.5
−1.5
(|x| + |y|)

1
3

1
3

dxdy
=
1
9
Z 1.5
−1.5
Z 1.5
−1.5
(|x| + |y|)dxdy
=
1
9

3
Z 1.5
−1.5
|x|dx + 3
Z 1.5
−1.5
|y|dy

=
1
3

2
Z 1.5
0
xdx + 2
Z 1.5
0
ydy

=
2
3
x2
2
1.5
0
+
y2
2
1.5
0
!
=
1
3

9
4
+
9
4

=
3
2
.
Problem 6
Let Xi be the random variable that denotes the face on the die roll i. Then the total number
of die is the random variable
Z =
10
X
i=1
Xi .
Then the expectation of Z is given by
E[Z] =
10
X
i=1
E[Xi] .
Now since Xi are uniformly distributed discrete random variables between 1, 2, 3, 4, 5, and
6, we have
E[Xi] =
1
6
(1 + 2 + 3 + 4 + 5 + 6) = 3.5 .
So we then have that E[Z] = 10 7
2

= 35.
Problem 7
Part (a): We want to know the expected number of objects chosen by both A and B. A
will select three objects, then when B selects his three objects it is like the problem where
we have three special items from ten and the number selected that are special (in this case

selected by person A) is given by a hypergeomeric random variable with parameters N = 10,
m = 3, and n = 3. So
P{X = i} =

m
i

N − m
n − i

N
n
i = 0, 1, 2, · · · , m
This distribution has an expected value given by
E[X] =
nm
N
=
9
10
= 0.9 .
Part (b): After A has selected his three, then B will select three from ten where three of
these ten are “special” i.e. the ones that are picked by A. Let X be the random variable
that specifies the number of A objects that B selects. Then X is a hypergeometric random
variable with parameters N = 10, m = 3, and n = 3. So
P{X = i} =

m
i

N − m
n − i

N
n
i = 0, 1, 2, · · · , n
Then if B selects X of A’s selections then (N −3)−(3−X) are not chosen by either A or B.
Lets check this. If X = 0, meaning that B selects none of A picks we have that N − 6 items
are not chosen by A or B. If X = 3, then all three of A’s picks are selected by B and the
number of unselected items is N − 3, so yes our formula seems true. Thus the expectation
of the number of unselected items is
E[(N − 3) − (3 − X)] = N − 6 + E[X] = N − 6 +
nm
6
= 10 − 6 +
9
10
= 4.9 .
Part (c): As in Part (b) if the number chosen by both A and B is a hypergeometric random
variable with parameters N = 10, m = 3, and n = 3. Then the number of elements chosen
by only one person is
(3 − X) + (3 − X) .
Where the first term is the the number of A’s selections not selected by B and the second
term is the number of B selections not selected by A. Thus our random variable is 6 − 2X,
so the expectation is given by
6 − 2E[X] = 6 − 2
mn
N

= 6 − 2

9
10

=
21
5
= 4.2 .
Problem 8
Then a new table is started if and only if person i does not have any friends in the room.
There will be no friends for person i with probability P{N = 0} = qi−1
= (1 − p)i−1
.

Following the hint, we can let Xi be an indicator random variable denoting whether or not
the ith person starts a new table. Then the total number of new tables is
T =
N
X
i=1
Xi .
So the expectation of the number of new tables is given by
E[T] =
N
X
i=1
E[Xi] =
N
X
i=1
P(Xi) ,
Where the probability P(Xi) is computed above so
E[T] =
N
X
i=1
(1 − p)i−1
=
N−1
X
i=1
(1 − p)i
=
(1 − p)N
− (1 − p)0
1 − p − 1
=
1 − (1 − p)N
p
.
Problem 9
Part (a): We want to calculate the expected number of empty bins. Let N be a random
variable denoting the number of empty bins. Then
N =
n
X
i=1
Ii ,
where Ii is an indicator random variable which is one if bin i is empty and is zero if bin i is
not empty. The expectation of N is given by
E[N] =
n
X
i=1
E[Ii] =
n
X
i=1
P(Ii) .
Now P(Ii) is the probability that bin i is empty, so
P(Ii) =

i − 1
i

i
i + 1

i + 1
i + 2

· · ·

n − 1
n

=
i − 1
n
for 1 i n
This is because bin i can only be filled when we insert the ith ball and it will remain empty
with probability i−1
i
. When we place the i+1st ball it will avoid the ith bin with probability

i
i+1
, etc. Thus
E[N] =
n
X
i=1
i − 1
n
=
1
n
n
X
i=1
(i − 1)
=
1
n
n−1
X
i=1
i =
1
n
n(n − 1)
2
=
n − 1
2
.
Part (b): We need to find the probability that none of the urns are empty. When we place
the nth ball it must be placed in the nth bin (and no lower bin) for all bins to be filled. This
happens with probability 1
n
. When placing the n − 1th ball it must go in the n − 1th bin
and this will happen with probability 1
n−1
. Continuing in this manner we can work our way
down to the first ball which must go in bin number one. Thus our probability is
p = 1

1
2

1
2

1
3

· · ·

1
n − 1

1
n

=
n
Y
k=1
1
k
.
Problem 10
Part (a): Let the random variable X denote the number of success in our three trials. Then
X can be decomposed as X =
P3
i=1 Xi where Xi is a Bernoulli random variable with a value
of one if trial i is a success or 0 otherwise. Then E[X] =
P2
i=1 E[Xi] = 1.8. If each trial i,
has the the same probability of success p then E[Xi] = p and the equation earlier becomes
3p = 1.8 or p = 0.6 ,
I would then expect that P{X = 3} =

3
3

p3
q0
= p3
= (0.6)3
. Now by definition
E[X] = 0P{X = 0} + 1P{X = 1} + 2P{X = 2} + 3P{X = 3} = 1.8 .
So if we assume that the three trials do not have the same probability of success we can
maximize P{X = 3} by taking all of the other probabilities to be zero. This means that
3P{X = 3} = 1.8 ⇒ P{X = 3} = 0.6 ,
to impose the unit sum condition we can take P{X = 4} = 0.4 and P{X = 1} = P{X =
2} = 0. With P{X = 3} = 0.6 provides a desired realization of our probability space.
Part (b): To make P{X = 3} as small as possible take it to be zero. We now need to
specify the remaining probabilities such that they sum to one i.e.
P{X = 0} + P{X = 1} + P{X = 2} = 1 ,
the expectation of X is correct
0P{X = 0} + 1P{X = 1} + 2P{X = 2} = 1.8 ,

and of course 0 ≤ P{X = i} ≤ 1, the expectation calculation given P{X = 1} = 1.8 −
2P{X = 2} which when put into the sum to unity constraint gives
P{X = 0} + (1.8 − 2P{X = 2}) + P{X = 2} = 1 ,
so
p{X = 0} − P{X = 2} = −0.8 .
One easy solution to this equation (there are multiple) is to take P{X = 2} = 0.8 then
P{X = 0} = 0 and P{X = 1} = 0.2 so a probability scenario that result in P{X = 3}
minimum is
P{X = 0} = 0 , P{X = 1} = 0.2 , P{X = 2} = 0.8 , P{X = 3} = 0 .
Problem 11
We will have a change over, if a head changes to a tail (this happen with probability 1 − p)
or a tail changes to a head (this happens with probability p). Lets let the number of change
over be represented by the random variable N. Then following the hint, we can decomposed
N into a sum of Bernoulli random variables Xi. This is
N =
n−1
X
i=1
Xi ,
Now the random variable Xi takes the value one if the coin changes from a head to a tail
or from a tail to a head and is equal to zero if it does not change type (i.e. stays heads or
tails). Then
E[N] =
n−1
X
i=1
E[Xi] =
n−1
X
i=1
P(Xi) .
To evaluate E[N] we need to compute P(Xi). This will be
p(1 − p) + (1 − p)p = 2p(1 − p) ,
which can be seen by conditioning on the type of the coin flip. That is
P(Xi) = P{Xi = 1|(T, T)}P(T, T) + P{Xi = 1|(T, H)}P(T, H)
+ P{Xi = 1|(H, T)}P(H, T) + P{Xi = 1|(H, H)}P(H, H)
= 0 + (1 − p)p + p(1 − p) + 0 = 2p(1 − p) .
Thus we have that
E[N] = 2(n − 1)p(1 − p) .

Problem 13
Lets assume that the average person lives to be 100 years old. Let Ai be the indicator
random variable denoting if the person holding card i has an age that matches the number
on the card he is holding. Then let N be the random variable representing the total number
of people who’s age matches the card that they are holding. Then
N =
1000
X
i=1
Ai .
So that E[N] is given by
E[N] =
1000
X
i=1
E[Ai] =
1000
X
i=1
P{Ai} .
Now if we assume that people of all ages (20 to 100) are represented in our sample. We have
P{Xi} = 0 if i = 1, 2, · · · , 19, 101, 102, · · · so the above equals
100
X
i=20
P{Ai} =
100
X
i=20

1
100 − 20 + 1

= 1 .
Since for the people holding the cards numbered 20, 21, 22, · · · , 100 each has a 1
100−20+1

chance of having the correct date on it.
Problem 14
Let Xm be the number of draws or iterations required to take the urn from m black balls to
m − 1 black balls. Let Xm−1 be the number of iterations needed to take the urn from m − 1
black balls to m − 2 black balls, etc. Then the total number of stages needed is given by
N =
1
X
i=m
Xi ,
so that E[N] =
P1
i=m E[Xi]. Now to complete this problem we compute each E[Xi] in tern.
Now E[Xm] is the expected number of draws to reduce the number of black balls by one (to
m − 1). This will happen with probability 1 − p. Thus Xm is a geometric random variable
with probability of success given by 1 − p. Thus
P{Xm = i} = pi−1
(1 − p) for i = 1, 2, · · ·
This variable Xm has an expected value of 1
1−p
. This result holds for every random variable
Xi. Thus we have
E[N] =
1
X
i=m

1
1 − p

=
m
1 − p
.

Problem 15
Let Eij be an indicator random variable denoting if the man i and j form a matched pair.
Then let N be the random variable denoting the number of matched pairs. Then
N =
X
(i,j)
Ei,j ,
so the expectation of N is given by
E[N] =
X
(i,j)
E[Ei,j] =
X
(i,j)
P(Ei,j) .
Now P(Ei,j) = 1
N
1
N−1

and there are

N
2

total pairs in the sum. Thus
E[N] =

N
2

1
N(N − 1)

=
N(N − 1)
2
1
N
1
N − 1
=
1
2
.
Problem 16
Let
X =

Z Z X
0 otherwise
,
which defines a function f(·) of our random variable (this function has x as a parameter).
Now using the definition of the expectation of a random variable we have for E[X] the
following expression
E[X] =
Z
xp(x)dx =
Z
f(z)p(z)dz
=
Z ∞
x
zp(z)dz =
Z ∞
x
z
1
√
2π
e−z2/2
dz .
To evaluate this integral let v = z
√
2
so that dv = dz
√
2
and dz =
√
2dv. Then the above
integral becomes
E[X] =
Z ∞
x/
√
2
√
2v
√
2π
e−v2 √
2dv =
√
2
√
π
Z ∞
x/
√
2
ve−v2
dv
=
√
2
√
π
e−v2
2(−1)
∞
x/
√
2
=
1
√
2π
e−x2/2
Problem 17
Part (a): If we are not given any information about our earlier guesses then we must pick
one of the n! orderings of the cards and just count the number of matches we have. Let

Ai be an indicator random variable determining if the cards at position i match. Then
let N be the random variable denoting the number of matches. Thus N =
Pn
i=1 Ai so
E[N] =
Pn
i=1 E[Ai] =
Pn
i=1 P(Ai). But P(Ai) = 1
n
, since at position i we have one chance
in n of finding a match. Thus we have that
E[N] =
X
i=1
nP(Ai) =
n
X
i=1
1
n
=
n
n
= 1 ,
as claimed.
Part (b): The best strategy is to obviously not reguess any of the cards that one is shown
but at each stage pick uniformly from among the possible remaining n − i unrevealed cards.
Thus with this prescription and the definition of Ai as above we have that
E[N] =
n
X
i=1
P(Ai) .
Now in this case we have that
P(A1) =
1
n
P(A2) =
1
n − 1
P(A3) =
1
n − 2
.
.
.
P(A2) =
1
2
P(A1) = 1
Thus we have that
E[N] =
1
n
+
1
n − 1
+
1
n − 2
+ · · · +
1
2
+ 1 ≈
Z n
1
dx
x
= ln(n) .
Problem 18 (counting matched cards)
Let Ai be the event that when we turn over card i it matches the required cards face. For
example A1 is the event that turning over card one reveals an ace, A2 is the event that
turning over the second card reveals a deuce etc. The the number of matched cards N is
give by the sum of these indicator random variable as
N =
52
X
i=1
Ai .
Taking the expectation of this result and using linearity requires us to evaluate E[Ai] =
P(Ai). For card i the probability that when we turn it over it matches the expected face is
given by
P(Ai) =
4
52
,

since there are four suites that could match a given face. Thus we then have for the expected
number of matching cards that
E[N] =
52
X
i=1
E[Ai] =
52
X
i=1
P(Ai) = 52 ·
4
52
= 4 .
Problem 19
Part (a): We will catch type i = 2, 3, · · · , r insects with probability
Pr
i=2 Pi = 1−P1. Thus
the probability that we catch i − 1 insects of type 2, 3, · · · , r before we catch one of of type
one is given by a geometric random variable. This means that if X is the random variable
representing the trial that resulted in catching a type one insect then
P{X = i} = (1 − P1)i−1
P1 .
We are interested in the expected value of the number of type 2, 3, · · · , r insects caught
before catching one of type one. Thus as a random variable N in terms of X is given by
N = X − 1. With this we see that
E[N] = E[X] − 1 =
1
P1
− 1 =
P1 − 1
P1
.
Part (b): We can compute the mean number of insect types before catching one of type one
by using conditional expectation. Let N be the random variable that denotes the different
number of insects caught before catching one of type one. Also let the random variable K be
the total number of insects caught when we catch our first type one insect. This means that
on catch K we catch our first type one insect. Then conditioning on this random variable
K we have
E[N] = E[E[N|K]] .
Now to evaluate E[N|K] we recognize that this is the expectation number of insects from
2, 3, · · · , r caught when we catch our type one insect on catch k. Then since for the catches
1, 2, 3, · · · , k − 1 we are selecting from the insect types 2, 3, · · · , r, each specific insect type
is caught with probability
P2
Pr
i=2 Pi
,
P3
Pr
i=2 Pi
, · · · ,
Pr
Pr
i=2 Pi
.
Now since r
X
r=2
Pi = 1 − P1 ,
the above terms are equivalent to
P2
1 − P1
,
P3
1 − P1
, · · ·
Pr
1 − P1
.

From Example 3d in this chapter, the expected number of different insects caught is given
by
E[N|K] = (r − 1) −
r
X
i=2

1 − P̃i
k−1
= (r − 1) −
r
X
i=2

1 −
Pi
1 − P1
k−1
.
Taking the outer expectation over the random variable K we have
E[N] =
∞
X
k=1
(r − 1) −
r
X
r=2

1 −
Pi
1 − P1
k−1
!
P{K = k} .
Since K is a geometric random variable, we have that
P{K = k} = (1 − P1)k−1
P1 ,
which gives E[N] as
E[N] =
∞
X
k=1
(r − 1)P{K = k} −
r
X
i=2

1 −
Pi
1 − P1
k−1
P1
= (r − 1) − P1
∞
X
k=1
r
X
i=2
(1 − P1 − Pi)k−1
= (r − 1) − P1
r
X
i=2
∞
X
k=1
(1 − P1 − Pi)k−1
= (r − 1) − P1
r
X
i=2
1
P1 + Pi
.
Problem 21 (more birthday problems)
Let Ai,j,k be an indicator random variable if persons i, j, and k have the same birthday and
no one else does. Then if we let N denote the random variable representing the number of
groups of three people all of whom have the same birthday we see that N is given by a sum
of these random variables as
N =
X
ijk
Ai,j,k .
Then taking the expectation of the above expression we have
E[N] =
X
ijk
E[Ai,j,k] .
Now there are

100
3

terms in the above sum (since there are one hundred total people
and our sum involves all subsets of three people), and the probability of each event Ai,j,k

happening is given by
P(Ai,j,k) =
1
3652

1 −
1
365
100−3
=
1
3652

364
365
97
since person j and person k’s birthdays must match that of person i, and the remaining 97
people must have different birthdays (the problem explicitly states we are looking for the
expected number days that are the birthday of exactly three people and not more). Thus
the total expectation of the number of groups of three people that have the same birthday
is then given by
E[N] =

100
3

1
3652

364
365
97
= 0.93014 ,
in agreement with the back of the book.
Part (b): Note: the following does not match the back of the book, if anyone
sees anything incorrect with this argument please let me know.
Let Ai be the event that the ith person has a distinct birthday, i.e. the event that the ith
person has a different birthday then all the other people. Let Ii be an indicator random
variable taking the value one if this event is true and zero otherwise. Then the number of
distinct birthdays is given by
X =
n
X
i=1
Ii ,
so the expected number of distinct birthdays in then
E[X] =
n
X
i=1
E[Ii] =
n
X
i=1
P(Ai) ,
Now
P(Ai) =

364
365
n−1
,
since none of the other n − 1 people can have the same birthday as person i. Thus
E[X] = nP(A1) = n

364
365
n−1
.
When n = 100 this becomes
100

364
365
99
= 76.21 .
Problem 22 (number of times to roll a fair die to get all six sides)
This is exactly like the coupon collecting problem where we have six coupons with a prob-
ability of obtaining any one of them given by 1/6. Then this problem is equivalent to

determining the expected number of coupons we need to collect before we get a complete
set. From Example 2i from the book we have the expected number of rolls X to be given by
E[X] = N

1 +
1
2
+ · · · +
1
N − 1
+
1
N

when N = 6 this becomes
E[X] = 6

1 +
1
2
+ · · · +
1
5
+
1
6

= 14.7 .
Problem 26
Part (a): The density for the random variable X(n) defined as X(n) = max(X1, X2, · · · , Xn)
where Xi is drawn from the distribution function F(x) (density function f(x)) is given by
fX(n)
(x) =
n!
(n − 1)!
(F(x))n−1
f(x) = nF(x)n−1
f(x) .
When Xi is a uniform random variable between [0, 1] we have f(x) = 1, and F(x) = x, so
that the above becomes
fX(n)
(x) = nxn−1
.
Then the expectation of this random variable is given by
E[X(n)] =
Z 1
0
x(nxn−1
)dx =
nxn+1
n + 1
1
0
=
n
n + 1
.
Part (b): The minimum random variable X(1) defined by X(1) = min(X1, X2, · · · , Xn) has
a distribution function given by
fX(1)
(x) = n(1 − F(x))n−1
f(x) .
Again when Xi is a uniform random variable our expectation is given by
E[X(1)] =
Z 1
0
xn(1 − x)n−1
dx
= n

x(1 − x)n
(−1)
n
1
0
−
Z 1
0
(1 − x)n
n
(−1)dx
#
=
Z 1
0
(1 − x)n
dx =
(1 − x)n+1
(−1)
n + 1
1
0
=
1
n + 1
.

Problem 30 (a squared expectation)
We find, by expanding the quadratic and using independence, that
E[(X − Y )2
] = E[X2
− 2XY + Y 2
] = E[X2
] − 2E[X]E[Y ] + E[Y 2
] .
In terms of the variance E[X2
] is given by E[X2
] = Var(X) + E[X]2
so the above becomes
E[(X − Y )2
] = Var(X) + E[X]2
− 2E[X]E[Y ] + Var(Y ) + E[Y ]2
= σ2
+ µ2
− 2µ2
+ σ2
+ µ2
= 2σ2
.
Problem 33 (evaluating expectations and variances)
Part (a): We find, expanding the quadratic and using the linearity property of expectations
that
E[(2 + X)2
] = E[4 + 4X + X2
] = 4 + 4E[X] + E[X2
] .
In terms of the variance, E[X2
] is given by E[X2
] = Var(X) + E[X]2
, both terms of which
we know from the problem statement. Using this the above becomes
E[(2 + X)2
] = 4 + 4(1) + (5 + 12
) = 14 .
Part (b): We find, using properties of the variance that
Var(4 + 3X) = Var(3X) = 9Var(X) = 9 · 5 = 45 .
Problem 40
If
f(x, y) =
1
y
e−(y+x/y)
for x 0 , y 0
then by the definition of the expectation we have that
E[X] =
Z ∞
x=0
xf(x)dx =
Z ∞
x=0
x
Z ∞
y=0
f(x, y)dydx
=
Z ∞
x=0
Z ∞
y=0

x
y

e−(y+ x
y )dydx
=
Z ∞
y=0
Z ∞
x=0

x
y

e−(y+ x
y )dxdy .
The last two lines are obtained by exchanging the order of the integration. To integrate this
expression with respect to x, let v be defined as v = x
y
, so that dv = dx
y
, and the above
expression becomes
E[X] =
Z ∞
y=0
Z ∞
v=0
ve−(y+v)
ydvdy
=
Z ∞
y=0
ye−y
Z ∞
v=0
ve−v
dvdy .

Now evaluating the v integral using integration by parts we have
Z ∞
0
ve−v
dv = −vev
|∞
0 +
Z ∞
0
e−v
dv
= −e−v ∞
0
= −(0 − 1) = 1 .
With this the expression for E[X] becomes
E[X] =
Z ∞
y=0
ye−y
1dy = 1 .
Now in the same way
E[Y ] =
Z ∞
y=0
y
Z ∞
x=0
f(x, y)dxdy =
Z ∞
y=0
y
Z ∞
x=0
1
y
e−(y+ x
y )dxdy .
To evaluate the x integral, let v = x
y
then dv = dx
y
and we have the above equal to
E[Y ] =
Z ∞
y=0
y
Z ∞
v=0
1
y
e−y
e−v
ydvdy
=
Z ∞
y=0
ye−y
Z ∞
v=0
e−v
dvdy
=
Z ∞
y=0
ye−y
−e−v ∞
v=0
dy
=
Z ∞
y=0
ye−y
dy
= − ye−y ∞
0
+
Z ∞
y=0
e−y
dy = 1 .
Finally to compute Cov(X, Y ) using the definition we require the calculation of E[XY ]. This
is given by
E[XY ] =
Z ∞
0
Z ∞
0
xy
1
y
e−(y+ x
y
)
dxdy
=
Z ∞
y=0
Z ∞
x=0
xe−(y+ x
y
)
dxdy .
To perform the x integration let v = x
y
so that dv = dx
y
E[XY ] =
Z ∞
y=0
Z ∞
v=0
yve−(y+v)
ydvdy
=
Z ∞
y=0
y2
e−y
Z ∞
v=0
ve−v
dvdy .
Since
R ∞
0
ve−v
dv = 1, the above equals
Z ∞
0
y2
e−y
dy = −y2
e−y ∞
0
+ 2
Z ∞
0
ye−y
dy = 2 .
Then
Cov(X, Y ) = E[XY ] − E[X]E[Y ] = 2 − 1(1) = 1 .

Problem 45
To be pairwise uncorrelated means that Cor(Xi, Xj) = 0 if i 6= j.
Part (a): We have
Cov(X1 + X2, X1 + X2) =
2
X
i=1
3
X
j=2
Cov(Xi, Xj)
= Cov(X1, X2) + Cov(X1, X3)
+ Cov(X2, X2) + Cov(X2, X3) .
Using the fact that that these variables are pairwise uncorrelated the right hand side of the
above equals
0 + 0 + 12
+ 0 = 1 .
The correlation between to random variables X and Y is (defined as)
ρ(X, Y ) =
Cov(X, Y )
p
Var(X)Var(Y )
follows once we have the two variances. We now compute these variances
Var(X1 + X2) = Var(X1) + Var(X2) + 2Cov(X1, X2)
= Var(X1) + Var(X2) = 1 + 1 = 2 .
In the same way Var(X2 + X3) = 2, so that
ρ(X1 + X2, X2 + X3) =
1
√
2
Cov(X1 + X2, X3 + X4) = Cov(X1, X3) + Cov(X1, X4)
+ Cov(X2, X3) + Cov(X2, X4) = 0 .
so obviously then ρ(X1 + X2, X3 + X4) = 0 regardless of the value of the variances.
Problem 48 (conditional expectation of die rolling)
Part (a): The probability that the first six is rolled on the nth roll is given by a geometric
random variable with parameter p = 1/6. Thus the expected number of rolls to get a six is
given by
E[X] =
1
p
= 6 .

Part (b): We want to evaluate E[X|Y = 1]. Since in this expectation we are told that the
first roll of our dice results in a five we have that
E[X|Y = 1] = 1 + E[X] = 1 +
1
p
= 1 + 6 = 7 ,
since after the first roll we again have that the number of rolls to get the first six is a
geometric random variable with p = 1/6.
Part (c): We want to evaluate E[X|Y = 5], which means that the first five happens on the
fifth roll. Thus the rolls 1, 2, 3, 4 all have a probability of 1/5 to show a six. After the fifth
roll, there are again six possible outcomes of the die so the probability of obtaining a six is
given by 1/6. Defining the event A to be the event that we do not roll a six in any of the
first four rolls (and implicitly given that the first five happens on the fifth roll) we see that
P(A) =

4
5
4
= 0.4096 ,
since with probability of 1/5 we will roll a six and with probability 4/5 we will not roll a six.
With this definition and using the definition of expectation we find that
E[X|Y = 5] = 1

1
5

+ 2

4
5

1
5
+ 3

4
5
2
1
5
+ 4

4
5
3
1
5
+
∞
X
k=6
k P(A)

5
6
k−6
1
6
!
.
We will evaluate this last sum numerically. This is done in the Matlab file chap 7 prob 48.m,
where we find that
[X|Y = 5] = 5.8192 ,
in agreement with the book.
Problem 49 (misshaped coins)
Note: This result does not match the back of the book. If anyone sees any errors
in what I have done please contact me. We desire to compute the conditional expected
number of heads in our ten flips. Let N be the random variable specifying the number of
heads obtained when we flip our coin ten ten times. Let E be the event that on the first
three flips we obtain two heads and one tail. Then E[N|E] can be computed by conditioning
on the misshapen coin chosen. Let A be the event that we select the coin with p = 0.4. Then
E[N|E] = E[N|A, E]P(A) + E[N|Ac
, E]P(Ac
) .
Assuming that each coin is selecting uniformly with probability 1
2
we need to compute
E[N|A, E]. The easiest way to do this is to notice that this is two plus the expectation
of a binomial random variable with parameters (n, p) = (7, 0.4). Since two of the first three
flips resulted in a head. Thus
E[N|A, E] = 2 + 0.4(7) =
24
5
.

In the same way
E[N|Ac
, E] = 2 + 0.7(7) =
69
10
.
Thus
E[N|E] =
1
2

48
10
+
69
10

=
117
20
.
Problem 50 (compute E[X2
|Y = y])
By definition, the requested expectation is given by
E[X2
|Y = y] =
Z ∞
0
x2
f(x|Y = y)dx .
Lets begin by computing f(x|Y = y), using the definition of this density in terms of the
joint density
f(x|y) =
f(x, y)
f(y)
.
Since we are given f(x, y) we begin by first computing f(y). We find that
f(y) =
Z ∞
0
f(x, y)dx =
Z ∞
0
e−x/y
e−y
y
dx
=
e−y
y
Z ∞
0
e−x/y
dx =
e−y
y
(−y) e−x/y ∞
0
= e−y
.
So that f(x|y) is given by
f(x|y) =
e−x/y
e−y
y
ey
=
e−x/y
y
.
With this expression we can evaluate our expectation above. We have (using integration by
parts several times)
E[X2
|Y = y] =
Z ∞
0
x2 e−x/y
y
dx
=
1
y
Z ∞
0
x2
e−x/y
dx
=
1
y

x2
(−y)e−x/y ∞
0
−
Z ∞
0
2x(−y)e−x/y
dx

= 2
Z ∞
0
xe−x/y
dx
= 2

x(−y)e−x/y ∞
0
−
Z ∞
0
(−y)e−x/y
dx

= 2y
Z ∞
0
e−x/y
dx
= 2y(−y) e−x/y ∞
0
= 2y2
.

Problem 51 (compute E[X3
|Y = y])
By definition, the requested expectation is given by
E[X3
|Y = y] =
Z
x3
f(x|Y = y)dx .
Lets begin by computing f(x|Y = y), using the definition of this density in terms of the
joint density
f(x|y) =
f(x, y)
f(y)
.
Since we are given f(x, y) we begin by first computing f(y). We find that
f(y) =
Z y
0
f(x, y)dx =
Z y
0
e−y
y
dx = e−y
.
So that f(x|y) is given by
f(x|y) =
e−y
y
ey
=
1
y
.
With this expression we can evaluate our expectation above. We have
E[X3
|Y = y] =
Z y
0
x3 1
y
dx =
1
y
x4
4
y
0
=
y3
4
.
Problem 52 (the average weight)
Let W denote the random variable representing the weight of a person selected from the
total population. Then we can compute E[W] by conditioning on the subgroups. Letting
Gi denote the event we are drawing from subgroup i, we have
E[W] =
r
X
i=1
E[W|Gi]P[Gi] =
r
X
i=1
wipi .
Problem 53 (the time to escape)
Let T be the random variable denoting the number of days until the prisoner reaches freedom.
We can evaluate E[T] by conditioning on the door selected. If we denote Di be the event
the prisoner selects door i then we have
E[T] = E[T|D1]P(D1) + E[T|D2]P(D2) + E[T|D3]P(D3) .
Each of the above expressions can be evaluated. For example if the prisoner selects the first
door then after two days he will be right back where he started and thus has in expectation
E[T] more days left. Thus
E[T|D1] = 2 + E[T] .

Using logic like this we see that E[T] can be expressed as
E[T] = E[T|D1]P(D1) + E[T|D2]P(D2) + E[T|D3]P(D3)
= (2 + E[T])(0.5) + (4 + E[T])(0.3) + (1)(0.2) .
Solving the above expression for E[T] we find that E[T] = 12.
Problem 56
Let M be the random variable representing the number of people who enter an elevator on
the ground floor. Then once we are loaded up with the M people then we can envision each
of the M people uniformly selecting one of the N floors to get off on. This is exactly the
same as counting the number of different coupons collected with probability of selecting each
type to be 1
N
. Thus we can compute the expected number of stops made by the elevator by
conditioning on the number of passengers loaded in the elevator initially and the result from
Example 3d (the expected number of distinct coupons when drawing M). For example, let
X be the random variable denoting the number of stops made when M passengers are on
board. Then we want to compute
E[X] = E[E[X|M]] .
Now E[X|M = m] is given by the result of Example 3d so that
E[X|M = m] = N −
N
X
i=1

1 −
1
N
m
= N − N

1 −
1
N
m
.
Thus the total expectation of X is then given by
E[X] =
∞
X
m=0
E[X|M = m]P{M = m}
=
∞
X
m=0

N − N

1 −
1
N
m
P{M = m}
= N
∞
X
m=0
P{M = m} − N
∞
X
m=0

1 −
1
N
m
P{M = m}
= N − N
∞
X
m=0

1 −
1
N
m
e−10
10m
m!
= N − Ne−10
∞
X
m=0
10 1 − 1
N
m
m!
= N − Ne−10
exp{10

1 −
1
N

}
= N

1 − exp{−10 + 10 −
1
N
}

= N

1 − e− 10
N

Problem 57
Let NA be the random variable denoting the number of accidents in a week. Then E[NA] = 5.
Let NI be the random variable denoting the number of injured when an accident occurs.
Let N be the total number of workers injured each week. Then E[N] can be calculated by
conditioning on the number of accidents in a given week NA as
E[N] = E[E[N|NA]] .
Now we are told that the number of workers injured in each accident is independent of the
number of accidents that occur. We then have that
E[E[N|NA]] = E[N|NA] · E[NA] = 2.5 · 5 = 12.5 .
Problem 58 (flipping a biased coin until a head and a tail appears)
Part (a): We reason as follows if the first flip lands heads then we will continue to flip
until a tail appears at which point we stop. If the first flip lands tails we will continue to
flip until a head appears. In both cases the number of flips required until we obtain our
desired outcome (a head and a tail) is a geometric random variable. Thus computing the
desired expectation is easy once we condition on the result of the first flip. Let H denote
the event that the first flip lands heads then with N denoting the random variable denoting
the number of flips until both a head and a tail occurs we have
E[N] = E[N|H]P{H} + E[N|Hc
]P{Hc
} .
Since P{H} = p and P{Hc
} = 1 − p the above becomes
E[N] = pE[N|H] + (1 − p)E[N|Hc
] .
Now we can compute E[N|H] and E[N|Hc
]. Now E[N|H] is one plus the expected number
of flips required to obtain a tail. The expected number of flips required to obtain a tail is
the expectation of a geometric random variable with probability of succuss 1 − p and thus
we have that
E[N|H] = 1 +
1
1 − p
.
The addition of the one in the above expression is due to the fact that we were required to
performed one flip to determining what the first flip was. In the same way we have
E[N|Hc
] = 1 +
1
p
.
With these two sub-results we have that E[N] is given by
E[N] = p +
p
1 − p
+ (1 − p) +
1 − p
p
= 1 +
p
1 − p
+
1 − p
p
.
Part (b): We can reason this probability as follows. Since once the outcome of the first coin
flip is observed we repeatedly flip our coin as many times as needed to obtain the opposite

face we see that we will end our experiment on a head only if the first coin flip is a tail.
Since this happens with probability 1 − p this must also be the probability that the last flip
lands heads.
Problem 61
Part (a): Conditioning on N we have that
P{M ≤ x} =
∞
X
n=1
P{M ≤ x|N = n}P{N = n} .
Now for a geometric random variable N with parameter p we have that P{N = n} =
p(1 − p)n−1
for n ≥ 1 so we have that
P{M ≤ x} =
∞
X
n=1
P{M ≤ x|N = n}p(1 − p)n−1
.
From the discussion in Chapter 6 we see that P{M ≤ x|N = n} = F(x)n
so the above
becomes
P{M ≤ x} =
∞
X
n=1
F(x)n
p(1 − p)n−1
= pF(x)
∞
X
n=1
F(x)n−1
(1 − p)n−1
= pF(x)
∞
X
n=0
(F(x)(1 − p))n
= pF(x)
1
1 − F(x)(1 − p)
=
pF(x)
1 − (1 − p)F(x)
.
Part (b): By definition we have P{M ≤ x|N = 1} = F(x)
Part (c): To evaluate P{M ≤ x|N 1} we can again condition on N to obtain
P{M ≤ x|N 1} =
∞
X
n=1
P{M ≤ x, N = n|N 1}
=
∞
X
n=2
P{M ≤ x, N = n|N 1}
=
∞
X
n=2
P{M ≤ x|N = n, N 1}P{N = n|N 1}
=
∞
X
n=2
P{M ≤ x|N = n}P{N = n|N 1} .

Now as before we have that P{M ≤ x|N = n} = F(x)n
and that
P{N = n|N 1} =
P{N = n, N 1}
P{N 1}
=
P{N = n}
1 − p
=
p(1 − p)n−1
1 − p
= p(1 − p)n−2
.
Thus we have that
P{M ≤ x|N 1} =
p
1 − p
∞
X
n=2
F(x)n
(1 − p)n−2
=
pF(x)2
1 − p
∞
X
n=0
(F(x)(1 − p))n
=
pF(x)2
1 − p

1
1 − F(x)(1 − p)

.
Part (d): Conditioning on N = 1 and N 1 we have that
P{M ≤ x} = P{M ≤ x|N = 1}P{N = 1} + P{M ≤ x|N 1}P{N 1}
= F(x)p +
pF(x)2
(1 − p)
1 − F(x)(1 − p)
= pF(x)

1 +
F(x)(1 − p)
1 − F(x)(1 − p)

=
pF(x)
1 − F(x)(1 − p)
.
This is the same as in Part (a)!.
Problem 62
Defining N(x) = min{n :
Pn
i=1 Ui x}
Part (a): Let n = 0 then P{N(x) ≥ 1} = 1 since we must have at least one term in our sum.
Lets also derive the expression for the case n = 1. We see that P{N(x) ≥ 2} = P{U1 x},
because the event that we need at least two random draws is equivalent to the event that the
first random draw is less than x. This is equivalent to the cumulative distribution function
for a uniform random variable so equals FU (a) = a and therefore
P{N(x) ≥ 2} = x .
Now lets assume (to be shown by induction) that
P{N(x) ≥ k + 1} =
xk
k!
for k ≤ n ,

We want to compute P{N(x) ≥ k + 2} which we will do by conditioning on the value U1.
Thus we have
P{N(x) ≥ k + 2} =
Z x
u1=0
P{N(x) ≥ k + 2|U1 = u1}P{U1 = u1}du1
=
Z x
u1=0
P{N(x) ≥ k + 2|U1 = u1}du1 .
To evaluate P{N(x) ≥ k + 2|U1 = u1} we note that it is the probability that we require
k + 2 or more terms in our sum to create a sum that is larger than x. Given that the first
random variable U1 is equal to u1. We know that because we require a at least k + 1 terms
that this value of u1 must be less than x. This puts the upper limit on the integral of x and
we see that the expression P{N(x) ≥ k + 2|U1 = u1} is equivalent to
P{N(x − u1) ≥ k + 1} =
(x − u1)k
k!
,
by the induction hypothesis. Our integral above becomes
Z x
0
(x − u1)k
k!
du1 = −
(x − u1)k+1
(k + 1)!
x
0
= −

0 −
xk+1
(k + 1)!

=
xk+1
(k + 1)!
which is what we were trying to prove.
With this expression we can evaluate the expectation of N(x) by using the identity that
E[N] =
∞
X
n=0
P{N ≥ n + 1} ,
which is proven in Problem 2 of the theoretical exercises. With the expression for P{N ≥
n + 1} above we find that
E[N] =
X
n≥0
xn
n!
= ex
,
as expected.
Problem 63 (Cov(X, Y ))
Warning: For some reason this solution does not match the answer given in the back of
the book. If anyone knows why please let me know. I have not had as much time as I would
have liked to go over this problem, careat emptor.
Part (a): With X and Y as suggested we have
Cov(X, Y ) = Cov
10
X
i=1
Xi,
3
X
j=1
Yj
!
=
10
X
i=1
3
X
j=1
Cov(Xi, Yj) .

Here Xi is a Bernoulli random variable specifying if red ball i is drawn or not. Since defined
in this way Xi and Yj are independent and the above factorization is valid. For two Bernoulli
random variables we have
Cov(Xi, Yj) = E[XiYj] − E[Xi]E[Yj] .
We have
E[XiYj] = P(XiYj) =
10
30
8
29
E[Xi] = P(Xi) =
10
30
E[Yj] = P(Yj) =
8
30
1
18
.
Thus Cov(Xi, Yj) = 80
30
1
29
− 1
30

.
Part (b): Cov(XY ) = E[XY ] − E[X]E[Y ]. Now to compute E[X], we recognized that X
is a hypergeomeric random variable with parameters N = 18, m = 10, and n = 12, so that
E[X] =
10(12)
18
=
20
3
.
To compute E[Y ] we recognize that Y is a hypergeometric random variable with parameter
N = 18, m = 8, n = 12 so
E[Y ] =
8(12)
18
=
16
3
.
Finally to compute E[XY ] we condition on X (or Y ) as suggested in the book as E[XY ] =
E[E[XY |Y ]]. Now
E[XY |Y = y] = E[Xy|Y = y] = yE[X|Y = y] .
The probability X|Y = y is a hypergeometric random variable with parameters N = 18 − y,
m = 10, n = 12 − y, for 0 ≤ y ≤ 12 and so has an expectation given by
10(12 − y)
18 − y
.
Thus we have E[XY |Y ] = Y

10(12−Y )
18−Y

, so that
E[XY ] =
X
Y
Y

10(12 − Y )
18 − Y

P{Y }
=
8
X
y=0
y

10(12 − y)
18 − y

8
y

18 − 8
12 − y

18
12

Problem 64
Part (a): We can compute this expectation by conditioning on the type of light bulb
selected. Let the event T1 be the event that we select the type one light bulb and T2 be the
event that we select the type two light bulb. Then
E[X] = E[X|T1]P(T1) + E[X|T2]P(T2) = µ1p + µ2(1 − p) .
Part (b): Again conditioning on the type of light bulb selected we have
E[X2
] = E[X2
|T1]P(T1) + E[X2
|T2]P(T2) .
Now for the these Gaussians we have in terms of the variables of the problem that E[X2
|T1] =
Var(X|T1) + E[X]2
= σ2
1 + µ2
1. So the value of E[X2
] the becomes
E[X2
] = p(σ2
1 + µ2
1) + (1 − p)(σ2
1 + µ2
1) .
Thus Var(X) is then given by
Var(X) = E[X2
] − E[X]2
= p(σ2
1 + µ2
1) + (1 − p)(σ2
1 + µ2
1) − (µ1p + µ2(1 − p))2
= p(1 − p)(µ2
1 + µ2
2) + pσ2
1 + (1 − p)σ2
2 − 2p(1 − p)µ1µ2 ,
after some simplification. Not that this problem can also be solved using the conditional
variance formula. The conditional variance formula is given by
Var(X) = E(Var(X|Y )) + Var(E[X|Y ]) .
Since E[X|T1] = µ1, and E[X|T1] = µ2, the variance of E[X|T] and the second term in the
conditonal variance formula is given by
Var(E[X|T]) = E[E[X|T]2
] − E[E[X|T]]2
= µ2
1p + µ2
2(1 − p) − (µ1p + µ2(1 − p))2
.
Also the random variable Var(X|Y ) can be computed by recognizing that
Var(X|T1) = σ2
1 and
Var(X|T2) = σ2
2 ,
so that
E[Var(X|T)] = σ2
1p + σ2
2(1 − p) .
Putting all of these pieces together we find that
Var(X) = σ2
1p + σ2
2(1 − p) + µ2
1p + µ2
2(1 − p) − (µ1p + µ2(1 − p))2
= pµ2
1(1 − p) + (1 − p)pµ2
2 + pσ2
1 + (1 − p)σ2
2 − 2p(1 − p)µ1µ2, ,
the same result as before.

Problem 65 (bad winter storms)
We can compute the expectation of the number of storms by conditioning on the type of
winter we will have. If we let G be the event that the winter is good and B be the event
that the winter is bad then we have (with N the random variable denoting the number of
winter storms) the following
E[N] = E[N|G]P(G) + E[N|B]P(B)
= 3(0.4) + 5(0.6) = 4.2 .
To compute the variance we will use the conditional variance formula given by
Var(N) = E[Var(N|Y )] + Var(E[N|Y ]) ,
where Y is the random variable denoting the type of winter. We will compute the first term
on the right hand side of this expression first. Since the variances given the type of storm
are known i.e.
Var(N|Y = G) = 3 and
Var(N|Y = B) = 5 ,
by the fact that a Poisson random variable has equal means and variances. Thus the expec-
tation of these variances can be calculated as
E[Var(N|Y )] = 3(0.4) + 5(0.6) = 4.2 .
Now to compute the second term in the conditional variance formula we recall that
E[N|Y = G] = 3 and
E[N|Y = B] = 5 ,
so that using the definition of the variance, the variance of the random variable E[N|Y ] is
given by
Var(E[N|Y = G]) = (3 − 4.2)2
(0.4) + (5 − 4.2)2
(0.6) = 0.96 .
Combining these two components we see that
Var(N) = 4.2 + 0.96 = 5.16 .
Problem 66 (our miners variance)
Following the example in the book we can compute E[X2
] in much the same way as in
example 5c. By conditioning on the door taken we have that
E[X2
] = E[X2
|Y = 1]P{Y = 1} + E[X2
|Y = 2]P{Y = 2} + E[X2
|Y = 3]P{Y = 3}
=
1
3
(E[X2
|Y = 1] + E[X2
|Y = 2] + E[X2
|Y = 3]) .

But now we have to compute E[X2
|Y ] for the various possible Y values. The easiest to
compute is E[X2
|Y = 1] which would equal 32
= 9 since when our miner selects the first
door he is able to leave the mine in three hours. The other two expectations are computed
using something like a “no memory” property of this problem. As an example if the miner
takes the second door Y = 2 then after five hours he returns back to the mine exactly where
he started. Thus the expectation of X2
, given that he takes the second door, is equal to the
expectation of (5+X)2
with no information as to the next door he may take. Mathematically,
expressing this we then have
E[X2
|Y = 2] = E[(5 + X)2
] and
E[X2
|Y = 3] = E[(7 + X)2
] .
Expanding the quadratic in the above expectations we find that
E[X2
|Y = 2] = E[25 + 10X + X2
] = 25 + 10E[X] + E[X2
] = 175 + E[X2
]
E[X2
|Y = 3] = E[49 + 14X + X2
] = 49 + 14E[X] + E[X2
] = 259 + E[X2
] .
Using the previously computed result that E[X] = 15. Thus when we put these expressions
in our expansion of E[X2
] above we find that
E[X2
] =
1
3
(9 + 175 + E[X2
] + 259 + E[X2
]) ,
or upon solving for E[X2
] gives E[X2
] = 443. We can then easily compute the variance of
X. We find that
Var(X) = E[X2
] − E[X]2
= 443 − 152
= 218 .
Problem 67 (gambling with the Kelly strategy)
Let En be the expected fortune after n gambles of a gambler who uses the kelly strategy.
Then we are told that E0 = x (in fact in this case this is his exact forture i.e. there no
expectation). Now we can compute in terms of En−1 by conditioning on whether we win or
loose. At time n − 1 we have a fortune of En−1 and we bet 2p − 1 of this fortune. Thus if
we win (which happens with probability p) we will then have En−1 + (2p − 1)En−1. While
if we loose (which happens with probabilty 1 − p) we will then have En−1 − (2p − 1)En−1.
Thus En our expected fortune at time n is then given by
En = (En−1 + (2p − 1)En−1)p + (En−1 − (2p − 1)En−1)(1 − p)
= En−1p + En−1(1 − p) + En−1{(2p − 1)p − (2p − 1)(1 − p)}
= En−1 + (2p − 1)2
En−1
= (1 + (2p − 1)2
)En−1 for n = 1, 2, · · · .
Writting this expression out for n = 1, 2, · · · and by using induction we see that En is given
by
En = (1 + (2p − 1)2
)n
E0 = (1 + (2p − 1)2
)n
x .

Problem 68 (Poisson accidents)
Let E2 be the event that the person has a number of accidents (per year) given by a Poisson
random variable with λ = 2 and E3 the event that the person has a number of accidents
(again per year) given by a Poisson random variable with λ = 3. Then the probability a
person has k accidents can be computed by conditioning on the type of person someone is
i.e. whether they are of E2 or of E3 type. We then find (if N is the random variable denoting
the number of accidents a person has this year)
P{N = k} = P{N = k|E2}P(E2) + P{N = k|E3}P(E3)
= 0.6

e−2
2k
k!

+ 0.4

e−3
3k
k!

.
Part (a): Evaluating the above for k = 0 we find that
P{N = 0} = 0.6e−2
+ 0.4e−3
= 0.101 .
Part (b): Evaluating the above for k = 3 we find that
P{N = 3} = 0.6

e−2
23
3!

+ 0.4

e−3
33
3!

= 0.1978 .
If we have no accidents in the previous year this information will change the probability that
a person is a type E2 or a type E3 person. Specifically, if Y0 is the information/event that
our person had no accidents in the previous year, the calculation we now want to evaluate is
P{N = k|Y0} = P{N = k|E2, Y0}P(E2|Y0) + P{N = k|E3, Y0}P(E3|Y0)
= P{N = k|E2}P(E2|Y0) + P{N = k|E3}P(E3|Y0) .
Where P(Ei|Y0) is the probability the person is of “type”, Ei, given the information about
no accidents. We are also assuming that N is conditionally independent of Y0 given Ei i.e.
P{N = k|Ei, Y0} = P{N = k|Ei}. We can compute the conditional probabilities P(Ei|Y0)
with Bayes’ rule. We find
P(E2|Y0) =
P(Y0|E2)P(E2)
P(Y0|E2)P(E2) + P(Y0|E3)P(E3)
,
and the same type formula for P(E3|Y0). Now we have computed the denominator of the
above expression in Part (a) above. Thus we find that
P(E2|Y0) =
(e−2
)(0.6)
P{N = 0}
= 0.803
P(E3|Y0) =
(e−3
)(0.4)
P{N = 0}
= 0.196 .
With these two expressions we can calculate the probability we obtain any number of ac-
cidents in the next year. Incorporating the information that the event Y0 happend that
P{N = k} is given by
P{N = k|Y0} = 0.803

e−2
2k
k!

+ 0.196

e−3
3k
k!

.

Evaluating this expression for k = 3 we find that P{N = 3|Y0} = 0.18881. The information
that the our person had no accidents in the previous year reduced the probability that they
will have three accidents this year (computed above) as one would expect. These calculations
can be found in the file chap 7 prob 68.m.
Problem 70
Part (a): We want to calculate P{F1 = H} or the probability that the first flip is heads.
We will do this by conditioning on the coin that is choosen. Let P{F1 = H|C = p} be the
probability the first flip is heads given that the choosen coin has p as its probabilty of heads.
Then
P{F1 = H} =
Z 1
0
P{F1 = H|C = p}P{C = p}dp .
Since we are assuming a uniform distribution of probabilities for our coins we have that the
above is given by Z 1
0
P{F1 = H|C = p}dp .
Now P{F1 = H|C = p} = p so the above becomes
Z 1
0
pdp =
1
2
.
Part (b): In this case let E be the event that the first two flips are both heads. Then in
exactly the same way as in Part (a) we have
P{E} =
Z 1
0
P{E|C = p}dp .
Now let P{E|C = p} = p2
so that the above becomes 1
3
.
Problem 71
In exactly the same way as for Problem 70 let E be the event that i heads occur given
that the coin selected has a probability of landing heads of p. Then conditioning on this
probability we have that
P{E} =
Z 1
0
P{E|C = P}dp .
But P{E|C = p} =

n
i

pi
(1 − p)n−i
and we have that
P{E} =
Z 1
0

n
i

pi
(1 − p)n−i
dp .

Remembering the definition of the Beta function and the hint provided in the book we see
that
P{E} =
n!
i!(n − i)!

i!(n − i)!
(n + 1)!

=
1
n + 1
,
as claimed.
Problem 72
Again following the framework provide in Problems 70 and 71 we can calculate these prob-
abilities by conditioning on the coin selected (and given that its corresponding probability
of heads is p). We have
P{N ≥ i} =
Z 1
0
P{N ≥ i|C = p}dp .
Now given the coin we are considering has probability p of obtaining heads
P{N ≥ i|C = p} = 1 − P{N i|C = p} = 1 −
i−1
X
n=1
P{N = n|C = p} .
Where P{N = n|C = p} is the probability that our first head appears on flip n. The random
variable N is geometric so we know that P{N = n|C = p} = p(1 − p)n−1
for n = 1, 2, · · ·
1 −
i−1
X
n=1
p(1 − p)n−1
.
Integrating this with respect to p we have that
P{N ≥ i} =
Z 1
0
1 −
i−1
X
n=1
p(1 − p)n−1
!
dp
= 1 −
i−1
X
n=1
Z 1
0
p(1 − p)n−1
dp
= 1 −
i−1
X
n=1
1!(n − 1)!
(n + 1)!
= 1 −
i−1
X
n=1
1
n(n + 1)
.
Using partial fractions to evaluate the sum above we have that
1
n(n + 1)
=
1
n
−
1
n + 1
,
from which we recognize that the above sum is of a telescoping type so that we find that
i−1
X
n=1
1
n(n + 1)
=

1
1
−
1
i

.

Thus in total we find that
P{N ≥ i} = 1 −

1 −
1
i

=
1
i
.
Part (b): We could follow the same procedure as in Part (a) by conditioning on the coin
selected and noting that P{N = i|C = p} = p(1 − p)i−1
or we could simply notice that
P{N = i} = P{N ≥ i} − P{N ≥ i + 1}
=
1
i
−
1
i + 1
=
1
i(i + 1)
.
Part (c): Given the probabilities computed in Part (b) the expression E[N] is easily com-
puted
E[N] =
∞
X
n=1
n

1
n(n + 1)

=
∞
X
n=1
1
n + 1
= ∞
Problem 75
From the chart in the book we see that X is a Poisson random variable with parameters
λ = 2 and that Y is a Binomial random variable with parameters (n, p) = (10, 3
4
).
Part (a): The moment generating function for X + Y is the product of the moment gener-
ating function for X and Y . Thus
MX+Y (t) = exp{2et
− 2}

3
4
et
+
1
4
10
,
Then P{X + Y = 2} is the third term in the Taylor expansion centered on et
i.e.
P{X + Y = 2} =
d2
d(et)2
MX+Y (t)
et=0
Computing the first derivative of the above (with respect to the variable is et
) we find
d
d(et)
MX+Y = 2 exp{2et
− 2}

3
4
et
+
1
4
10
+ 10 exp{2et
− 2}

3
4
et
+
1
4
9
3
4

.

So that the second derivative is given by
d2
d(et)2
MX+Y = 4 exp{2et
− 2}

3
4
et
+
1
4
10
+ 20

3
4

exp{2et
− 2}

3
4
et
+
1
4
9
+ 20

3
4

exp{2et
− 2}

3
4
et
+
1
4
9
+ 90

3
4
2
exp{2et
− 2}

3
4
et
+
1
4
8
.
Evaluating this expression at et
= 0 gives
d2
MX+Y
d(et)2
= 4e−2

1
4
10
+ 40

3
4

e−2

1
4
9
+ 90

3
4
2
e−2

1
4
8
,
which can easily be further evaluated.
Part (b): Now P{XY = 0} can be computed by summing the probabilities of the mutually
exclusive individual terms that could result in the product XY being zero. We find
P{XY = 0} = P{(X = 0, Y = 0)} + P{(X = 0, Y 6= 0)} + P{(X 6= 0, Y = 0)} .
Now the first of these is given by
P{(X = 0, Y = 0)} = e−2

1
4
10
+ e−2
1 −

1
4
10
!
+ (1 − e−2
)

1
4
10
.
Part (c): Now E[XY ] = E[X]E[Y ] since X and Y are independent. Since X is a Poisson
random variable we know that E[X] = 2 and since Y is a binomial random variable we know
that E[Y ] = 10 3
4

= 15
2
. So that
E[XY ] = 15 .
Problem 2
Following the hint we have that
E[|X − a|] =
Z
|x − a|f(x)dx = −
Z a
−∞
(x − a)f(x)dx +
Z ∞
a
(x − a)f(x)dx .
Then taking the derivative of this expression with respect to a we have that
dE[|X − a|]
da
= 0 −
Z a
−∞
(−1)f(x)dx − 0 −
Z ∞
a
f(x)dx
=
Z a
−∞
f(x)dx −
Z ∞
a
f(x)dx ,

Setting this expression equal to zero gives that a must satisfy
Z a
−∞
f(x)dx =
Z ∞
a
f(x)dx .
Which is the exact definition of the median of the distribution f(·). That is a is the point
where one half of the probability is to left of a and where one half of the probability is to
the right of a.
Problem 6 (the integral of the complement of the distribution function)
We desire to prove that
E[X] =
Z ∞
0
P{X t}dt .
Following the hint in the book define the random variable X(t) as
X(t) =

1 if t X
0 if t ≥ X
Then integrating the variable X(t) we see that
Z ∞
0
X(t)dt =
Z X
0
1dt = X .
Thus taking the expectation of both sides we have
E[X] = E
Z ∞
0
X(t)dt

.
This allows us to use the assumed identity that we can pass the expectation inside the
integration as
E
Z ∞
0
X(t)dt

=
Z ∞
0
E[X(t)]dt ,
so applying this identity to the expression we have for E[X] above we see that E[X] =
R ∞
0
E[X(t)]dt. From the definition of X(t) we have that E[X(t)] = P{X t} and we then
finally obtain the fact that
E[X] =
Z ∞
0
P{X t}dt ,
as we were asked to prove.
Problem 10 (the expectation of a sum of random variables)
We begin by defining R(k) to be
R(k) ≡ E
Pk
i=1 Xi
Pn
i=1 Xi
#
.

Then we see that R(k) satisfies a recursive expression given by
R(k) − R(k − 1) = E

Xk
Pn
i=1 Xi

for 2 ≤ k ≤ n .
To further simplify this we would like to evaluate the expectation on the right hand side
of the above. Now by the assumed independence of all Xi’s the expectation on the right
handside of the above is independent of k, and is a constant C. Thus it can be evaluated by
considering
1 = E
Pn
k=1 Xk
Pn
i=1 Xi

=
n
X
k=1
E

Xk
Pn
i=1 Xi

= nC .
Which when we solve for C gives C = 1/n or in terms of the original expectations
E

Xk
Pn
i=1 Xi

=
1
n
for 1 ≤ k ≤ n .
Thus using our recursive expression R(k) = R(k − 1) + 1/n, we see that since
R(1) = E

X1
Pn
i=1 Xi

=
1
n
,
that
R(2) =
1
n
+
1
n
=
2
n
.
Continuing our iterations in this way we find that
R(k) = E
Pk
i=1 Xi
Pn
i=1 Xi
#
=
k
n
for 1 ≤ k ≤ n .
Problem 11
Let Xi denote the Bernoulli indicator random variable that is one if outcome i never occurs
in all n trials and is zero if it does occur. Then
X =
r
X
i=1
Xi .
The expected number of outcomes that never occur is given by E[X] =
Pr
i=1 E[Xi]. But
E[Xi] = P(Xi) = (1 − Pi)n
, since with probability 1 − Pi the ith event won’t happen with
one draw. Thus
E[X] =
r
X
i=1
(1 − Pi)n
.

To find that maximum or minimum of this expression with respect to the Pi we can’t simply
take the derivative of E[X] and set it equal to zero because that won’t enforce the constraint
that
Pr
i=1 Pi = 1. To enforce this constraint we introduce a Lagrangian multiplier λ and a
Lagrangian L defined by
L ≡
r
X
i=1
(1 − Pi)n
+ λ
r
X
i=1
Pi − 1
!
.
Then taking the derivatives of L with respect to Pi and λ and setting all of these expressions
equal to zero we get
∂L
∂Pi
= n(1 − Pi)n−1
(−1) + λ = 0 for i = 1, 2, · · · , r
∂L
∂λ
=
r
X
i=1
Pi − 1 = 0 .
It is this system that we solve for P1, P2, · · · , Pr and λ. We can solve the first equation for
Pi in terms of λ and obtain the following
Pi = 1 −

λ
n
1
n−1
.
When this is put in the constraint equation ∂L
∂λ
= 0 gives
r
X
i=1
1 −

λ
n
1
n−1
!
− 1 = 0 .
Solving this for λ gives
λ = n

1 −
1
r
n−1
.
Putting this value of into the expression we derived earlier for Pi = 1
r
. We can determine if
this solution is a minimum for E[X] by computing the second derivative of this expression.
Specifically
∂E[X]
∂Pi
= −n(1 − Pi)n−1
∂2
E[X]
∂Pi∂Pj
= n(n − 1)(1 − Pi)n−2
δij .
So that the matrix ∂2E[X]
∂Pi∂Pj
is diagonal with positive entries and is therefore positive definite.
Thus the values Pi = 1
r
corresponds to a minimum of E[X].
Problem 12
Part (a): Let Ii be an indicator random variable that is one if trial i results in a success
and is zero if trial i results in a failure. Then defining X =
Pn
i=1 Ii we see that X represents

the random variable that denotes the total number of successes. Then
E[X] =
n
X
i=1
E[Ii] =
n
X
i=1
P(Ii = 1) =
n
X
i=1
Pi .
Part (b): Since

X
2

is the number of paired events that occur, we have that

X
2

=
X
ij
IiIj .
Taking the expectation of both sides gives
E[

X
2

] = E[
X(X − 1)
2
] =
X
ij
P{Ii = 1, Ij = 1} =
X
ij
PiPj ,
by using independence of the events Ii = 1 and Ij = 1. Expanding the quadratic in the
expectation on the left hand side we find that
E[X2
] − E[X] = 2
X
ij
PiPj ,
so that E[X2
] is given by
E[X2
] =
n
X
i=1
Pi + 2
n
X
i=1
i−1
X
j=1
PiPj .
with the definition that
P0
j=1(·) = 0.
Using these expressions the variance of X is given by
Var(X) = E[X2
] − E[X]2
=
n
X
i=1
Pi + 2
n
X
i=1
i−1
X
j=1
PiPj −
n
X
i=1
Pi
!2
=
n
X
i=1
Pi + 2
n
X
i=1
i−1
X
j=1
PiPj −
n
X
i=1
P2
i − 2
n
X
i=1
i−1
X
j=1
PiPj
=
n
X
i=1
Pi −
n
X
i=1
P2
i .
The independence assumption makes no difference in Part (a) but in Part (b) to evaluate
the probability P{Ii = 1, Ij = 1} we explicitly invoked independence.

Problem 13 (record values)
Part (a): Let Rj be an indicator random variable denoting whether or not the j-th random
variable (from n) is a record value. This is that Rj = 1 if and only if Xj is a record value i.e.
Xj ≥ Xi for all 1 ≤ i ≤ j, and Xj is zero otherwise. Then the number N of record values is
given by summing up these indicator
N =
n
X
j=1
Rj .
Taking the expectation of this expression we find that
E[N] =
n
X
j=1
E[Rj] =
n
X
j=1
P{Rj} .
Now P{Rj} is the probability that Xj is the maximum from among all Xi samples where
1 ≤ i ≤ j. Since each Xi is equally likely to be the maximum we have that
P{Rj} = P{Xj = max1≤i≤j(Xi)} =
1
j
,
and the expected number of record values is given by
E[N] =
n
X
j=1
1
j
,
as claimed.
Part (b): From the discussion in the text if N is a random variable denoting the number
of record values that occur then we have

N
2

=
X
ij
RiRj .
Thus taking the expectation and expanding the expression

N
2

in the above we have
E[N2
− N] = E

2
X
ij
RiRj
#
= 2
X
ij
P(Ri, Rj) .
Now P(Ri, Rj) is the probability that Xi and Xj are record values. Since there is no con-
straint on Rj if Ri is a record value this probability is given by
P(Ri, Rj) =
1
j
1
i
.
Thus we have that
E[N2
] = E[N] + 2
n−1
X
i=1
n
X
j=i+1
1
j
1
i
=
n
X
j=1
1
j
+ 2
n−1
X
i=1
1
i
n
X
j=i+1
1
j
,

so that the variance is given by
Var(N) = E[N2
] − E[N]2
=
n
X
j=1
1
j
+ 2
n−1
X
i=1
1
i
n
X
j=i+1
1
j
−
n
X
j=1
1
j
!2
=
n
X
j=1
1
j
+ 2
n−1
X
i=1
1
i
n
X
j=i+1
1
j
!
−
n
X
j=1
1
j2
− 2
n−1
X
i=1
1
i
n
X
j=i+1
1
j
!
=
n
X
j=1
1
j
−
n
X
j=1
1
j2
.
where we have used the fact that (
P
i ai)2
=
P
i a2
i + 2
P
ij aiaj, thus
Var(N) =
n
X
j=1
1
j
−
1
j2
=
n
X
j=1
j − 1
j2
,
as claimed.
Problem 14
We begin by first computing the variance of the number of coupons needed to amass a full
set. Following Example 2i from the book the total number of coupons that are collected, X,
can be decomposed as sum of random variables Xi which are recognized as the number of
additional coupons needed after i distinct types have been obtained to obtain a new type.
Now
Var
N−1
X
i=0
Xi
!
=
N−1
X
i=0
Var(Xi) − 2
X
ij
Cov(Xi, Xj) .
Here Var(Xi) is the variance of a geometric random variable with parameter N−i
N
. This is
given by
1 − N−i
N

N−i
N
2 =
N2
(N − i)2

N − N + i
N

=
Ni
(N − i)2
.
Since Xi and Xj are pairwise independent, introducing the value of Xi does not affect the
value of Xj. Thus
Var(X) =
N−1
X
i=0
Ni
(N − i)2
,
as we were to show.
Problem 15
Part (a): Define Xi to be an indicator random variable such that if trial i is a success
then Xi = 1 otherwise Xi = 0. Then if X is a random variable representing the number of

successes from all n trials we have that
X =
X
i
Xi ,
taking the expectation of both sides we find that E[X] =
P
i E[Xi] =
P
i Pi. Thus an
expression for the mean µ is given by
µ =
X
i
Pi .
Part (b): Using the result from the book we have that

X
2

=
X
ij
XiXj ,
so that taking the expectation of the above gives
E[

X
2

] =
1
2
E[X2
− X] =
X
ij
E[XiXj] .
But the expecation of XiXj is given by (using independence of the trials Xi and Xj)
E[XiXj] = P{XiXj} = P{Xi}P{Xj}. Thus the above expecation becomes
E[X2
] = E[X] + 2
X
ij
PiPj = µ + 2
n−1
X
i=1
Pi
n
X
j=i+1
Pj .
From which we can compute the variance of X as
Var(X) = E[X2
] − E[X]2
= µ + 2
n−1
X
i=1
Pi
n
X
j=i+1
Pj −
n
X
i=1
Pi
!2
= µ + 2
n−1
X
i=1
Pi
n
X
j=i+1
Pj −
n
X
i=1
P2
i − 2
n−1
X
i=1
Pi
n
X
j=i+1
Pj
=
n
X
i=1
Pi(1 − Pi) .
To find the values of Pi that maximize this variance we use the method of Lagrange multi-
plers. Consider the following Lagrangian
L =
n
X
i=1
Pi(1 − Pi) + λ
n
X
i=1
Pi − 1
!
.
Taking the derivatives of this expression with respect to Pi and λ gives
∂L
∂Pi
= 1 − Pi − Pi + λ for 1 ≤ i ≤ n
∂L
∂λ
=
n
X
i=1
Pi − 1 .

The first equation gives for Pi (in terms of λ) the expression that Pi = 1+λ
2
which when put
into the second constraint gives
λ =
2
n
− 1 =
2 − n
n
.
Which means that
Pi =
1
n
.
To determine if this maximizes or minimizes the functional Var(X) we need to consider the
second derivative of the Var(X) expression, i.e.
∂2
Var(X)
∂Pi∂Pj
= −2δij ,
with δij the Kronecker delta. Thus the matrix of second derivatives is negative definite
implying that our solutions Pi = 1
n
will maximize the variance.
Part (c): To select a choice of Pi’s that minimizes this variance we note that Var(X) = 0
if Pi = 0 or Pi = 1 for every i. In this case the random variable X is a constant.
Problem 17
Define the random variable Y as Y ≡ λX1 + (1 − λ)X2. Then the variance of Y is given by
Var(Y ) = Var(λX1 + (1 − λ)X2)
= λ2
Var(X1) + (1 − λ)2
Var(X2)
+ 2
X
ij
Cov(λXi, (1 − λ)Xj) .
Since X1 and X2 are independent their covariance is zero so the above becomes
Var(Y ) = λ2
σ2
1 + (1 − λ)2
σ2
2 .
To make this variance as small as possible we desire to minimize this function with respect
to λ. Taking the derivative of this expression with respect to λ and setting it equal to zero
gives
2λσ2
1 + 2(1 − λ)(−1)σ2
2 = 0 ,
which when we solve for λ gives the following
λ =
σ2
2
σ2
1 + σ2
2
.
A second derivative gives the expression 2σ2
1 + 2σ2
2 a positive quantity and shows that at
this value of λ E[Y ] is indeed a minimum. This value of λ weights the samples X1 and X2
explicitly as
Y =

σ2
2
σ2
1 + σ2
2

X1 +

σ2
1
σ2
1 + σ2
2

X2 ,
which we see is in inverse proportion to the variance of the individual Xi, so if X1 has a
small variance we weight the value of X2 less than that of X1.

Problem 18
Part (a): The distribution of Ni +Nj is binomial with probability of success given by pi +pj.
Part (b): Since a binomial distribution has a variance given by npq we have that
Var(Ni) = mpi(1 − pi)
Var(Nj) = mpj(1 − pj)
Var(Ni + Nj) = m(pi + pj)(1 − pi − pj) .
So that the expression
Var(Ni + Nj) = Var(Ni) + Var(Nj) + 2Cov(Ni, Nj) ,
becomes
m(pi + pj)(1 − pi − pj) − mpi(1 − pi) − mpj(1 − pj) = 2Cov(Ni, Nj) .
This simplifies to
Cov(Ni, Nj) = −mpipj ,
as claimed.
Problem 19
Expanding the given expression we have that
Cov(X + Y, X − Y ) = Cov(X, X) − Cov(X, Y ) + Cov(Y, X) − Cov(Y, Y )
= Cov(X, X) − Cov(Y, Y ) .
If X and Y are identically distributed then Cov(X, X) = Cov(Y, Y ) and the above expression
is zero.
Problem 20
To solve this problem we will use the definition of conditional variance which is defined by
Cov(X, Y |Z) = E[(X − E[X|Z])(Y − E[Y |Z])] .
Part (a): By expanding the expression inside the expectation above we have
(X − E[X|Z])(Y − E[Y |Z]) = XY − XE[Y |Z] − Y E[X|Z] + E[X|Z]E[Y |Z] .

and the fact that Γ(k) = (k − 1)! when k is an integer, we find that the expectation of X(i)
is given by
E[X(i)] =
n!
(i − 1)!(n − i)!

Γ(i + 1)Γ(n − i + 1)
Γ(i + n − i + 2)

=
n!
(i − 1)!(n − i)!

i!(n − i)!
(n + 1)!

=
i
n + 1
.
In the same way we have
E[X2
(i)] =
n!
(i − 1)!(n − i)!
Z 1
0
xi+1
(1 − x)n−i
dx
=
n!
(i − 1)!(n − i)!

(i + 1)!(n − i)!
(n + 2)!

=
i(i + 1)
(n + 1)(n + 2)
.
Combining these two we have finally that
Var(X(i)) =
i(i + 1)
(n + 1)(n + 2)
−
i2
(n + 1)2
=
i(n + 1 − i)
(n + 1)2(n + 2)
for i = 1, 2, · · · , n
Part (b): Since the denominator of Var(X(i)) for all i is a constant, to minimize (or maxi-
mize) this expression we can study the numerator i(n+1−i). Then the minimum/maximum
for this expression occurs at i = 1 or n or the index where d
di
(i(n + 1 − i)) = 0. Taking this
derivative we find that the first order necessary condition is
n + 1 − i − i = 0 ,
or that i = n+1
2
. Note this is effectively the sample median, i.e. if n is odd this is an integer
otherwise this is non-integer. Since the second derivative of this expression is given by
d2
(i(n + 1 − i))
di2
= −2 0 ,
The value of Var(X(i)) at i = n+1
2
corresponds to a local maximum and has a value given by

n + 1
2

n + 1 −

n + 1
2

=

n + 1
2
2
.
This is to be compared to the value of the numerator i(n + 1 − i) when i = 1 or n both of
which equal n. Thus Var(X(1)) = Var(X(n)) and the maximum and minimum statistic (i = 1
and i = n) have the smallest variance while the “median” element i = i+1
2
(or the nearest
integer) has the largest variance.

Problem 22
We begin by remembering the definition of the correlation coefficient between two random
variables X and Y
ρ(X, Y ) =
Cov(X, Y )
p
Var(X)
p
Var(Y )
.
Since Y + a + bX we have that Var(Y ) = Var(a + bX) = b2
Var(X), and Cov(X, Y ) =
Cov(X, X) = bVar(X). With these ρ becomes
ρ(X, Y ) =
bVar(X)
p
Var(X)|b|
p
Var(X)
=
b
|b|
=

−1 b 0
+1 b 0
.
Problem 23
To compute ρ(Y, Z) we need to compute Cov(Y, Z). Since Y = a + bZ + cZ2
, we see that
Cov(Y, Z) = aCov(1, Z) + bCov(Z, Z) + cCov(Z2
, Z)
= 0 + b + cCov(Z2
, Z) .
Now from Problem 54 in this Chapter we know that Cov(Z2
, Z) = 0, Var(Z) = 1, and we
can compute Var(Y ) as
Var(Y ) = Var(a + bZ + cZ2
)
= Var(bZ + cZ2
)
= E[(bZ + cZ2
)2
] − E[(bZ + cZ2
)]2
.
Now (bZ + cZ2
)2
= b2
Z2
+ 2bcZ3
+ c3
Z4
, so the expectation of this expression becomes
b2
· 1 + c2
E[Z4
]. Now to compute E[Z4
] when Z is a standard normal we can use the
definition of expectation and evaluate
E[Z4
] =
1
√
2π
Z ∞
−∞
z4
e− 1
2
z2
dz .
Introducte the variable v = 1
2
z2
, so that dv = zdz, and z =
√
2
√
v so that our integral above
becomes (using the evenness of the integrand and doubling the integral)
E[Z4
] =
2
√
2π
Z ∞
0
4v2
e−v dv
√
2v1/2
=
4
√
π
Z ∞
0
v3/2
e−v
dv .
Remembering the definition of the Gamma function Γ(x) ≡
R ∞
0
vx−1
e−x
dx, we see that the
above is equal to 4
√
π
Γ(5
2
) and from the identities Γ(x + 1) = xΓ(x) and Γ(1
2
) =
√
π we have
that
Γ(
5
2
) =
3
2
Γ(
3
2
) =
3
2
1
2
Γ(
1
2
) =
3
√
π
4
.
Thus our expectation becomes E[Z4
] = 3. WWX: I need to finish this problem.

Problem 24
Following the hint we see that
0 E[(tX + Y )2
] = E[t2
X2
+ 2tXY + Y 2
] = t2
E[X2
] + 2tE[XY ] + E[Y 2
] ,
so that the roots (in the variable t) of this equation must be imaginary and we must have
that “b2
− 4ac 0” which using the expressions for this problem becomes
(2E[XY ])2
− 4E[X2
]E[Y 2
] 0 ,
or
E[XY ]2
≤ E[X2
]E[Y 2
] ,
as claimed.
Problem 54
We have that Cov(Z, Z2
) = E[Z3
] − E[Z]E[Z2
]. Since Z is a standard normal random
variable we know that E[Z] = 0 and E[Z3
] = 0. Both of these can be seen from the identity
of integrating an odd function over an symmetric integral. Thus Cov(Z, Z2
) = 0.

Chapter 8 (Limit Theorems)
Chapter 8: Problems
Problem 1 (bounding the probability we are between two numbers)
We are told that µ = 20 and σ2
= 20 so that
P{0 X 40} = P{−20 X − 20 20} = 1 − P{|X − 20| 20} .
Now by Chebyshev’s inequality
P{|X − µ| ≥ k} ≤
σ2
k2
,
we know that
P{|X − 20| 20} ≤
20
202
= 0.05 .
This implys that (negating both sides that)
−P{|X − 20| 20} −0.05 ,
so that 1−P{|X −20| 20} 0.95. In summary then we have that P{0 X 40} 0.95.
Problem 2 (distribution of test scores)
We are told, that if X is the students score in taking this test then E[X] = 75.
Part (a): Then by Markov’s inequality we have
P{X ≥ 85} ≤
E[X]
85
=
75
85
=
15
17
.
If we also know the variance of X is given by VarX = 25, then we can use the one-sided
Markov inequality given by
P{X − µ ≥ a} ≤
σ2
σ2 + a2
.
With µ = 75, a = 10, σ2
= 25 this becomes
P{X ≥ 85} ≤
25
25 + 102
=
1
5
.
Part (b): Using Chernoff’s inequality given by
P{|X − µ| ≥ kσ} ≤
1
k2
,

we have (since we want 5k = 10 or k = 2) that
P{|X − 75| ≥ 2 × 5} ≤
1
22
= 0.25 ,
Thus
P{|X − 75| ≤ 10} = 1 − P{|X − 75| ≥ 10} = 1 −
1
4
=
3
4
.
Part (c): We desire to compute
P{75 − 5 ≤
1
n
n
X
i=1
xi ≤ 75 + 5} = P{|
1
n
n
X
i=1
xi − 75| ≤ 5}
Defining X =
Pn
i=1 Xi, we have that µ = E[X] = 75 and Var(X) = 1
n2 × nVar(X) = 25
n
. So
to use Chernoff’ inequality on this problem we desire a k such that k

5
√
n

= 5 so k =
√
n
and then Chernoff’s bound gives
P{|
1
n
n
X
i=1
xi − 75| 5} ≤
1
n
.
So to make P{|1
n
Pn
i=1 xi − 75| 5} ≤ 0.1 we must take
1
n
≤ 0.1 ⇒ n ≥ 10 .
Problem 3 (an example with the central limit theorem)
We want to compute n such that
P{
1
n
Pn
i=1 Xi − 75
5/
√
n
≤
5
5/
√
n
} ≥ 0.9 .
Now by the central limit theorem the expression
1
n
Pn
i=1 Xi − 75
5/
√
n
,
we have that the above can be written (first removing the absolute values)
P{
1
n
Pn
i=1 Xi − 75
5/
√
n
≤
√
n} = 1 − 2P{
1
n
Pn
i=1 Xi − 75
5/
√
n
≤
√
n}
= 1 − 2Φ(−
√
n) .
Setting this equal to 0.9 gives Φ(−
√
n) = 0.05, or when we solve for n we obtain
n (−Φ−1
(0.05))2
= 2.7055 .
In the file chap 8 prob 3.m we use the Matlab command norminv to compute this value.
We see that we should take n ≥ 3.

Problem 4 (sums of Poisson random variables)
Part (a): The Markov inequality is P{X ≥ a} ≤ E[X]
a
, so if X =
P20
i=1 Xi then E[X] =
P20
i=1 E[Xi] = 20, and the Markov inequality becomes in this case
P{X ≥ 15} ≤
20
15
=
4
3
.
Note that since all probabilities must be less than one, this bound is not informative.
Part (b): We desire to compute (using the central limit theorem) P{
P20
i=1 Xi 15}. Thus
the desired probability is given by (since σ =
p
Var(Xi) = 1)
P{
P20
i=1 Xi − 20
√
20

15 − 20
√
20
} = 1 − P{Z −
5
√
20
}
= 0.8682 .
This calculation can be found in chap 8 prob 4.m.
Problem 5 (rounding to integers)
Let R =
P50
i=1 Ri be the approximate sum where each Ri is the rounded variable and let
X =
P50
i=1 Xi be the exact sum. We desire to compute P{|X − R| 3}, which can be
simplified to give
P{|X − R| 3} = P
( 50
X
i=1
Xi −
50
X
i=1
Ri 3
)
= P
( 50
X
i=1
(Xi − Ri) 3
)
.
Now Xi −Ri are independent uniform random variables between [−0.5, 0.5] so the above can
be evaluated using the central limit theorem. For this sum of random variables the mean of
the individual random variables Xi − Ri is zero while the standard deviation σ is given by
σ2
=
(0.5 − (−0.5))2
12
=
1
12
.
Thus by the central limit theorem we have that
P
( 50
X
i=1
(Xi − Ri) 3
)
= P
( P50
i=1(Xi − Ri)
50/
√
12

3
50/
√
12
)
= 2P
(P50
i=1(Xi − Ri)
50/
√
12

−3
50/
√
12
)
= 2Φ(
−3
50/
√
12
) = 0.8353 .
This calculation can be found in chap 8 prob 5.m.

Problem 6 (rolling a die until our sum exceeds 800)
The sum of n die rolls is given by X =
Pn
i=1 Xi with Xi a random variable taking values
1, 2, 3, 4, 5, 6 all with probability of 1/6. Then
µ = E[
n
X
i=1
E[Xi] = nE[Xi] =
n
6
(1 + 2 + 3 + 4 + 5 + 6) =
7
2
n
In addition, because of the independence of our Xi we have that Var(X) = nVar(Xi). For
the individual random variables Xi we have that Var(Xi) = E[X2
i ]−E[Xi]2
. For die we have
E[X2
i ] =
1
6
(1 + 4 + 9 + 16 + 25 + 36) =
91
6
.
so that our variance is given by
Var(Xi) =
91
6
−

7
2
2
= 2.916 .
Now the probability we want to calculate is given by P{X 300}, which we can maniuplate
into a form where we can apply the central limit theorm. We have
P

X − 7n
2
√
2.916
√
n

300 − 7n
2
√
2.916
√
n

Now if n = 80 we have the above given by
P

X − 7
2
· 80
√
2.916
√
80

300 − 7
2
· 80
√
2.916
√
80

= 1 − P{Z 1.309} = 1 − Φ(1.309) = 0.0953 .
Problem 7 (working bulbs)
The total lifetime of all the bulbs is given by
T =
100
X
i=1
Xi ,
where Xi is an exponential random variable with mean five hours. Then since the random
variable T is the sum of independent identically distributed random variables we can use the
central limit theorm to derive estimates about T. For example we know that
Pn
i=1 Xi − nµ
σ
√
n
,
is approximatly a standard normal. Thus to evaluate (since σ2
= 25) we have that
P{T 525} = P

T − 100(5)
10(5)

525 − 500
50

= 1 − P{Z 1/2}
= 1 − Φ(0.5) = 1 − 0.6915 = 0.3085 .

Problem 8 (working bulbs with replacement times)
Our expression for the total time that there is a working bulb in problem 7 without any
replacment time is given by
T =
100
X
i=1
Xi .
If there is a random time required to replace each bulb then we our random variable T must
now include this randomness and becomes
T =
100
X
i=1
Xi +
99
X
i=1
Ui .
Again we desire to evaluate P{T ≤ 550}. To evaluate this let
T =
99
X
i=1
(Xi + Ui) + X100 ,
which motivates us to define the random variables Vi as
Vi =

Xi + Ui i = 1, · · · , 99
X100 i = 100
Then T =
P100
i=1 Vi and the Vi’s are all independent. Below we will introduce the variables
µi and σi to be the mean and the standard deviation respectivly of the random variable Vi.
Taking the expectation of T we find
E[T] =
100
X
i=1
E[Vi] =
99
X
i=1
(E[Xi] + E[Ui]) + E[X100]
= 100 · 5 + 99

1
4

= 524.75 .
In the same way the variance of this summation is also given by
Var(T) =
99
X
i=1
(Var(Xi) + Var(Ui)) + Var(X100)
= 100 · 5 + 99 ·
1
4

1
12

= 502.0625 .
By the central limit theorm we have that
P
( 100
X
i=1
Vi ≤ 550
)
= P
(P100
i=1(Vi − µi)
pPn
i=1 σ2
i
≤
550 −
P100
i=1 µi
pPn
i=1 σ2
i
)
.
Where the variables µi and σi the means and standard deviations of the variables Vi. Cal-
culating the expression on the right handside of the inequality above i.e.
550 −
P100
i=1 µi
pPn
i=1 σ2
i
,

we find it equal to 550−524.75
√
502.0625
= 1.1269. Therefore we see that
P
( 100
X
i=1
Vi ≤ 550
)
≈ Φ(1.1269) = 0.8701 ,
using the Matlab function normcdf.
Problem 9 (how large n needs to be)
Warning: This result does not match the back of the book. If anyone can find
anything incorrect with this problem please let me know.
A gamma random variable with parameters (n, 1) is equivalent to a sum of n exponential
random variables each with parameter λ = 1. i.e. X =
Pn
i=1 Xi, with each Xi an exponential
random variable with λ = 1. This result is discussed in Example 3b Page 282 Chapter 6 in
the book. Then the requested problem seems equivalent to computing n such that
P
Pn
i=1 Xi
n
− 1 0.01

0.01 .
which we will do by converting this into an expression that looks like the central limit theorem
and then evaluate. Recognizing that X is a sum of exponential with parameters λ = 1, we
have that
µ = E[X] = E[
n
X
i=1
Xi] =
n
X
i=1
E[Xi] =
n
X
i=1
1
λ
= n .
In the same way since Var(Xi) = 1
λ2 = 1, we have that
σ2
= Var(X) =
n
X
i=1
Var(Xi) = n .
Then the central limit theorem applied to the random variable X claims that as n → ∞, we
have
P
Pn
i=1 Xi − n
√
n
a

= Φ(a) − Φ(−a) .
or taking the requested probabilistic statement and converting it we find that
P
Pn
i=1 Xi
n
− 1 0.01

= 1 − P
Pn
i=1 Xi
n
− 1 ≤ 0.01

= 1 − P
Pn
i=1 Xi − n
n
≤ 0.01

= 1 − P
Pn
i=1 Xi − n
√
n
≤ 0.01
√
n

≈ 1 − (Φ(0.01
√
n) − Φ(−0.01
√
n)) .

From the following identity on the cumulative distribution of a normal random variable we
have that Φ(x) − Φ(−x) = 1 − 2Φ(−x), so that the above equals
1 − (1 − 2Φ(−0.01
√
n)) = 2Φ(−0.01
√
n) .
To have this be less that 0.01 requires a value of n such that
2Φ(−0.01
√
n) ≤ 0.01 .
Solving for n then gives n ≥ (−100Φ−1
(0.005))2
= (257.58)2
.
Problem 11 (a simple stock model)
Given the recurrence relationship Yn = Yn−1 + Xn for n ≥ 1, with Y0 = 100, we see that a
solution to this is given by
Yn =
n
X
k=1
Xk + Y0 .
If we assume that the Xk’s are independent identically distributed random variables with
mean 0 and variance σ2
, we are asked to evaluate
P{Y10 105} .
Which we will do by transforming this problem into something that looks like an application
of the central limit theorem. We find that
P{Y10 105} = P{
10
X
k=1
Xk 5}
= P
(P10
k=1 Xk − 10 · (0)
√
10

5 − 10 · (0)
√
10
)
= 1 − P
(P10
k=1 Xk − 10 · (0)
√
10

5
√
10
)
≈ 1 − Φ(
5
√
10
) = 0.0569 .
Problem 12
The total life of our 100 components is given by L =
P100
i=1 Xi with Xi exponentially dis-
tributed with rate λi = 1
10+ i
10
= 10
100+i
. We want to estimate the following probability
P {L 1200} = P
( 100
X
i=1
Xi 1200
)
.

From the properties of exponential random variables the mean of each Xi is given by µi =
1
λi
= 10 + i
10
and the variance is Var(Xi) = 1
λ2
i
= 10 + i
10
2
. Then to compute the above the
probability with respect to L we transforme the right handside in the usual manner. The
central limit theorem for independent random variables gives
P



P100
i=1(Xi − µi)
qP100
i=1 σ2
i
≤ a



→ Φ(a) as n → ∞ .
So the above probability can be calculated as
P{L 1200} = 1 − P



P100
i=1(Xi − µi)
qP100
i=1 σ2
i
≤
1200 −
P100
i=1 µi
qP100
i=1 σ2
i



≈ 1 − Φ

1200 −
P100
i=1 µi
qP100
i=1 σ2
i


If we change the distribution of Xi such that Xi is uniform over (0, 20 + i
5
) we then from
properties of uniform random variable we know that
µi = 10 +
i
10
σ2
i =
20 + i
5
− 0
2
12
=
1
3

10 +
i
10
2
.
which is different from the previous variance calculation by the reduction of each individual
variance by three. Thus the
P100
i=1 σ2
i is also reduced by 1
3
from the earlier result. This
propagates through the calculation and we see that
P{L 1200} = 1 − Φ

1200 −
P100
i=1 µi
√
3
qP100
i=1 σ2
i

 ,
with µi and σi in the above evaluated for the exponential variable.
Problem 13
Warning: Here are some notes I had on this problem. I’ve not had the time to check these
in as much detail as I would have liked. Caveat emptor.
Part (a): Let Xi be the score of the ith student. Then since Xi is drawn from a distribution
with mean 74 and standard deviation of 14. Then the average test scores for this class of 25
is given by
A =
1
25
25
X
i=1
Xi .

Then the probability that A exceeds 80 is P{A ≥ 80} = 1 − P{A ≤ 80}. From the central
limit theorem we see that the probability P{A ≤ 80} can be expressed in terms of the
standard normal. Specifically
P{A ≤ 80} = P{
1
25
25
X
i=1
Xi ≤ 80} = P{
25
X
i=1
Xi ≤ 25(80)}
= P
(P25
i=1 Xi − 25(74)
14
√
25
≤
25(80) − 25(74)
14
√
25
)
= Φ

25(6)
14(5)

= Φ

15
7

Part (c): We have µ = 74 and σ = 14. S25 = 1
25
P25
i=1 Xi and S64 = 1
64
P64
i=1 Yi. From the
central limit theorem we know that
1
n
Pn
i=1 Xi − µ

σ
√
n
∼ N (0, 1) ,
so that
1
n
n
X
i=1
Xi ∼ N (µ,
σ2
n
) .
Thus S25 ∼ N (74, 142
25
) and S64 ∼ N (74, 142
64
). Now
V = S64 − S25 ∼ N (0,
142
25
+
142
64
) ,
so that
P{V ≥ 2.2} = 1 − P{V ≤ 2.2}
= 1 − P



V
q
142
25
+ 142
64
≤
2.2
q
142
25
+ 142
64



= 1 − Φ

 2.2
q
142
25
+ 142
64

 .
Problem 14
Let Xi be the random variable denoting the mean lifetime of the ith component. We are told
that E[Xi] = 100 and Var(Xi) = 302
. We assume that we have n components of this type in
stock and can assume that each is replaced immediately when the previous one breaks. We
then desire to compute the value of n such that
P
( n
X
i=1
Xi 2000
)
0.95 ,

or equivalently
P
( n
X
i=1
Xi 2000
)
1 − 0.95 = 0.05 .
Now this can be done by using the central limit theorem for independent random variables.
We have that
P
Pn
i=1(Xi − 100)
30n1/2

2000 − 100n
30n1/2

→ Φ

2000 − 100n
30
√
n

.
Thus we should select n such that
Φ

2000 − 100n
30
√
n

≈ 0.05 ,
which is a nonlinear function that needs to be solved to find the smallest value of n.
Problem 15
Let C be the random variable that denotes the total yearly claim for our 10,000 policy
holders. Then C =
P10000
i=1 Xi with Xi the random claim made by the ith policy holder. We
desire to evaluate P{C 2.7 106
} or
P
(10000
X
i=1
Xi 2.7 106
)
or 1 − P
(10000
X
i=1
Xi 2.7 106
)
.
Using the central limit theorem for independent random variables we have that
P



P10000
i=1 Xi −
P10000
i=1 240
qP10000
i=1 8002

2.7 106
− 240 104
800
√
104

= Φ

2.7 106
− 240 104
800
√
104

,
which can easily be evaluated.
Problem 16
If we assume that the number N of men-women pairs is approximately normally distributed
then we desire to calculate P{N 30}. The mean of men-women pairs I would expect to be
1
2
(100) = 50, with a variance of “npq” or 1
2
1
2

(100) = 25. Thus we can evaluate the above
using
P{N 30} = 1 − P{N 30}
= 1 − P

N − 50
5

−20
5

= 1 − Φ(−4) ≈ 1 .
I would expect this to not be a good approximation.

Problem 18
Let Y denoted the random variable representing the number of fish that must be caught to
obtain at least one of these types.
Part (a): The probability to catch any one given fish is 1
4
.
Problem 19 (expectations of functions of random variables)
For each of the various parts we will apply Jensen’s inequality E[f(X)] ≥ f(E[X]) which
requires f(x) to be convex i.e. f′′
(x) ≥ 0. Now since we are told that E[X] = 25 we can
compute the following.
Part (a): For the function f(x) = x3
, we have that f′′
(x) = 6x ≥ 0 since we are told that
X is a nonnegative random variable. Thus Jensen’s inequality gives
E[X3
] ≥ 253
= 15625 .
Part (b): For the function f(x) =
√
x, we have that f′
(x) = 1
2
√
x
, and f′′
(x) = − 1
4
√
x
0.
Thus f(x) is not a convex function but −f(x) is. Applying Jensen’s inequality to −f(x)
gives E[−
√
X] ≥ −
√
25 = −5 or
E[
√
X] ≤ 5 .
Part (c): For the function f(x) = log(x), we have that f′
(x) = 1
x
, and f′′
(x) = − 1
x2 0.
Thus f(x) is not a convex function but −f(x) is. Applying Jensen’s inequality to −f(x)
gives E[− log(X)] ≥ − log(25) or
E[log(X)] ≤ log(25) .
Part (d): For the function f(x) = e−x
, we have that f′′
(x) = e−x
0. Thus f(x) is a
convex function. Applying Jensen’s inequality to f(x) gives
E[e−X
] ≥ eE[X]
= e25
.
Problem 20 (E[X] ≤ (E[X2
]1/2
≤ (E[X3
])1/3
≤ · · · )
Now Jensens’ inequality is that if f(x) is a convex function then E[f(x)] ≥ f(E[X]). If f is
invertable then this is equivalent to
E[X] ≤ f−1
(E[f(X)]) ,

which will be functional equation we will use to derive the requested results. Now to show
the first stage of the inequality sequence let f(x) = x2
, then f′′
(x) = 2 0 so f(·) is convex
and f−1
(x) = x1/2
. An application of the above functional expression gives
E[X] ≤ (E[X2
])1/2
.
To show that E[X] ≤ E[X3
]1/3
one could perform the same logic with the function f(x) = x3
.
To show the expression E[X2
]1/2
≤ E[X3
]1/3
, we will apply Jensen’s inequlity a second time.
For this second application let Y = f(X) then E[f(X)] = E[Y ] ≤ g−1
(E[g(Y )]) for any
convex g(·). Thus
E[f(X)] ≤ g−1
(E[g(f(X))]) .
On defining Y = f(X) (and X = f−1
(Y )) we have that
E[f−1
(Y )] ≤ f−1
(g−1
(E[g(Y )])) ,
or
f(E[f−1
(Y )]) ≤ g−1
(E[g(Y )]) .
Thus beginning with the function pair f(x) = x and g(x) = x2
we have
E[Y ] ≤ (E[Y 2
])1/2
,
or the first inequality. For the second inequality we can take a function pair to consist of
f−1
(x) = x2
(so that f(x) = x1/2
) and g(x) = x3
(so that g−1
(x) = x1/3
) then we have that
(E[Y 2
])1/2
≤ (E[Y 3
])1/3
.
For the third inequality we can take f and g, such that f−1
(x) = x3
(so that f(x) = x1/3
)
and g(x) = x4
(so that g−1
(x) = x1/4
). Then
(E[Y 3
])1/3
≤ (E[Y 4
])1/4
,
These relationships can be continued in general.
Problem 5
We desire to prove the following, if we define Bn(x) as
Bn(x) =
n
X
k=0
f(
k
n
)

n
k

xk
(1 − x)n−k
,
then we want to show that limn→∞ Bn(x) = f(x). To do this begin by defining X1, X2, · · · , Xk
to be independent random variables each with mean x, then E

f X1+X2+X3+···+Xn
n

can
be evaluated by first noting that if Xi are Bernoulli random variables with mean x then
X1 + X2 + · · · + Xn is a Binomial random variable with parameters (n, x) and thus
Pr{
n
X
i=1
Xi = k} =

n
k

xk
(1 − x)n−k
,

so that
E

f
1
n
n
X
i=1
Xi
!#
=
n
X
k=0
f

k
n

Pr
( n
X
i=1
Xi = k
)
,
by the definition of expectation. Continuing we have that
E

f
1
n
n
X
i=1
Xi
!#
=
n
X
k=0
f

k
n

n
k

xk
(1 − x)n−k
.
Now if we can show that when we define Zn = 1
n
Pn
i=1 Xi that
Pr{|Zn − x| ǫ} → 0 as n → ∞ ,
then from Theoretical exercise number 4 we have that
E[f(Zn)] → f(z) as n → ∞ ,
and we have proven the the famous Weierstrass theorem from analysis. Now from the central
limit theorem we have that the random variable
1
n
Pn
i=1 Xi − µ

σ
√
n
=
1
n
Pn
i=1 Xi − x

σ
√
n
,
tends to the standard normal as n → ∞. With this result we have that the probability that
we desire to bound
P
(
1
n
n
X
i=1
Xi − x
)
is equivalent to
P



1
n
Pn
i=1 Xi − x

σ
√
n

ǫ

σ
√
n




.
By the central limit this is equal to 2Φ(− ǫ
σ
√
n). Since as n → ∞ we have that − ǫ
σ
√
n → −∞
so that
Φ(−
ǫ
σ
√
n) → 0 ,
and we have the condition required in problem number 4 and have proven the desired result.
Problem 1 (an alternate Chebyshev inequality)
Now the Chebyshev inequality is given by
P{|X − µ| ≥ k} ≤
σ2
k2
.
Defining k = σκ the above becomes
P{|X − µ| ≥ σκ} ≤
σ2
σ2κ2
=
1
κ2
,
which is the desired inequality.

Problem 10
Using the Chernoff bound of P{X ≤ a} ≤ e−ta
M(t), we recall that if X is a Poisson random
variable its moment generating function is given by M(t) = eλ(et−1)
so
P{X ≤ i} ≤ e−ti
eλ(et−1)
for t 0 ,
To minimized the right hand side of this expression is equivalent to minimizing −ti+λ(et
−1).
Taking the derivative with respect to t and setting it equal to zero we have
−i + λet
= 0 .
Solving for t gives t = ln(i/λ). Since i λ this t is negative as required. Putting this into
the expression above gives
P{X ≤ i} ≤ e−i ln(i/λ)
eλ(eln(i/λ)−1)
= eln(( i
λ )
−i
)
eλ(i/λ−1)
= e−λ (λe)i
ii
.
Problem 12 (an upper bound on the complemetary error function)
From the definition of the normal density we have that
P{X a} =
Z ∞
a
1
√
2π
e−x2/2
dx ,
which we can simplify by the following change of variable. Let v = x − a (then dv = dx)
P{X a} =
Z ∞
0
1
√
2π
e−(v+a)2/2
dv
=
Z ∞
0
1
√
2π
e−(v2+2va+a2)/2
dv
=
e− a2
2
√
2π
Z ∞
0
e− v2
2 e−va
dv
≤
e− a2
2
√
2π
Z ∞
0
e− v2
2 dv ,
since e−va
≤ 1 for all v ∈ [0, ∞) and a 0. Now because of the identity
Z ∞
0
e− v2
2 dv =
r
π
2
,
we see that the above becomes
P{X a} ≤
1
2
e− a2
2 .

Problem 13 (a problem with expectations)
We are assuming that if E[X] 0 and θ 6= 0 such that E[eθX
] = 1, and want to show that
θ 0. To do this recall Jensen’s inequality which for a convex function f and an arbitrary
random variable Y is given by
E[f(Y )] ≥ f(E[Y ]) .
If we let the random variable Y = eθX
and the function f(y) = − ln(y), then Jensen’s
inequality becomes (since this function f is convex)
−E[θX] ≥ − ln(E[eθX
]) ,
or using the information from the problem we have
θE[X] ≤ ln(1) = 0 .
Now since E[X] 0 by dividing by this expression we have θ 0 as was to be shown.

Chapter 9 (Additional Topics in Probability)
Chapter 9: Problems
Problem 2 (helping Al cross the highway)
At the point where Al wants to cross the highway the number of cars that cross is a Poisson
process with rate λ = 3, the probability that k cars appear in t time is given by
P{N = k} =
e−λt
(λt)k
k!
.
Thus Al will have no problem in the case when no cars come during her crossing. If her
crossing time takes s second this will happen with probability
P{N = 0} = e−λs
= e−3s
.
Note that this is the density function for a Poisson random variable (or the cumulative
distribution function of a Poisson random variable with n = 0). This expression is tabulated
for s = 2, 5, 10, 20 seconds in chap 9 prob 2.m.
Problem 3 (helping a nimble Al cross the highway)
Following the results from Problem 2, Al will cross unhurt, with probability
P{N = 0} + P{N = 1} = e−λs
+ e−λs
(λs) = e−3s
+ 3se−3s
.
Note that this is the cumulative distribution function for a Poisson random variable. This
expression is tabulated for s = 5, 10, 20, 30 seconds in chap 9 prob 3.m.

Chapter 10 (Simulation)
Chapter 10: Problems
Problem 2 (simulating a specified random variable)
Assuming our random variable has a density given by
f(x) =

e2x
−∞ x 0
e−2x
0 x ∞
Lets compute the cumulative distribution F(x) for this density function. This is needed if
we simulate from f using the inverse transformation method. We find that
F(x) =
Z x
−∞
e2ξ
dξ for − ∞ x 0
=
e2ξ
2
x
−∞
=
1
2
e2x
.
and that
F(x) =
1
2
+
Z x
0
e−2ξ
dξ for 0 x ∞
=
1
2
+
e−2ξ
(−2)
x
0
= 1 −
1
2
e−2x
.
Then to simulate from the density f(·) we require the inverse of this cumulative probability
density function. Since our F is given in terms of two different domains we will compute
this inverse function in the same way. If 0 y 1
2
, then the equation we need to invert i.e.
y = F(x) is equivalent to
y =
1
2
e2x
or x =
1
2
ln(2y) for 0 y
1
2
While if 1
2
y 1 then y = F(x) is equivalent to
y = 1 −
1
2
e−2x
,
or by solving for x we find that
x = −
1
2
ln(2(1 − y)) for
1
2
y 1 .
Thus combining these two results we find that
F−1
(y) =
1
2
ln(2y) 0 y 1
2
−1
2
ln(2(1 − y)) 1
2
y 1
Thus our simulation method would repeatedly generate uniform random variables U ∈ (0, 1)
and apply F−1
(U) (defined above) to them computing the corresponding y’s. These y’s are
guaranteed to be derived from our density function f.

References
[1] J. Bewersdorff. Luck, Logic, and White Lies. AK Peters, 2004.

A Solution Manual For A First Course In Probability

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A Solution Manual For A First Course In Probability