Lecture Biostatistics and general concept about probability

5E Note 4
Biostatistics
Chapter 2 Probability and Discrete Probability
Distributions
Experiment, Event, Sample space, Probability, Counting rules,
Conditional probability, Bayes’s rule, random variables, mean,
variance

5E Note 4
Review
I. What’s in last lecture?
Descriptive Statistics – Numerical Measures.
Chapter 2.
II. What's in this and the next two lectures?
Experiment, Event, Sample space, Probability,
Counting rules, Conditional probability,
Bayes’s rule, random variables, mean,
variance. Read Chapter 3.

5E Note 4
Descriptive and Inferential Statistics
Statistics can be broken into two basic types:
• Descriptive Statistics (Chapter 2):
We have already learnt this topic
• Inferential Statistics (Chapters 7-13):
Methods that making decisions or predictions
about a population based on sampled data.
• What are Chapters 3-6?
Probability

5E Note 4
Why Learn Probability?
• Nothing in life is certain. In everything we do, we
gauge the chances of successful outcomes, from
business to medicine to the weather
• A probability provides a quantitative description of
the chances or likelihoods associated with various
outcomes
• It provides a bridge between descriptive and
inferential statistics
Population Sample
Probability
Statistics

5E Note 4
Probabilistic vs Statistical Reasoning
• Suppose I know exactly the proportions of
car makes in California. Then I can find the
probability that the first car I see in the
street is a Ford. This is probabilistic
reasoning as I know the population and
predict the sample
• Now suppose that I do not know the
proportions of car makes in California, but
would like to estimate them. I observe a
random sample of cars in the street and then
I have an estimate of the proportions of the
population. This is statistical reasoning

Note 5 of 5E
What is Probability?
• In Chapters 2, we used graphs and
numerical measures to describe data sets
which were usually samples.
• We measured “how often” using
Relative frequency = f/n
Sample
And “How often”
= Relative frequency
Population
Probability
• As n gets larger,

Note 5 of 5E
Basic Concepts
• An experiment is the process by which
an observation (or measurement) is
obtained.
• An event is an outcome of an experiment,
usually denoted by a capital letter.
–The basic element to which probability
is applied
–When an experiment is performed, a
particular event either happens, or it
doesn’t!

Note 5 of 5E
Experiments and Events
• Experiment: Record an age
–A: person is 30 years old
–B: person is older than 65
• Experiment: Toss a die
–A: observe an odd number
–B: observe a number greater than 2

Note 5 of 5E
Basic Concepts
• Two events are mutually exclusive if,
when one event occurs, the other cannot,
and vice versa.
•Experiment: Toss a die
–A: observe an odd number
–B: observe a number greater than 2
–C: observe a 6
–D: observe a 3
Not Mutually
Exclusive
Mutually
Exclusive
B and C?
B and D?

Note 5 of 5E
Basic Concepts
• An event that cannot be decomposed is called
a simple event.
• Denoted by E with a subscript.
• Each simple event will be assigned a
probability, measuring “how often” it occurs.
• The set of all simple events of an experiment
is called the sample space, S.

Note 5 of 5E
Example
• The die toss:
• Simple events: Sample space:
1
2
3
4
5
6
E1
E2
E3
E4
E5
E6
S ={E1, E2, E3, E4, E5, E6}
S
• E1
• E6
• E2
• E3
• E4
• E5

Note 5 of 5E
Basic Concepts
• An event is a collection of one or more
simple events.
•The die toss:
–A: an odd number
–B: a number > 2
S
A ={E1, E3, E5}
B ={E3, E4, E5,E6}
B
A
• E1
• E6
• E2
• E3
• E4
• E5

Note 5 of 5E
The Probability
of an Event
• The probability of an event A measures “how
often” A will occur. We write P(A).
• Suppose that an experiment is performed n
times. The relative frequency for an event A is
n
f
n

occurs
A
times
of
Number
n
f
A
P
n
lim
)
(


• If we let n get infinitely large,

Note 5 of 5E
The Probability
of an Event
• P(A) must be between 0 and 1.
–If event A can never occur, P(A) = 0. If
event A always occurs when the
experiment is performed, P(A) =1.
• The sum of the probabilities for all
simple events in S equals 1.
• The probability of an event A is found
by adding the probabilities of all the
simple events contained in A.

Note 5 of 5E
– Suppose that 10% of the U.S. population has
red hair. Then for a person selected at
random,
Finding Probabilities
• Probabilities can be found using
–Estimates from empirical studies
–Common sense estimates based on
equally likely events.
P(Head) = 1/2
P(Red hair) = .10
• Examples:
–Toss a fair coin.

Note 5 of 5E
Using Simple Events
• The probability of an event A is equal to
the sum of the probabilities of the simple
events contained in A
• If the simple events in an experiment are
equally likely, you can calculate
events
simple
of
number
total
A
in
events
simple
of
number
)
( 

N
n
A
P A

Note 5 of 5E
Example 1
Toss a fair coin twice. What is the probability
of observing at least one head?
H
1st Coin 2nd Coin Ei P(Ei)
H
T
T
H
T
HH
HT
TH
TT
1/4
1/4
1/4
1/4
P(at least 1 head)
= P(E1) + P(E2) + P(E3)
= 1/4 + 1/4 + 1/4 = 3/4

Note 5 of 5E
Example 2
A bowl contains three M&Ms®
, one red, one
blue and one green. A child selects two M&Ms
at random. What is the probability that at least
one is red?
1st M&M 2nd M&M Ei P(Ei)
RB
RG
BR
BG
1/6
1/6
1/6
1/6
1/6
1/6
P(at least 1 red)
= P(RB) + P(BR)+ P(RG)
+ P(GR)
= 4/6 = 2/3
m
m
m
m
m
m
m
m
m
GB
GR

Note 5 of 5E
Example 3
The sample space of throwing a pair of dice is

Note 5 of 5E
Example 3
Event Simple events Probability
Dice add to 3 (1,2),(2,1) 2/36
Dice add to 6 (1,5),(2,4),(3,3),
(4,2),(5,1)
5/36
Red die show 1 (1,1),(1,2),(1,3),
(1,4),(1,5),(1,6)
6/36
Green die show 1 (1,1),(2,1),(3,1),
(4,1),(5,1),(6,1)
6/36

Note 5 of 5E
Counting Rules
• Sample space of throwing 3 dice has
216 entries, sample space of throwing
4 dice has 1296 entries, …
• At some point, we have to stop listing
and start thinking …
• We need some counting rules

Note 5 of 5E
The mn Rule
• If an experiment is performed in two stages,
with m ways to accomplish the first stage and
n ways to accomplish the second stage, then
there are mn ways to accomplish the
experiment.
• This rule is easily extended to k stages, with
the number of ways equal to
n1 n2 n3 … nk
Example: Toss two coins. The total number of
simple events is:
2  2 = 4

Note 5 of 5E
Examples
Example: Toss three coins. The total number of
simple events is: 2  2  2 = 8
Example: Two M&Ms are drawn from a dish
containing two red and two blue candies. The total
number of simple events is:
6  6 = 36
Example: Toss two dice. The total number of
simple events is:
m
m
4  3 = 12
Example: Toss three dice. The total number of
simple events is: 6  6  6 = 216

Note 5 of 5E
Permutations
• The number of ways you can arrange
n distinct objects, taking them r at a time
is
Example: How many 3-digit lock combinations
can we make from the numbers 1, 2, 3, and 4?
.
1
!
0
and
)
1
)(
2
)...(
2
)(
1
(
!
where
)!
(
!






n
n
n
n
r
n
n
Pn
r
24
)
2
)(
3
(
4
!
1
!
4
4
3 


P
The order of the choice is
important!

Note 5 of 5E
Examples
Example: A lock consists of five parts and
can be assembled in any order. A quality
control engineer wants to test each order for
efficiency of assembly. How many orders are
there?
120
)
1
)(
2
)(
3
)(
4
(
5
!
0
!
5
5
5 


P
The order of the choice is
important!

Note 5 of 5E
Combinations
• The number of distinct combinations of n
distinct objects that can be formed,
taking them r at a time is
Example: Three members of a 5-person committee must
be chosen to form a subcommittee. How many different
subcommittees could be formed?
)!
(
!
!
r
n
r
n
Cn
r


10
1
)
2
(
)
4
(
5
1
)
2
)(
1
)(
2
(
3
1
)
2
)(
3
)(
4
(
5
)!
3
5
(
!
3
!
5
5
3 




C
The order of
the choice is
not important!

Note 5 of 5E
Example
• A box contains six M&Ms®
, four red
and two green. A child selects two M&Ms at
random. What is the probability that exactly
one is red?
The order of
the choice is
not important!
m
m
m
m
m m
Ms.
&
M
2
choose
to
ways
15
)
1
(
2
)
5
(
6
!
4
!
2
!
6
6
2 


C
M.
&
M
green
1
choose
to
ways
2
!
1
!
1
!
2
2
1 

C
M.
&
M
red
1
choose
to
ways
4
!
3
!
1
!
4
4
1 

C 4  2 =8 ways to
choose 1 red and 1
green M&M.
P(exactly one
red) = 8/15

Note 5 of 5E
Example
A deck of cards consists of 52 cards, 13 "kinds"
each of four suits (spades, hearts, diamonds, and
clubs). The 13 kinds are Ace (A), 2, 3, 4, 5, 6, 7,
8, 9, 10, Jack (J), Queen (Q), King (K). In many
poker games, each player is dealt five cards from
a well shuffled deck.
hands
possible
960
,
598
,
2
1
)
2
)(
3
)(
4
(
5
48
)
49
)(
50
)(
51
(
52
)!
5
52
(
!
5
!
52
are
There 52
5 



C

Note 5 of 5E
Example
Four of a kind: 4 of the 5 cards are the same
“kind”. What is the probability of getting
four of a kind in a five card hand?
and
There are 13 possible choices for the kind of
which to have four, and 52-4=48 choices for
the fifth card. Once the kind has been
specified, the four are completely determined:
you need all four cards of that kind. Thus
there are 13×48=624 ways to get four of a
kind.
The probability=624/2598960=.000240096

Note 5 of 5E
Example
One pair: two of the cards are of one kind,
the other three are of three different kinds.
What is the probability of getting one pair
in a five card hand?
kind
that
of
cards
four
the
of
two
of
choices
possible
6
are
there
choice,
given the
pair;
a
have
which to
of
kind
for the
choices
possible
13
are
There
4
2 
C

Note 5 of 5E
Example
There are 12 kinds remaining from
which to select the other three cards in
the hand. We must insist that the kinds
be different from each other and from
the kind of which we have a pair, or we
could end up with a second pair, three
or four of a kind, or a full house.

Note 5 of 5E
Example
422569
.
98960
1098240/25
y
probabilit
The
1,098,240.
64
220
6
13
is
hands
pair"
one
"
of
number
the
Therefore
three.
all
of
suits
for the
choices
64
4
of
total
a
cards,
three
those
of
each
of
suit
for the
choices
4
are
There
cards.
three
remaining
the
of
kinds
pick the
to
ways
220
are
There
3
12
3









C

Note 5 of 5E
S
Event Relations
The beauty of using events, rather than simple events, is
that we can combine events to make other events using
logical operations: and, or and not.
The union of two events, A and B, is the event that
either A or B or both occur when the experiment is
performed. We write
A B
A B
B
A 

Note 5 of 5E
S
A B
Event Relations
The intersection of two events, A and B, is
the event that both A and B occur when the
experiment is performed. We write A B.
B
A 
• If two events A and B are mutually
exclusive, then P(A B) = 0.

Note 5 of 5E
S
Event Relations
The complement of an event A consists of
all outcomes of the experiment that do not
result in event A. We write AC
.
A
AC

Note 5 of 5E
Example
Select a student from the classroom and
record his/her hair color and gender.
–A: student has brown hair
–B: student is female
–C: student is male
What is the relationship between events B and C?
• AC
:
• BC:
• BC:
Mutually exclusive; B = CC
Student does not have brown hair
Student is both male and female = 
Student is either male and female = all students = S

Note 5 of 5E
Calculating Probabilities for
Unions and Complements
• There are special rules that will allow you to
calculate probabilities for composite events.
• The Additive Rule for Unions:
• For any two events, A and B, the probability
of their union, P(A B), is
)
(
)
(
)
(
)
( B
A
P
B
P
A
P
B
A
P 




A B

Note 5 of 5E
Example: Additive Rule
Example: Suppose that there were 120
students in the classroom, and that they
could be classified as follows:
Brown Not Brown
Male 20 40
Female 30 30
A: brown hair
P(A) = 50/120
B: female
P(B) = 60/120
P(AB) = P(A) + P(B) – P(AB)
= 50/120 + 60/120 - 30/120
= 80/120 = 2/3 Check: P(AB)
= (20 + 30 + 30)/120

Note 5 of 5E
Example: Two Dice
A: red die show 1
B: green die show 1
P(AB) = P(A) + P(B) – P(AB)
= 6/36 + 6/36 – 1/36
= 11/36

Note 5 of 5E
A Special Case
When two events A and B are
mutually exclusive, P(AB) = 0
and P(AB) = P(A) + P(B).
Brown Not Brown
Male 20 40
Female 30 30
A: male with brown hair
P(A) = 20/120
B: female with brown hair
P(B) = 30/120
P(AB) = P(A) + P(B)
= 20/120 + 30/120
= 50/120
A and B are mutually
exclusive, so that

Note 5 of 5E
Example: Two Dice
A: dice add to 3
B: dice add to 6
A and B are mutually
exclusive, so that
P(AB) = P(A) + P(B)
= 2/36 + 5/36
= 7/36

Note 5 of 5E
Calculating Probabilities
for Complements
• We know that for any event A:
–P(A AC
) = 0
• Since either A or AC
must occur,
P(A AC
) =1
• so that P(A AC
) = P(A)+ P(AC
) = 1
P(AC
) = 1 – P(A)
A
AC

Note 5 of 5E
Example
Brown Not Brown
Male 20 40
Female 30 30
A: male
P(A) = 60/120
B: female
P(B) = ?
P(B) = 1- P(A)
= 1- 60/120 = 60/120
A and B are
complementary, so that
Select a student at random from
the classroom. Define:

Note 5 of 5E
Calculating Probabilities for
Intersections
In the previous example, we found P(A  B)
directly from the table. Sometimes this is
impractical or impossible. The rule for calculating
P(A  B) depends on the idea of independent and
dependent events.
Two events, A and B, are said to be
independent if the occurrence or
nonoccurrence of one of the events does
not change the probability of the
occurrence of the other event.

Note 5 of 5E
Conditional Probabilities
The probability that A occurs, given
that event B has occurred is called
the conditional probability of A
given B and is defined as
0
)
(
if
)
(
)
(
)
|
( 

 B
P
B
P
B
A
P
B
A
P
“given”

Note 5 of 5E
Example 1
Toss a fair coin twice. Define
–A: head on second toss
–B: head on first toss
HT
TH
TT
1/4
1/4
1/4
1/4
P(A|B) = ½
P(A|not B) = ½
HH
P(A) does not
change, whether
B happens or
not…
A and B are
independent!

Note 5 of 5E
Example 2
A bowl contains five M&Ms®
, two red and
three blue. Randomly select two candies, and
define
–A: second candy is red.
–B: first candy is blue.
m
m
m
m
m
P(A|B) =P(2nd
red|1st
blue)= 2/4 = 1/2
P(A|not B) = P(2nd
red|1st
red) = 1/4
P(A) does change,
depending on
whether B happens
or not…
A and B are
dependent!

Note 5 of 5E
Example 3: Two Dice
Toss a pair of fair dice. Define
– A: red die show 1
– B: green die show 1
P(A|B) = P(A and B)/P(B)
=1/36/1/6=1/6=P(A)
P(A) does not
change, whether
B happens or
not…
A and B are
independent!

Note 5 of 5E
Example 3: Two Dice
Toss a pair of fair dice. Define
– A: add to 3
– B: add to 6
P(A|B) = P(A and B)/P(B)
=0/36/5/6=0
P(A) does change
when B happens
A and B are dependent!
In fact, when B happens,
A can’t

Note 5 of 5E
Defining Independence
• We can redefine independence in terms of
conditional probabilities:
Two events A and B are independent if and
only if
P(A|B) = P(A) or P(B|A) = P(B)
Otherwise, they are dependent.
• Once you’ve decided whether or not two
events are independent, you can use the
following rule to calculate their
intersection.

Note 5 of 5E
The Multiplicative Rule for
Intersections
• For any two events, A and B, the
probability that both A and B occur is
P(A B) = P(A) P(B given that A occurred)
= P(A)P(B|A)
• If the events A and B are independent, then
the probability that both A and B occur is
P(A B) = P(A) P(B)

Note 5 of 5E
Example 1
In a certain population, 10% of the people can be
classified as being high risk for a heart attack. Three
people are randomly selected from this population.
What is the probability that exactly one of the three are
high risk?
Define H: high risk N: not high risk
P(exactly one high risk) = P(HNN) + P(NHN) + P(NNH)
= P(H)P(N)P(N) + P(N)P(H)P(N) + P(N)P(N)P(H)
= (.1)(.9)(.9) + (.9)(.1)(.9) + (.9)(.9)(.1)= 3(.1)(.9)2
= .243

Note 5 of 5E
Example 2
Suppose we have additional information in the
previous example. We know that only 49% of the
population are female. Also, of the female patients, 8%
are high risk. A single person is selected at random. What
is the probability that it is a high risk female?
Define H: high risk F: female
From the example, P(F) = .49 and P(H|F) = .08.
Use the Multiplicative Rule:
P(high risk female) = P(HF)
= P(F)P(H|F) =.49(.08) = .0392

Note 5 of 5E
The Law of Total Probability
P(A) = P(A  S1) + P(A  S2) + … + P(A  Sk)
= P(S1)P(A|S1) + P(S2)P(A|S2) + … + P(Sk)P(A|Sk)
Let S1 , S2 , S3 ,..., Sk be mutually exclusive
and exhaustive events (that is, one and only
one must happen). Then the probability of
any event A can be written as

Note 5 of 5E
The Law of Total Probability
A
A Sk
A  S1
S2….
S1
Sk
P(A) = P(A  S1) + P(A  S2) + … + P(A  Sk)
= P(S1)P(A|S1) + P(S2)P(A|S2) + … + P(Sk)P(A|Sk)

Note 5 of 5E
Bayes’ Rule
Let S1 , S2 , S3 ,..., Sk be mutually exclusive and
exhaustive events with prior probabilities P(S1),
P(S2),…,P(Sk). If an event A occurs, the posterior
probability of Si, given that A occurred is
,...k
,
i
S
A
P
S
P
S
A
P
S
P
A
S
P
i
i
i
i
i 2
1
for
)
|
(
)
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Proof
i
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S
A
P
S
P
S
A
P
S
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P
AS
P
A
S
P
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A
P
S
P
AS
P
S
P
AS
P
S
A
P








Note 5 of 5E
We know:
P(F) =
P(M) =
P(H|F) =
P(H|M) =
Example
From a previous example, we know that 49% of the
population are female. Of the female patients, 8% are
high risk for heart attack, while 12% of the male patients
are high risk. A single person is selected at random and
found to be high risk. What is the probability that it is a
male? Define H: high risk F: female M: male
61
.
)
08
(.
49
.
)
12
(.
51
.
)
12
(.
51
.
)
|
(
)
(
)
|
(
)
(
)
|
(
)
(
)
|
(





F
H
P
F
P
M
H
P
M
P
M
H
P
M
P
H
M
P
.12
.08
.51
.49

Note 5 of 5E
Example
Suppose a rare disease infects one out of
every 1000 people in a population. And
suppose that there is a good, but not perfect,
test for this disease: if a person has the
disease, the test comes back positive 99% of
the time. On the other hand, the test also
produces some false positives: 2% of
uninfected people are also test positive. And
someone just tested positive. What are his
chances of having this disease?

Note 5 of 5E
We know:
P(A) = .001 P(Ac
) =.999
P(B|A) = .99 P(B|Ac
) =.02
Example
Define A: has the disease B: test positive
0472
.
02
.
999
.
99
.
001
.
99
.
001
.
)
|
(
)
(
)
|
(
)
(
)
|
(
)
(
)
|
(







 c
A
B
P
c
A
P
A
B
P
A
P
A
B
P
A
P
B
A
P
We want to know P(A|B)=?

Note 5 of 5E
Example
A survey of job satisfaction2
of teachers was
taken, giving the following results
2
“Psychology of the Scientist: Work Related Attitudes of U.S. Scientists”
(Psychological Reports (1991): 443 – 450).
Satisfied Unsatisfied Total
College 74 43 117
High School 224 171 395
Elementary 126 140 266
Total 424 354 778
Job Satisfaction
L
E
V
E
L

Note 5 of 5E
Example
If all the cells are divided by the total number
surveyed, 778, the resulting table is a table of
empirically derived probabilities.
College 0.095 0.055 0.150
High School 0.288 0.220 0.508
Elementary 0.162 0.180 0.342
Total 0.545 0.455 1.000
L
E
V
E
L
Job Satisfaction

Note 5 of 5E
Example
For convenience, let C stand for the event that
the teacher teaches college, S stand for the
teacher being satisfied and so on. Let’s look at
some probabilities and what they mean.
is the proportion of teachers who are
college teachers.
P(C) 0.150

satisfied with their job.
P(S) 0.545

college teachers and who are satisfied
with their job.
P(C S) 0.095


College 0.095 0.055 0.150
High School 0.288 0.220 0.508
Elementary 0.162 0.180 0.342
Total 0.545 0.455 1.000
L
E
V
E
L
Job Satisfaction

Note 5 of 5E
Example
is the proportion of teachers who
are college teachers given they are
satisfied. Restated: This is the
proportion of satisfied that are
college teachers.
P(C S)
P(C | S)
P(S)
0.095
0.175
0.545

 

is the proportion of teachers who
are satisfied given they are college
teachers. Restated: This is the
proportion of college teachers that
are satisfied.
P(S C)
P(S | C)
P(C)
P(C S) 0.095
P(C) 0.150
0.632

 



College 0.095 0.055 0.150
High School 0.288 0.220 0.508
Elementary 0.162 0.180 0.342
Total 0.545 0.455 1.000
L
E
V
E
L
Job Satisfaction

Note 5 of 5E
Example
P(C S) 0.095
P(C) 0.150 and P(C | S) 0.175
P(S) 0.545
   

College 0.095 0.055 0.150
High School 0.288 0.220 0.508
Elementary 0.162 0.180 0.342
Total 0.545 0.455 1.000
L
E
V
E
L
Job Satisfaction
P(C|S)  P(C) so C and S are dependent events.
Are C and S independent events?

Note 5 of 5E
Example
College 0.095 0.055 0.150
High School 0.288 0.220 0.508
Elementary 0.162 0.180 0.658
Total 0.545 0.455 1.000
L
E
V
E
L
Job Satisfaction
P(C) = 0.150, P(S) = 0.545 and
P(CS) = 0.095, so
P(CS) = P(C)+P(S) - P(CS)
= 0.150 + 0.545 - 0.095
= 0.600
P(CS)?

Note 5 of 5E
Tom and Dick are going to take
a driver's test at the nearest DMV office. Tom
estimates that his chances to pass the test are
70% and Dick estimates his as 80%. Tom and
Dick take their tests independently.
Define D = {Dick passes the driving test}
T = {Tom passes the driving test}
T and D are independent.
P (T) = 0.7, P (D) = 0.8
Example

Note 5 of 5E
What is the probability that at most one of
the two friends will pass the test?
Example
P(At most one person pass)
= P(Dc
 Tc
) + P(Dc
 T) + P(D  Tc
)
= (1 - 0.8) (1 – 0.7) + (0.7) (1 – 0.8) + (0.8) (1 – 0.7)
= .44
P(At most one person pass)
= 1-P(both pass) = 1- 0.8 x 0.7 = .44

Note 5 of 5E
What is the probability that at least one of
the two friends will pass the test?
Example
P(At least one person pass)
= P(D  T)
= 0.8 + 0.7 - 0.8 x 0.7
= .94
P(At least one person pass)
= 1-P(neither passes) = 1- (1-0.8) x (1-0.7) = .94

Note 5 of 5E
Suppose we know that only one of the two
friends passed the test. What is the
probability that it was Dick?
Example
P(D | exactly one person passed)
= P(D  exactly one person passed) / P(exactly one
person passed)
= P(D  Tc
) / (P(D  Tc
) + P(Dc
 T) )
= 0.8 x (1-0.7)/(0.8 x (1-0.7)+(1-.8) x 0.7)
= .63

Note 5 of 5E
Random Variables
• A quantitative variable x is a random variable if
the value that it assumes, corresponding to the
outcome of an experiment is a chance or random
event.
• Random variables can be discrete or
continuous.
• Examples:
x = SAT score for a randomly selected student
x = number of people in a room at a randomly
selected time of day
x = number on the upper face of a randomly
tossed die

Note 5 of 5E
Probability Distributions for
Discrete Random Variables
The probability distribution for a discrete
random variable x resembles the relative
frequency distributions we constructed in
Chapter 2. It is a graph, table or formula that
gives the possible values of x and the
probability p(x) associated with each value.
1
)
(
and
1
)
(
0
have
must
We



 x
p
x
p

Note 5 of 5E
Example
Toss a fair coin three times and
define x = number of heads.
1/8
1/8
1/8
1/8
1/8
1/8
1/8
1/8
P(x = 0) = 1/8
P(x = 1) = 3/8
P(x = 2) = 3/8
P(x = 3) = 1/8
HHH
HHT
HTH
THH
HTT
THT
TTH
TTT
x
3
2
2
2
1
1
1
0
x p(x)
0 1/8
1 3/8
2 3/8
3 1/8
Probability
Histogram for x

Note 5 of 5E
Example
Toss two dice and define
x = sum of two dice. x p(x)
2 1/36
3 2/36
4 3/36
5 4/36
6 5/36
7 6/36
8 5/36
9 4/36
10 3/36
11 2/36
12 1/36

Note 5 of 5E
Probability Distributions
Probability distributions can be used to describe
the population, just as we described samples in
Chapter 2.
–Shape: Symmetric, skewed, mound-shaped…
–Outliers: unusual or unlikely measurements
–Center and spread: mean and standard
deviation. A population mean is called m and a
population standard deviation is called s.

Note 5 of 5E
The Mean
and Standard Deviation
Let x be a discrete random variable with
probability distribution p(x). Then the
mean, variance and standard deviation of x
are given as
2
2
2
:
deviation
Standard
)
(
)
(
:
Variance
)
(
:
Mean











x
p
x
x
xp

Note 5 of 5E
Example
Toss a fair coin 3 times and
record x the number of heads.
x p(x) xp(x) (x-m)2
p(x)
0 1/8 0 (-1.5)2
(1/8)
1 3/8 3/8 (-0.5)2
(3/8)
2 3/8 6/8 (0.5)2
(3/8)
3 1/8 3/8 (1.5)2
(1/8)
5
.
1
8
12
)
( 


 x
xp

)
(
)
( 2
2
x
p
x 
 


688
.
75
.
75
.
28125
.
09375
.
09375
.
28125
.
2










Note 5 of 5E
Example
The probability distribution for x the
number of heads in tossing 3 fair
coins.
• Shape?
• Outliers?
• Center?
• Spread?
Symmetric;
mound-shaped
None
m = 1.5
s = .688
m

Note 5 of 5E
Key Concepts
I. Experiments and the Sample Space
1. Experiments, events, mutually exclusive events,
simple events
2. The sample space
II. Probabilities
1. Relative frequency definition of probability
2. Properties of probabilities
a. Each probability lies between 0 and 1.
b. Sum of all simple-event probabilities equals 1.
3. P(A), the sum of the probabilities for all simple events in A

Note 5 of 5E
Key Concepts
III. Counting Rules
1. mn Rule; extended mn Rule
2. Permutations:
3. Combinations:
IV. Event Relations
1. Unions and intersections
2. Events
a. Disjoint or mutually exclusive: P(A Ç B) = 0
b. Complementary: P(A) = 1 - P(AC
)
)!
(
!
!
)!
(
!
r
n
r
n
C
r
n
n
P
n
r
n
r





Note 5 of 5E
Key Concepts
3. Conditional probability:
4. Independent and dependent events
5. Additive Rule of Probability:
6. Multiplicative Rule of Probability:
7. Law of Total Probability
8. Bayes’ Rule
)
(
)
(
)
|
(
B
P
B
A
P
B
A
P


)
(
)
(
)
(
)
( B
A
P
B
P
A
P
B
A
P 




)
|
(
)
(
)
( A
B
P
A
P
B
A
P 


Note 5 of 5E
Key Concepts
V. Discrete Random Variables and Probability
Distributions
1. Random variables, discrete and continuous
2. Properties of probability distributions
3. Mean or expected value of a discrete random
variable:
4. Variance and standard deviation of a discrete
random variable:
1
)
(
and
1
)
(
0 


 x
p
x
p
2
2
2
:
deviation
Standard
)
(
)
(
:
Variance







 x
p
x
)
(
:
Mean x
xp




Lecture Biostatistics and general concept about probability

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