Probabilistic Systems Analysis- Edu Assignment Help

Probabilistic Systems Analysis
1. Express each of the following events in terms of the events A, B and C as
well as the operations of complementation, union and intersection:
(a) at least one of the events A, B, C occurs;
(b) at most one of the events A, B, C occurs;
(c) none of the events A, B, C occurs;
(d) all three events A, B, C occur;
(e) exactly one of the events A, B, C occurs;
(f) events A and B occur, but not C;
(g) either event A occurs or, if not, then B also does not occur. In each case
draw the corresponding Venn diagrams.
2. You flip a fair coin 3 times, determine the probability of the below events.
Assume all sequences are equally likely.
(a) Three heads: HHH
(b) The sequence head, tail, head: HTH
(c) Any sequence with 2 heads and 1 tail
(d) Any sequence where the number of heads is greater than or equal to the
number of tails
Problems
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3. Bob has a peculiar pair of four-sided dice. When he rolls the dice, the
probability of any particular outcome is proportional to the sum of the results
of each die. All outcomes that result in a particular sum are equally likely.
(a) What is the probability of the sum being even?
(b) What is the probability of Bob rolling a 2 and a 3, in any order?
4. Alice and Bob each choose at random a number in the interval [0, 2]. We
assume a uniform probability law under which the probability of an event is
proportional to its area. Consider the following events:
A : The magnitude of the difference of the two numbers is greater than 1/3.
B : At least one of the numbers is greater than 1/3.
C : The two numbers are equal.
D : Alice’s number is greater than 1/3.
Find the probabilities P(B), P(C), and P(A ∩ D).
5. Mike and John are playing a friendly game of darts where the dart board is a
disk with radius of 10in.
Whenever a dart falls within 1in of the center, 50 points are scored. If the
point of impact is between 1 and 3in from the center, 30 points are scored, if
it is at a distance of 3 to 5in 20 points are scored and if it is further that 5in, 10
points are scored.
Assume that both players are skilled enough to be able to throw the dart
within the boundaries of the board.

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Mike can place the dart uniformly on the board (i.e., the probability
of the dart falling in a given region is proportional to its area).
(a) What is the probability that Mike scores 50 points on one
throw?
(b) What is the probability of him scoring 30 points on one throw?
(c) John is right handed and is twice more likely to throw in the
right half of the board than in the left half. Across each half, the
dart falls uniformly in that region. Answer the previous questions
for John’s throw.
6. Prove that for any three events A, B and C, we have
P(A ∩ B ∩ C) ≥ P(A) + P(B) + P(C) − 2.
G1†. Consider an experiment whose sample space is the real line.
(a) Let {an} be an increasing sequence of numbers that converges to a and
{bn} a decreasing sequence that converges to b. Show that
lim P([an, bn]) = P([a, b]).
n→∞
Here, the notation [a, b] stands for the closed interval {x | a ≤ x ≤ b}. Note:
This result seems intuitively obvious. The issue is to derive it using the axioms
of probability theory

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(b) Let {an} be a decreasing sequence that converges to a and {bn} an
increasing sequence that converges to b. Is it true that
lim P([an, bn]) = P([a, b])?
n→∞
Note: You may use freely the results from the problems in the text in your proofs
Solutions
1. (a) A ∪ B ∪ C
(b) (A ∩ Bc ∩ Cc) ∪ (Ac ∩ B ∩ Cc) ∪ (Ac ∩ Bc ∩ C) ∪ (Ac ∩ Bc ∩ Cc)
(c) (A ∪ B ∪ C)c = Ac ∩ Bc ∩ Cc
(d) A ∩ B ∩ C
(e) (A ∩ Bc ∩ Cc) ∪ (Ac ∩ B ∩ Cc) ∪ (Ac ∩ Bc ∩ C)
(f) A ∩ B ∩ Cc
(g) A ∪ (Ac ∩ Bc)

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2. Since all outcomes are equally likely we apply the discrete uniform
probability law to solve the problem. To solve for any event we simply count
the number of elements in the event and divide by the total number of
elements in the sample space.
There are 2 possible outcomes for each flip, and 3 flips. Thus there are 23 = 8
elements (or sequences) in the sample space.
(a) Any sequence has probability of 1/8. Therefore P({H, H, H}) = 1/8 .
(b) This is still a single sequence, thus P({H, T, H}) = 1/8 .
(c) The event of interest has 3 unique sequences, thus P({HHT, HTH, THH}) =
3/8 .
(d) The sequences where there are more heads than tails are A : {HHH, HHT,
HTH, THH}. 4 unique sequences gives us P(A) = 1/2 .
3. The easiest way to solve this problem is to make a table of some sort,
similar to the one below.

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5. (a) The probability of Mike scoring 50 points is proportional to the
area of the inner disk. Hence, it is equal to απR2 = απ, where α is a
constant to be determined. Since the probability of landing the
dart on the board is equal to one, απ102 = 1, which implies that α
= 1/(100π). Therefore, the probability that Mike scores 50 points is
equal to π/(100π) = 0.01
(b) In order to score exactly 30 points, Mike needs to place the dart
between 1 and 3 inches from the origin. An easy way to compute
this probability is to look first at that of scoring more than 30
points, which is equal to απ32 = 0.09. Next, since the 30 points ring
is disjoint from the 50 points disc, probability of scoring more than
30 points is equal to the probability of scoring 50 points plus that of
scoring exactly 30 points. Hence, the probability of Mike scoring
exactly 30 points is equal to 0.09 − 0.01 = 0.08
(c) For the part (a) question. The probability of John scoring 50 points
is equal to the probability of throwing in the right half of the board
and scoring 50 points plus that of throwing in the left half and
scoring 50 points.

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The first term in the sum is proportional to the area of the right half
of the inner disk and is equal to απR2/2 = απ/2, where α is a
constant to be determined.
Similarly, the probability of him throwing in the left half of the board
and scoring 50 points is equal to βπ/2, where β is a constant (not
necessarily equal to α). In order to determine α and β, let us
compute the probability of throwing the dart in the right half of the
board. This probability is equal to
Since that probability is equal to 2/3, α = 1/(75π). In a similar
fashion, β can be determined to be 1/(150π). Consequently, the
total probability is equal to 1/150 + 1/300 = 0.01
For the part (b), The probability of scoring exactly 30 points is equal
to that of scoring more than 30 points minus that of scoring exactly
50. By applying the same type of analysis as in (b) above, the
probability is found to be equal to 0.08
These numbers suggest that John and Mike have similar skills, and
are equally likely to win the game. The fact that Mike’s better
control (or worst, depending on how you look at it) of the direction
of his throw does not increase his chances of winning can be
explained by the observation that both players’ control over the
distance from the origin is identical.

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6. See the textbook, Problem 1.11 page 55, which proves the general
version of Bonferroni’s inequality.
(b) No. Consider the following example. Let an = a + n 1 , bn = b − n
1 for all n. Then {an} is a decreasing sequence that converges to a,
and {bn} is an increasing sequence that converges to b. If we define
a probability law that places non-zero probability only on points a
and b, then limn→∞ P([an, bn]) = 0, but P([a, b]) = 1.

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This example is closely related to the continuity property of
probabilities. In this case, if we define An = [an, bn], then A1, A2, . .
. is “monotonically increasing,” i.e., An ⊂ An+1, but A = (∪∞ n An) =
(a, b), which is an open interval whose probability is 0 under our
probability law.

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