Introduction to Probability 2nd Edition by Dimitri P. Bertsekas and John N. Tsitsiklis – Complete Problem Solutions Manual for Probability Theory and Statistics

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Introduction to
Probability 2nd
Edition
Problem Solutions


Dimitri P. Bertsekas and John N. Tsitsiklis
Massachusetts Institute of Technology




Athena Scientific, Belmont, Massachusetts




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CHAPT ER 1




Solution to Problem 1.1. We have

A = {2, 4, 6}, B = {4, 5, 6},

so A ∪ B = {2, 4, 5, 6},
and (A ∪ B)c = {1, 3}.


On the other hand,

Ac ∩ Bc = {1, 3, 5} ∩ {1, 2, 3} = {1, 3}.

Similarly, we have A ∩ B = {4, 6}, and

(A ∩ B)c = {1, 2, 3, 5}.

On the other hand,

Ac ∪ Bc = {1, 3, 5} ∪ {1, 2, 3} = {1, 2, 3, 5}.


Solution to Problem 1.2. (a) By using a Venn diagram it can be seen that for
any sets S and T , we have
S = (S ∩ T ) ∪ (S ∩ Tc).
(Alternatively, argue that any x must belong to either T or to T c , so x belongs
to S if and only if it belongs to S ∩ T or to S ∩ Tc.) Apply this equality with S =
Ac and T = B, to obtain the first relation

Ac = (Ac ∩ B) ∪ (Ac ∩ Bc).

Interchange the roles of A and B to obtain the second relation.
(b) By De Morgan’s law, we have

(A ∩ B)c = Ac ∪ Bc,

and by using the equalities of part (a), we obtain

(A∩B)c = (Ac∩B)∪(Ac∩Bc) ∪ (A∩Bc)∪(Ac∩Bc) = (Ac∩B)∪(Ac∩Bc)∪(A∩Bc).

(c) We have A = {1, 3, 5} and B = {1, 2, 3}, so A ∩ B = {1, 3}. Therefore,

(A ∩ B)c = {2, 4, 5, 6},



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and
Ac ∩ B = {2}, Ac ∩ Bc = {4, 6}, A ∩ Bc = {5}.
Thus, the equality of part (b) is verified.
Solution to Problem 1.5. Let G and C be the events that the chosen
student is a genius and a chocolate lover, respectively. We have P(G) = 0.6,
P(C) = 0.7, and P(G ∩ C) = 0.4. We are interested in P(Gc ∩ Cc), which is obtained
with the following calculation:

P(Gc ∩Cc) = 1—P(G∪C) = 1— P(G)+P(C)—P(G∩C) = 1—(0.6+0.7—0.4) = 0.1.

Solution to Problem 1.6. We first determine the probabilities of the six
possible outcomes. Let a = P({1}) = P({3}) = P({5}) and b = P({2}) =
P({4}) = P({6}).
We are given that b = 2a. By the additivity and normalization axioms, 1 = 3a +
3b = 3a + 6a = 9a. Thus, a = 1/9, b = 2/9, and P({1, 2, 3}) = 4/9.
Solution to Problem 1.7. The outcome of this experiment can be any finite
sequence of the form (a1, a2, . . . , an), where n is an arbitrary positive integer, a1,
a2, . . . , an—1 belong to {1, 3}, and an belongs to {2, 4}. In addition, there are
possible outcomes in which an even number is never obtained. Such outcomes
are infinite sequences (a1, a2, . . .), with each element in the sequence belonging to
{1, 3}. The sample space consists of all possible outcomes of the above two types.
Solution to Problem 1.8. Let pi be the probability of winning against the
opponent played in the ith turn. Then, you will win the tournament if you win
against the 2nd player (probability p2) and also you win against at least one of the
two other players [probability p1 + (1 — p1)p3 = p1 + p3 — p1p3]. Thus, the
probability of winning the tournament is
p2(p1 + p3 — p1p3).

The order (1, 2, 3) is optimal if and only if the above probability is no less than
the probabilities corresponding to the two alternative orders, i.e.,

p2(p1 + p3 — p1p3) ≥ p1(p2 + p3 —

p2p3), p2(p1 + p3 — p1p3) ≥ p3(p2 + p1

— p2p1).
It can be seen that the first inequality above is equivalent to p2 ≥ p1, while the second
inequality above is equivalent to p2 ≥ p3.
Solution to Problem 1.9. (a) Since Ω = ∪n Si, we have
n

A= (A ∩ Si),
i=1

while the sets A ∩ Si are disjoint. The result follows by using the additivity axiom.
(b) The events B ∩ Cc, Bc ∩ C, B ∩ C, and Bc ∩ Cc form a partition of Ω, so by
part (a), we have

P(A) = P(A ∩ B ∩ Cc) + P(A ∩ Bc ∩ C) + P(A ∩ B ∩ C) + P(A ∩ Bc ∩ Cc). (1)

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The event A ∩ B can be written as the union of two disjoint events as follows:

A ∩ B = (A ∩ B ∩ C) ∪ (A ∩ B ∩ Cc),

so that
P(A ∩ B) = P(A ∩ B ∩ C) + P(A ∩ B ∩ Cc). (2)
Similarly
P(A ∩ C) = P(A ∩ B ∩ C) + P(A ∩ Bc ∩ C). (3)

,


Combining Eqs. (1)-(3), we obtain the desired result.
Solution to Problem 1.10. Since the events A ∩ Bc and Ac ∩ B are disjoint, we
have using the additivity axiom repeatedly,

P (A∩Bc)∪(Ac ∩B) = P(A∩Bc)+P(Ac ∩B) = P(A)—P(A∩B)+P(B)—P(A∩B).

Solution to Problem 1.14. (a) Each possible outcome has probability 1/36.
There are 6 possible outcomes that are doubles, so the probability of doubles is
6/36 = 1/6.
(b) The conditioning event (sum is 4 or less) consists of the 6 outcomes
}
(1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (3, 1) ,

2 of which are doubles, so the conditional probability of doubles is 2/6 = 1/3.
(c) There are 11 possible outcomes with at least one 6, namely, (6, 6), (6, i), and (i,
6), for i = 1, 2, . . . , 5. Thus, the probability that at least one die is a 6 is 11/36.
(d) There are 30 possible outcomes where the dice land on different numbers.
Out of these, there are 10 outcomes in which at least one of the rolls is a 6. Thus,
the desired conditional probability is 10/30 = 1/3.
Solution to Problem 1.15. Let A be the event that the first toss is a head
and let B be the event that the second toss is a head. We must compare the
conditional probabilities P(A ∩ B | A) and P(A ∩ B | A ∪ B). We have

P (A ∩ B) ∩ A P(A ∩ B)
P(A ∩ B | A) = = ,

and
P (A ∩ B) ∩ (A ∪ B) P(A ∩ B)
P(A ∩ B | A ∪ B) = = .
P(A ∪ B) P(A ∪ B)
Since P(A ∪ B) ≥ P(A), the first conditional probability above is at least as large,
so Alice is right, regardless of whether the coin is fair or not. In the case where
the coin is fair, that is, if all four outcomes HH, HT , TH, TT are equally likely,
we have

P(A ∩ B) 1/4 1 P(A ∩ B) 1/4 1
= = , = = .
P(A) 1/2 2 P(A ∪ 3/4 3
B)

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