Solution Manual for Fundamentals of Probability with Stochastic Processes (4th Edition, 2019) by Saeed Ghahramani

Covers All 12 Chapters




SOLUTIONS + TEST BANK

, C ontents
◮ 1 Axioms of Probability 1
1.2 Sample Space and Events 1
1.4 Basic Theorems 4
1.7 Random Selection of Points from Intervals 10
Review Problems 13

◮ Companion for Chapter 1 19

1B Applications of Probability to Genetics 19


◮
2 Combinatorial Methods 21
2.2 Counting Principle 21

2.3 Permutations 24
2.4 Combinations 27
2.5 Stirling’ Formula 42
Review Problems 42

◮ 3 Conditional Probability and Independence 47
3.1 Conditional Probability 47
3.2 The Multiplication Rule 52
3.3 Law of Total Probability 55
3.4 Bayes’ Formula 60
3.5 Independence 66
Review Problems 76
80

◮ Companion for Chapter 3

3B More on Applications of Probability to Genetics 80




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, Contents 3iii



◮ Distribution Functions and 85
4 Discrete Random Variables
4.2 Distribution Functions 85
4.3 Discrete Random Variables 89
4.4 Expectations of Discrete Random Variables 94
4.5 Variances and Moments of Discrete Random Variables 100
4.6 Standardized Random Variables 107
Review Problems 107
◮ 5 Special Discrete Distributions 110
5.1 Bernoulli and Binomial Random Variables 110
5.2 Poisson Random Variable 117
5.3 Other Discrete Random Variables 125
Review Problems 132
◮ 6 Continuous Random Variables 137
6.1 Probability Density Functions 137
6.2 Density Function of a Function of a Random Variable 141
6.3 Expectations and Variances 144
Review Problems 151
◮ 7 Special Continuous Distributions 153
7.1 Uniform Random Variable 153
7.2 Normal Random Variable 158
7.3 Exponential Random Variables 166
7.4 Gamma Distribution 171
7.5 Beta Distribution 175
7.6 Survival Analysis and Hazard Function 180
Review Problems 183
◮ 8 Bivariate Distributions 187
8.1 Joint Distribution of Two Random Variables 187
8.2 Independent Random Variables 200
8.3 Conditional Distributions 209
8.4 Transformations of Two Random Variables 218
Review Problems 230




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, Contents 4iii

◮ 9 Multivariate Distributions 238
9.1 Joint Distribution of n > 2 Random Variables 238
9.2 Order Statistics 248
9.3 Multinomial Distributions 253
Review Problems 255
◮ 10 More Expectations and Variances 260
10.1 Expected Values of Sums of Random Variables 260
10.2 Covariance 265
10.3 Correlation 274
10.4 Conditioning on Random Variables 276
10.5 Bivariate Normal Distribution 289
Review Problems 292
◮ Companion for Chapter 10 299
10B Pattern Appearance 299
◮ Sums of Independent Random 300
11 Variables and Limit Theorems
11.1 Moment-Generating Functions 300
11.2 Sums of Independent Random Variables 308
11.3 Markov and Chebyshev Inequalities 314
11.4 Laws of Large Numbers 318
11.5 Central Limit Theorem 321
Review Problems 326
◮ 12 Stochastic Processes 330
12.2 More on Poisson Processes 330
12.3 Markov Chains 335
12.4 Continuous-Time Markov Chains 353
Review Problems 362
◮ Companion for Chapter 12 367
12B Brownian Motion 367




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,Chapter 1


A xioms of Probability
1.2 SAMPLE SPACE AND EVENTS

1. {M, I, S, P} is a sample space for this experiment, and {I} is the event that the out- come is
a vowel.

2. A sample space is S = {0, 1, 2, . . . , 57}. The desired event is E = {3, 4, 5, 6, 7, 8}.
3. E is the event of at least two heads.

4. E is the event that one die shows three times as many dots as the other. F is the event that
the sum of the outcomes is exactly 6.

5. For 1 ≤ i, j ≤ 3, by (i, j) we mean that Vann’s card number is i, and Paul’s card number is j.
Clearly, A = (1, 2), (1, 3), (2, 3) and B = (2, 1), (3, 1), (3, 2) .
(a) Since A ∩ B = ∅, the events A and B are mutually exclusive.
(b) None of (1, 1), (2, 2), (3, 3) belongs to A ∪ B. Hence A ∪ B not being the sample
space shows that A and B are not complements of one another.

6. S = {RRR, RRB, RBR, RBB, BRR, BRB, BBR, BBB}.
7. {x : 0 < x < 20}; {1, 2, 3, . . . , 19}.
8. Denote the dictionaries by d1, d2; the third book by a. The answers are
{d1d2a, d1ad2, d2d1a, d2ad 1,ad1d2, ad2d1} and {d1d2a, ad1d2}.
9. EF : One 1 and one even.
Ec F : One 1 and one odd.
Ec F c : Both even or both belong to {3, 5}.
10. S = {QQ, QN, QP, QD, DN, DP, NP, NN, PP }. (a) {QP };
(b) {DN, DP, NN }; (c) ∅.
11. S =x: 7x≤: 7x≤≤x7≤1 9 1 ; ∪ x: 7 3 ≤ x ≤ 8 1 ∪ x: 8 3 ≤ x ≤ 9 1 .
4 6 4 4 4 6




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, Section 1.2 Sample Space and Events 22

12. (a) The sample space is S = {A1A2, AcA2, A1Ac, AcAc }.
1 2 1 2

(b) The event that the system is notcoperativec atc the random time is
E = {A A2 , A1A , A Ac }.
1 2 1 2

13. S is a sample space for the experiment of flipping a coin until two tails appear con-
secutively.

14. E ∪ F ∪ G = G: If E or F occurs, then G occurs.
EFG = G: If G occurs, then E and F occur.

15. For 1 ≤ i ≤ 3, 1 ≤ j ≤ 3, by aibj we mean passenger a gets off at hotel i and passenger b
gets off at hotel j. The answers are {aibj : 1 ≤ i ≤ 3, 1 ≤ j ≤ 3} and
{a1b1, a 2b2, a3b3}, respectively.
16. Let a, ℓ, and f represent the outcomes in which the subject identifies almond, lemon, and flax
as his or her favorite color, respectively. Let ∼a, ∼ℓ, and ∼f represent the outcomes in
which the subject does not identify almond, lemon, and flax as his or her favorite color,
respectively. The sample space of the experiment is

S = aℓf, (∼a)ℓf, a(∼ℓ)f, aℓ(∼f),(∼a)(∼ℓ)f,(∼a)ℓ(∼f),a(∼ℓ)(∼f),
(∼a)(∼ℓ)(∼f) .
17. Let x, y, and z be the demand, in thousands, in a random month, for band saws,
reciprocating saws, and hole saws, respectively. The sample space is
S = (x, y, z): 30 ≤ x ≤ 36, 28 ≤ y ≤ 33, 300 ≤ z ≤ 600 .
The desired event is
(x, y, z) ∈ S : x ≥ 33, z < 435 .
18. Let ad be the outcome that Alexia is dead in five years, and let aℓ be the outcome that she lives
at that time. Define rd and rℓ similarly. A sample space for this experiment is S = (ad, rd), (ad,
rℓ), (aℓ, rd), (aℓ, rℓ) . The event that at that time only one of them lives is E = (ad, rℓ), (aℓ, rd) .
19. (a) (E ∪ F)(F ∪ G) = (F ∪ E)(F ∪ G) = F ∪ EG.
(b) Using part (a), we have
(E∪F )(Ec ∪F )(E∪F c ) = (F ∪EEc )(E ∪F c ) = F (E∪F c ) = FE ∪FF c = FE.
20. (a) ABc C c ; (b) A ∪ B ∪ C; (c) Ac Bc Cc ; (d) ABCc ∪ ABc C ∪ Ac BC;
(e) ABcCc ∪ Ac Bc C ∪ Ac BCc ; (f) (A − B) ∪ (B − A) = (A ∪ B) − AB.




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, Section 1.2 Sample Space and Events 32

21. The event that the device is operative at that random time is
n
E= Ai = A1A2 · · · An.
i=1
22. If B = ∅, the relation is obvious. If the relation is true for every event A, then it is true for S,
the sample space, as well. Thus
S = (B ∩ Sc ) ∪ (Bc ∩ S) = ∅ ∪ Bc = Bc ,
showing that B = ∅.
23. Parts (a) and (d) are obviously true; part (c) is true by DeMorgan’s law; part (b) is false:
throw a four-sided die; let F = {1, 2, 3}, G = {2, 3, 4}, E = {1, 4}.
24. Introducing a rectangular coordinate system with origin at the center of dartboard, we have that
a sample space for the point at which a dart hits the board is S = (x, y) : x2 + y 2 < 81 .

25. (a) The sample space is S = H, TH, TTH, TTTH , ... . (b) The desired event is
E = H, TTH, T T T T H, T T T T T T H, ... .
26. (a) ∞ An; (b) 37n=1 An. ∞
27. Clearly,n=1
E1 ⊃ E2 ⊃ E3 ⊃ · · · ⊃ Ei ⊃ · · · . Hence i=1 Ei
= E1 = (−1/2, 1/2). Now the only point
that belongs to all E i’s is 0. For any other point, x, x ∈ (−1, 1),
there is an i for which x ∈/ (−1/2 i , 1/2i). So ∞ i=1 Ei = {0}.

28. Straightforward.

29. Straightforward.

30. Straightforward.

31. Let a1, a2, and a3 be the first, the second, and the third volumes of the dictionary. Let a4, a5,
a6, and a7 be the remaining books. Let A = {a1, a2, . . . , a7}; the answers are
S = x 1x 2 x 3x 4x 5 x 6x 7 : xi ∈ A, 1 ≤ i ≤ 7, and xi = xj if i = j
and
x1x2x3x4x5x6x7 ∈ S : xixi+1xi+2 = a1a2a3 for some i, 1 ≤ i ≤ 5 ,
respectively.
32. The sample space is
S= 31 Ai1Ai2Ai3 ∪ Ac Ai2Ai3 ∪ Ai1Ac Ai3 ∪ Ai1Ai2Ac c c
i3
i1 i2 i3 ∪ Ai1Ai2 A
i=1 c c c c
∪ A Ai2A ∪ Ai1A A ∪ Ac Ac Ac .
i1 i3 i2 i3 i1 i2 i3




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, Section 1.4 Basic Theorems 4
∞ ∞
33. An.
m=1 n=m
n−1
34. Let B1 = A1 , B2 = A2 − A1 , B3 = A3 − (A1 ∪ A2), . . . , Bn = An − Ai,
. . .. i=1



1.4 BASIC THEOREMS

1. No; P (sum 11) = 2/36 while P(sum 12) = 1/36.
2. Since P (AB) = 0, we have 1 ≥ P A ∪ B = P (A) + P (B).
3. 0.33 + 0.07 = 0.40.

4. No, they are not consistent. The first statement implies that the probability of success is
15/16, while the second statement implies that it is 1/16.

5. Let E be the event that an earthquake will damage the structure next year. Let H be the
event that a hurricane will damage the structure next year. We are given that P (E) = 0.015,
P(H) = 0.025, and P (EH) = 0.0073. Since
P(E ∪ H) = P(E) + P(H) − P(EH) = 0.015 + 0.025 − 0.0073 = 0.0327,
the probability that next year the structure will be damaged by an earthquake and/or a
hurricane is 0.0327. The probability that it is not damaged by any of the two natural
disasters is 0.9673.
6. Clearly, she made a mistake. Since EF ⊆ E, we must have P (EF ) ≤ P (E).
However, in Tina’s calculations, P (EF ) = 3 1
> = P(E).
8 4
7. We are interested in the probability of the event A ∪ B − AB. Since AB ⊆ A ∪ B, we have P
A ∪ B − AB = P A ∪ B − P (AB) = 0.8 − 0.3 = 0.5.
8. Let A be the event that a randomly selected applicant has a high school GPA of at least
3.0. Let B be the event that this applicant’s SAT score is 1200 or higher. We have
P A ∪ B) = P(A) + P(B) − P (AB) = 0.38 + 0.30 − 0.15 = 0.53.
Therefore, 53% of all applicants are admitted to the college.

9. Let J, B, and T be the events that Jacqueline, Bonnie, and Tina win, respectively. We are
given that P (B) = (2/3)P (J) and P (B) = (4/3)P (T ). Therefore, P (J) = (3/2)P (B) and P (T ) =
(3/4)P (B). Now P (J) + P (B) + P (T ) = 1 implies that
3 3
P (B) + P (B) + P (B) = 1.
2 4
This gives P(B) = 4/13. Thus P(J) = 6/13, and P(T ) = 3/13.




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, Section 1.4 Basic Theorems 5


10. Let M be the event that the next motorcycle policy holder of this company who gets into an
accident is married. Let F be the event that the person is female, and let U be the event that
he or she is under 40. The desired probability is calculated as follows:
5000 − 4100 1100 − 550 1 =0.05.
P(FUM c ) = P (FU ) − P(FUM ) = − =
7000 7000 20
11. The probability that at least two cards must be drawn to obtain a face card is equal to the
probability that the first card drawn is not a face card, which is 40/52 = 10/13 = 0.769.

12. Let A be the event of a randomly selected driver having an accident during the next 12
months. Let B be the event that the person is male. By Theorem 1.7, the desired probability
is
P(A) = P (AB) + P (ABc ) = 0.12 + 0.06 = 0.18.

13. Let A be the event that a randomly selected investor invests in traditional annuities. Let B be
the event that he or she invests in the stock market. Then P (A) = 0.75, P (B) = 0.45, and P
(A ∪ B) = 0.85. Since,
P(AB) = P (A) + P(B) − P (A ∪ B) = 0.75 + 0.45 − 0.85 = 0.35,
35% invest in both stock market and traditional annuities.

14. The probability that the first horse wins is 2/7. The probability that the second horse wins is
3/10. Since the events that the first horse wins and the second horse wins are mutually
exclusive, the probability that either the first horse or the second horse will
win is
2 3 41
+ = .
7 10 70
15. No, it is not, because not all states have the same number of representatives. For
example, Delaware has only one representative, Massachusetts has 9, and California has
53. So the probabilities that each of these states is selected are 1/435, 9/435, and 53/435,
respectively. Obviously, under Walter’s method it is not equally likely that each US state is
selected. If he had proposed a method to draw a senator from the list of the 100 voting
members of the US senate and then select the state to which that member belonged, then
the state selected would be random, because all states have the same number of senators
in the congress.

16. The probability that he or she reaches the age of 60 is
0.2089 + 0.2710 + 0.1683 + 0.0277 + 0.0052 = 0.06811.

The probability that he or she lives to be at least 20 but dies before the age 50 is

0.0218 + 0.0317 + 0.0634 = 0.1169.




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, Section 1.4 Basic Theorems 6
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