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Solutions Manual for Fundamentals of Probability with Stochastic Processes 3rd Edition by Saeed Ghahramani - Tutor website
Complete downloadable Solutions Manual for Fundamentals of Probability with Stochastic Processes 3rd Edition by Saeed Ghahramani. INSTRUCTOR RESOURCE INFORMATION TITLE: Fundamen…
Solutions Manual for Fundamentals of Probability with Stochastic Processes 3rd Edition by Saeed Ghahramani
Instructor's Solutions Manual
FUNDAMENTALS
OF PROBABILITY
WITH STOCHASTIC PROCESSES
FOURTH EDITION
SAEED GHAHRAMANI
Western New England University
Springfield, Massachusetts, USA
A CHAPMAN & HALL BOOK
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Contents
◮ 11 Sums of Independent Random
- ◮ 1 Axioms of Probability
- 1 Sample Space and Events
- 1 Basic Theorems
- 1 Random Selection of Points from Intervals - Review Problems
- ◮ Companion for Chapter
- 1B Applications of Probability to Genetics
- ◮ 2 Combinatorial Methods
- 2 Counting Principle
- 2 Permutations
- 2 Combinations
- 2 Stirling’ Formula - Review Problems
- ◮ 3 Conditional Probability and Independence
- 3 Conditional Probability
- 3 The Multiplication Rule
- 3 Law of Total Probability
- 3 Bayes’ Formula
- 3 Independence - Review Problems
- ◮ Companion for Chapter
- 3B More on Applications of Probability to Genetics
- ◮ 9 Multivariate Distributions Contents iv
- 9 Joint Distribution ofn> 2 Random Variables
- 9 Order Statistics
- 9 Multinomial Distributions - Review Problems
- ◮ 10 More Expectations and Variances
- 10 Expected Values of Sums of Random Variables
- 10 Covariance
- 10 Correlation
- 10 Conditioning on Random Variables
- 10 Bivariate Normal Distribution - Review Problems - ◮ Companion for Chapter
- 10B Pattern Appearance - Variables and Limit Theorems
- 11 Moment-Generating Functions
- 11 Sums of Independent Random Variables
- 11 Markov and Chebyshev Inequalities
- 11 Laws of Large Numbers
- 11 Central Limit Theorem - Review Problems - ◮ 12 Stochastic Processes
- 12 More on Poisson Processes
- 12 Markov Chains
- 12 Continuous-Time Markov Chains - Review Problems - ◮ Companion for Chapter
- 12B Brownian Motion
Chapter 1
Axioms of Probability
1 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 isS={ 0 , 1 , 2 ,..., 57 }. The desired event isE={ 3 , 4 , 5 , 6 , 7 , 8 }.
3. Eis the event of at least two heads.
4. Eis the event that one die shows three times as many dots as the other 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 isi, and Paul’s card
number isj. Clearly,A=
{
(1,2),(1,3),(2,3)
}
andB=
{
(2,1),(3,1),(3,2)
}
.
(a) SinceA∩B=∅, the eventsAandBare mutually exclusive.
(b) None of(1,1),(2,2),(3,3)belongs toA∪B. HenceA∪Bnot being the sample space shows thatAandBare 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 byd 1 ,d 2 ; the third book bya. The answers are
{d 1 d 2 a, d 1 ad 2 ,d 2 d 1 a, d 2 ad 1 ,ad 1 d 2 ,ad 2 d 1 }and{d 1 d 2 a, ad 1 d 2 }.
9. EF: One 1 and one even.
EcF: One 1 and one odd. EcFc: Both even or both belong to{ 3 , 5 }.
10. S={QQ, QN, QP, QD, DN, DP, NP, NN, P P}. (a) {QP};
(b) {DN, DP, NN}; (c) ∅.
11. S=
{
x:7≤x≤ 916
}
{ ;
x:7≤x≤ 714
}
∪
{
x:7 34 ≤x≤ 814
}
∪
{
x:8 34 ≤x≤ 916
}
.
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Section 1 Sample Space and Events 3
21. The event that the device is operative at that random time is
E=
⋂n
i=
Ai=A 1 A 2 ···An.
22. IfB=∅, the relation is obvious. If the relation is true for every eventA, then it is
true forS, the sample space, as well. Thus
S=(B∩Sc)∪(Bc∩S)=∅∪Bc=Bc,
showing thatB=∅.
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; letF={ 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 isS=
{
(x, y): x 2 +y 2 < 81
}
.
25.(a) The sample space isS=
{
H,TH,TTH,TTTH,...
}
. (b) The desired event is E=
{
H,TTH,TTTTH,TTTTTTH,...
}
.
26.(a)
⋃∞
n=1An; (b)
⋃ 37
n=1An.
27. Clearly,E 1 ⊃E 2 ⊃E 3 ⊃···⊃Ei⊃···. Hence
⋃∞
i=1Ei=E 1 =(− 1 / 2 , 1 /2). Now the only point that belongs to allEi’s is 0. For any other point,x,x∈(− 1 ,1), there is anifor whichx/∈(− 1 / 2 i, 1 / 2 i).So
⋂∞
i=1Ei={ 0 }.
28. Straightforward.
29. Straightforward.
30. Straightforward.
31. Leta 1 ,a 2 , anda 3 be the first, the second, and the third volumes of the dictionary. Let
a 4 ,a 5 ,a 6 , anda 7 be the remaining books. LetA={a 1 ,a 2 ,...,a 7 }; the answers are S=
{
x 1 x 2 x 3 x 4 x 5 x 6 x 7 :xi∈A, 1 ≤i≤ 7 ,andxi =xjifi =j
}
and { x 1 x 2 x 3 x 4 x 5 x 6 x 7 ∈S:xixi+1xi+2=a 1 a 2 a 3 for somei, 1 ≤i≤ 5
}
,
respectively.
32. The sample space is
S=
⋃ 31
i=
(
Ai 1 Ai 2 Ai 3 ∪Aci 1 Ai 2 Ai 3 ∪Ai 1 Aci 2 Ai 3 ∪Ai 1 Ai 2 Aci 3 ∪Aci 1 Aci 2 Ai 3
∪Aci 1 Ai 2 Aci 3 ∪Ai 1 Aci 2 Aci 3 ∪Aci 1 Aci 2 Aci 3
)
.
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Section 1 Basic Theorems 4
33.
⋂∞
m=
⋃∞
n=mAn.
34. LetB 1 =A 1 ,B 2 =A 2 −A 1 ,B 3 =A 3 −(A 1 ∪A 2 ),...,Bn=An−
⋃n− 1 i=1Ai, ....
1 Basic Theorems
1. No;P(sum 11)=2/ 36 whileP(sum 12)=1/ 36.
2. SinceP(AB)=0, we have 1 ≥P
(
A∪B
)
=P(A)+P(B).
3. 0 .33 + 0 = 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. LetEbe the event that an earthquake will damage the structure next year. LetH
be the event that a hurricane will damage the structure next year. We are given that P(E)=0. 015 ,P(H)=0. 025 , andP(EH)=0. 0073. Since
P(E∪H)=P(E)+P(H)−P(EH)=0 + 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. The probability that it is not damaged by any of the two natural disasters is 0.
6. Clearly, she made a mistake. SinceEF ⊆ E, we must haveP(EF) ≤ P(E).
However, in Tina’s calculations,P(EF)=
3
8
>
1
4
=P(E).
7. We are interested in the probability of the eventA∪B−AB. SinceAB⊆A∪B,
we haveP
(
A∪B−AB
)
=P
(
A∪B
)
−P(AB)=0. 8 − 0 .3=0. 5.
8. LetAbe the event that a randomly selected applicant has a high school GPA of at
least 3. LetBbe the event that this applicant’s SAT score is 1200 or higher. We have
P
(
A∪B)=P(A)+P(B)−P(AB)=0 + 0. 30 − 0 .15 = 0. 53.
Therefore, 53% of all applicants are admitted to the college.
9. LetJ,B, andT be the events that Jacqueline, Bonnie, and Tina win, respectively.
We are given thatP(B)=(2/3)P(J)andP(B)=(4/3)P(T). Therefore,P(J)= (3/2)P(B)andP(T)=(3/4)P(B).NowP(J)+P(B)+P(T)=1implies that 3 2
P(B)+P(B)+
3
4
P(B)=1.
This givesP(B)=4/ 13. ThusP(J)=6/ 13 , andP(T)=3/ 13.
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