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Additionally, a 404 Not Found error was encountered while trying to use an ErrorDocument to handle the request. Apache/2.4.41 (Ubuntu) Server at 25years.mitchellinstitute.org Port 443 solution manual fundamentals of probability with stochastic processes Ghahramani 4th edition pdfPublished on Dec 11, 2021 Authors: Saeed Ghahramani Published: CRC 2019 Edition: 4th Pages: 375 Type: pdf Size: 5MB bookmarket0098 Instructor's Solutions Manual FUNDAMENTALS OF PROBABILITY WITH STOCHASTIC PROCESSES FOURTH EDITION SAEED GHAHRAMANI Western New England UniversitySpringfield, Massachusetts, USAA CHAPMAN & HALL BOOKFOLFNKHUHWRGRZQORDG Contents◮ 11 Sums of Independent Random
Chapter 1Axioms of Probability1 Sample Space and Events1. {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 eventthat 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 cardnumber 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 }.FOLFNKHUHWRGRZQORDG Section 1 Sample Space and Events 3 21. The event that the device is operative at that random time isE=⋂n i= Ai=A 1 A 2 ···An. 22. IfB=∅, the relation is obvious. If the relation is true for every eventA, then it istrue 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) isfalse: 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, wehave 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) ⋃ 37n=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. Leta 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 isS=⋃ 31i= (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 ).FOLFNKHUHWRGRZQORDG 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 Theorems1. 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 successis 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. LetHbe 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)= 38>14=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 atleast 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)+34P(B)=1.This givesP(B)=4/ 13. ThusP(J)=6/ 13 , andP(T)=3/ 13. FOLFNKHUHWRGRZQORDG |
Fundamentals of probability with stochastic processes 4th edition solutions
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