Transcription of Solution Manual for: Introduction to Probability …
1 Solution Manual for: Introduction to Probability Models: Eighth Editionby Sheldon M. L. Weatherwax October 26, 2008 IntroductionChapter 1: Introduction to Probability TheoryChapter 1: ExercisesExercise 8 (Bonferroni s inequality)From the inclusion/exclusion identity for two sets we haveP(E F) =P(E) +P(F) P(EF).SinceP(E F) 1, the above becomesP(E) +P(F) P(EF) (EF) P(E) +P(F) 1,which is known as Bonferroni s inequality. From the numbersgiven we find thatP(EF) + 1 = 1: The possible values for the sum of the values when twodie are 10 (Boole s inequality)We begin by decomposing the countable union of setsAiA1 A2 A3..into a countable union of disjoint setsCj.
2 Define these disjoint sets asC1=A1C2=A2\A1C3=A3\(A1 A2)C4=A4\(A1 A2 A3)..Cj=Aj\(A1 A2 A3 Aj 1)Then by constructionA1 A2 A3 =C1 C2 C3 ,and theCj s are disjoint, so that we havePr(A1 A2 A3 ) = Pr(C1 C2 C3 ) = jPr(Cj).Since Pr(Cj) Pr(Aj), for eachj, this sum is bounded above by jPr(Aj),Problem 11 (the Probability the sum of the die isi)We can explicitly enumerate these probabilities by counting the number of times each oc-currence happens, in Table 1 we have placed the sum of the two die in the center of eachsquare. Then by counting the number of squares where are sum equals each number fromtwo to twelve, we haveP2=136, P7=636=16P3=236=118, P8=536P4=336=112, P9=436=19P5=436=19, P10=336=112P6=536, P11=236=118, P12= 13 (winning at craps)From Problem 11 we have computed the individual probabilities for various sum of tworandom die.
3 Following the hint, letEibe the event that the initial die sum toiand thatthe player wins. We can compute some of these probabilities immediatelyP(E2) =P(E3) =P(E12) = 0, andP(E7) =P(E11) = 1. We now need to computeP(Ei) fori= 4,5,6,8,9, following the hint defineEi,nto be the event that the player initial sum isiand winson then-thsubsequentroll. ThenP(Ei) = n=1P(Ei,n),since if we win, it must be either on the first, or second, or third, etc rollafter the initialroll. We now need to calculate theP(Ei,n) probabilities for eachn. As an example of thiscalculation first lets computeP(E4,n) which means that we initially roll a sum of four andthe player wins on then-th subsequent roll.
4 We will win if we roll a sum of a four or looseif we roll a sum of a seven, while if roll anything else we continue, so to win whenn= 1 wesee thatP(E4,1) =1 + 1 + 136=112,since to get a sum of four we can roll pairs consisting of (1,3), (2,2), and (3,1).To computeP(E4,2) the rules of craps state that we will win if a sum of four comesup (withprobability112) and loose if a sum of a seven comes up (with probability636=16) and continueplaying if anything else is rolled. This last event (continued play) happens with probability1 112 16= (E4,2) =(34)112=116. Here the first34is the Probability we don t roll a four or aseven on then= 1 roll and the second112comes from rolling a sum of a four on the secondroll (wheren= 2).
5 In the same way we have forP(E4,3) the followingP(E4,3) =(34) the first two factors of34are from the two rolls that keep us in the game , and thefactor of112, is the roll that allows us to win. Continuing in this in this manner we see thatP(E4,4) =(34)3112,and in general we find thatP(E4,n) =(34)n 1112forn computeP(Ei,n) for otheri, the derivations just performed, only change in the probabil-ities required to roll the initial sum. We thus find that for other initial rolls (heavily usingthe results of Problem 24) thatP(E5,n) =19(1 19 16)n 1=19(1318)n 1P(E6,n) =536(1 536 16)n 1=536(2536)n 1P(E8,n) =536(1 536 16)n 1=536(2536)n 1P(E9,n) =19(1 19 16)n 1=19(1318)n 1P(E10,n) =112(1 112 16)n 1=112(34)n computeP(E4) we need to sum the results above.
6 We have thatP(E4) =112 n 1(34)n 1=112 n 0(34)n=1121(1 34)= that this also gives the Probability forP(E10). ForP(E5) we findP(E5) =25, whichalso equalsP(E9). ForP(E6) we find thatP(E6) =511, which also equalsP(E8). Then ourprobability of winning craps is given by summing all of the above probabilities weighted bythe associated priors of rolling the given initial roll. We find by definingIito be the eventthat the initial roll isiandWthe event that we win at craps thatP(W) = 0P(I2) + 0P(I3) +13P(I4) +49P(I5) +59P(I6)+ 1P(I7) +59P(I8) +49P(I9) +13P(I10) + 1P(I11) + 0P(I12).Using the results of Exercise 25 to evaluateP(Ii) for eachiwe find that the above summationgivesP(W) =244495= calculations are performed in the Matlab 15 (some set identities)We want to prove thatE= (E F) (E Fc).
7 We will do this using the standard proofwhere we show that each set in the above is a subset of the other. We begin withx ifx F,xwill certainly be inE F, while ifx / Fthenxwill be inE Fc. Thusin either case (x Forx / F)xwill be in the set (E F) (E Fc).Ifx (E F) (E Fc) thenxis in eitherE F,E Fc, or both by the definition ofthe union operation. Nowxcannot be in both sets or else it would simultaneously be inFandFc, soxmust be in one of the two sets only. Being in either set means thatx Eandwe have that the set (E F) (E Fc) is a subset ofE. Since each side is a subset of theother we have shown set prove thatE F=E (Ec F), we will begin by lettingx E F, thusxis an elementofEor an element ofFor of both.
8 Ifxis inEat all then it is in the setE (Ec F). Ifx / Ethen it must be inFto be inE Fand it will therefore be inEc F. Again bothsides are subsets of the other and we have shown set 23 (conditioning on a chain of events)This result follows for the two set caseP{A B}=P{A|B}P{B}by grouping the sequenceofEi s in the appropriate manner. For example by grouping the intersection asE1 E2 En 1 En= (E1 E2 En 1) Enwe can apply the two set result to obtainP{E1 E2 En 1 En}=P{En|E1 E2 En 1}P{E1 E2 En 1}.Continuing now to pealEn 1from the setE1 E2 En 1we have the second probabilityabove equal toP{E1 E2 En 2 En 1}=P{En 1|E1 E2 En 2}P{E1 E2 En 2}.
9 Continuing to peal off terms from the back we eventually obtain the requested {E1 E2 En 1 En}=P{En|E1 E2 En 1} P{En 1|E1 E2 En 2} P{En 2|E1 E2 En 3}.. P{E3|E1 E2} P{E2|E1} P{E1}.Exercise 30 (target shooting with Bill and George)warning! not the event that the duck is hit , by either Bill or George sshot. LetBandGbe theevents that Bill (respectively George) hit the target. Thenthe outcome of the experimentwhere both George and Bill fire at the target (assuming that their shots work independentlyis)P(Bc, Gc) = (1 p1)(1 p2)P(Bc, G) = (1 p1)p2P(B, Gc) =p1(1 p2)P(B, G) = (a):We desire to computeP(B, G|H) which equalsP(B, G|H) =P(B, G, H)P(H)=P(B, G)P(H)NowP(H) = (1 p1)p2+p1(1 p2) +p1p2so the above Probability becomesp1p2(1 p1)p2+p1(1 p2) +p1p2=p1p2p1+p2 (b):We desire to computeP(B|H) which equalsP(B|H) =P(B, G|H) +P(B, Gc|H).
10 Since the first termP(B, G|H) has already been computed we only need to computeP(B, Gc|H).As before we find it to beP(B, Gc|H) =p1(1 p2)(1 p1)p2+p1(1 p2) + the total result becomesP(B|H) =p1p2+p1(1 p2)(1 p1)p2+p1(1 p2) +p1p2=p1p1+p2 33 (independence in class)LetSbe a random variable denoting the sex of the randomly selected person. TheScantake on the valuesmfor male andffor female. LetCbe a random variable representingdenoting the class of the chosen student. TheCcan take on the valuesffor freshman andsfor sophomore. We want to select the number of sophomore girls such that the randomvariablesSandCare independent. Letndenote the number of sophomore girls.
