prob-solved_2ndedition(概率导论第2版答案)

IntroductiontoProbability2ndEditionProblemSolutions(lastupdated:10/22/13)cDimitriP.BertsekasandJohnN.TsitsiklisMassachusettsInstituteofTechnologyc,Belmont,Massachusetts1CHAPTER1SolutiontoProblem1.1.WehaveA=f2;4;6g;B=f4;5;6g;soA[B=f2;4;5;6g,and(A[B)c=f1;3g:Ontheotherhand,Ac\Bc=f1;3;5g\f1;2;3g=f1;3g:Similarly,wehaveA\B=f4;6g,and(A\B)c=f1;2;3;5g:Ontheotherhand,Ac[Bc=f1;3;5g[f1;2;3g=f1;2;3;5g:SolutiontoProblem1.2.(a)ByusingaVenndiagramitcanbeseenthatforanysetsSandT,wehaveS=(S\T)[(S\Tc):(Alternatively,arguethatanyxmustbelongtoeitherTortoTc,soxbelongstoSifandonlyifitbelongstoS\TortoS\Tc.)ApplythisequalitywithS=AcandT=B,toobtaintherstrelationAc=(Ac\B)[(Ac\Bc):InterchangetherolesofAandBtoobtainthesecondrelation.(b)ByDeMorgan'slaw,wehave(A\B)c=Ac[Bc;andbyusingtheequalitiesofpart(a),weobtain(A\B)c=(Ac\B)[(Ac\Bc)
[(A\Bc)[(Ac\Bc)
=(Ac\B)[(Ac\Bc)[(A\Bc):(c)WehaveA=f1;3;5gandB=f1;2;3g,soA\B=f1;3g.Therefore,(A\B)c=f2;4;5;6g;2andAc\B=f2g;Ac\Bc=f4;6g;A\Bc=f5g:Thus,theequalityofpart(b)isveried.SolutiontoProblem1.5.LetGandCbetheeventsthatthechosenstudentisageniusandachocolatelover,respectively.WehaveP(G)=0:6,P(C)=0:7,andP(G\C)=0:4.WeareinterestedinP(Gc\Cc),whichisobtainedwiththefollowingcalculation:P(Gc\Cc)=1P(G[C)=1P(G)+P(C)P(G\C)
=1(0:6+0:70:4)=0:1:SolutiontoProblem1.6.Werstdeterminetheprobabilitiesofthesixpossibleoutcomes.Leta=P(f1g)=P(f3g)=P(f5g)andb=P(f2g)=P(f4g)=P(f6g).Wearegiventhatb=2a.Bytheadditivityandnormalizationaxioms,1=3a+3b=3a+6a=9a.Thus,a=1=9,b=2=9,andP(f1;2;3g)=4=9.SolutiontoProblem1.7.Theoutcomeofthisexperimentcanbeanynitesequenceoftheform(a1;a2;:::;an),wherenisanarbitrarypositiveinteger,a1;a2;:::;an1belongtof1;3g,andanbelongstof2;4g.Inaddition,therearepossibleoutcomesinwhichanevennumberisneverobtained.Suchoutcomesareinnitesequences(a1;a2;:::),witheachelementinthesequencebelongingtof1;3g.Thesamplespaceconsistsofallpossibleoutcomesoftheabovetwotypes.SolutiontoProblem1.8.Letpibetheprobabilityofwinningagainsttheopponentplayedintheithturn.Then,youwillwinthetournamentifyouwinagainstthe2ndplayer(probabilityp2)andalsoyouwinagainstatleastoneofthetwootherplayers[probabilityp1+(1p1)p3=p1+p3p1p3].Thus,theprobabilityofwinningthetournamentisp2(p1+p3p1p3):Theorder(1;2;3)isoptimalifandonlyiftheaboveprobabilityisnolessthantheprobabilitiescorrespondingtothetwoalternativeorders,i.e.,p2(p1+p3p1p3)p1(p2+p3p2p3);p2(p1+p3p1p3)p3(p2+p1p2p1):Itcanbeseenthattherstinequalityaboveisequivalenttop2p1,whilethesecondinequalityaboveisequivalenttop2p3.SolutiontoProblem1.9.(a)Since=[ni=1Si,wehaveA=n[i=1(A\Si);whilethesetsA\Siaredisjoint.Theresultfollowsbyusingtheadditivityaxiom.(b)TheeventsB\Cc,Bc\C,B\C,andBc\Ccformapartitionof,sobypart(a),wehaveP(A)=P(A\B\Cc)+P(A\Bc\C)+P(A\B\C)+P(A\Bc\Cc):(1)3TheeventA\Bcanbewrittenastheunionoftwodisjointeventsasfollows:A\B=(A\B\C)[(A\B\Cc);sothatP(A\B)=P(A\B\C)+P(A\B\Cc):(2)Similarly,P(A\C)=P(A\B\C)+P(A\Bc\C):(3)CombiningEqs.(1)-(3),weobtainthedesiredresult.SolutiontoProblem1.10.SincetheeventsA\BcandAc\Baredisjoint,wehaveusingtheadditivityaxiomrepeatedly,P(A\Bc)[(Ac\B)
=P(A\Bc)+P(Ac\B)=P(A)P(A\B)+P(B)P(A\B):SolutiontoProblem1.14.(a)Eachpossibleoutcomehasprobability1/36.Thereare6possibleoutcomesthataredoubles,sotheprobabilityofdoublesis6=36=1=6.(b)Theconditioningevent(sumis4orless)consistsofthe6outcomes(1;1);(1;2);(1;3);(2;1);(2;2);(3;1) ;2ofwhicharedoubles,sotheconditionalprobabilityofdoublesis2=6=1=3.(c)Thereare11possibleoutcomeswithatleastone6,namely,(6;6),(6;i),and(i;6),fori=1;2;:::;5.Thus,theprobabilitythatatleastonedieisa6is11/36.(d)Thereare30possibleoutcomeswherethedicelandondierentnumbers.Outofthese,thereare10outcomesinwhichatleastoneoftherollsisa6.Thus,thedesiredconditionalprobabilityis10=30=1=3.SolutiontoProblem1.15.LetAbetheeventthatthersttossisaheadandletBbetheeventthatthesecondtossisahead.WemustcomparetheconditionalprobabilitiesP(A\BjA)andP(A\BjA[B).WehaveP(A\BjA)=P(A\B)\A
P(A)=P(A\B)P(A);andP(A\BjA[B)=P(A\B)\(A[B)
P(A[B)=P(A\B)P(A[B):SinceP(A[B)P(A),therstconditionalprobabilityaboveisatleastaslarge,soAliceisright,regardlessofwhetherthecoinisfairornot.Inthecasewherethecoinisfair,thatis,ifallfouroutcomesHH,HT,TH,TTareequallylikely,wehaveP(A\B)P(A)=1=41=2=12;P(A\B)P(A[B)=1=43=4=13:AgeneralizationofAlice'sreasoningisthatifA,B,andCareeventssuchthatBCandA\B=A\C(forexample,ifABC),thentheeventAisatleast4aslikelyifweknowthatBhasoccurredthanifweknowthatChasoccurred.Alice'sreasoningcorrespondstothespecialcasewhereC=A[B.SolutiontoProblem1.16.Inthisproblem,thereisatendencytoreasonthatsincetheoppositefaceiseitherheadsortails,thedesiredprobabilityis1/2.Thisis,however,wrong,becausegiventhatheadscameup,itismorelikelythatthetwo-headedcoinwaschosen.Thecorrectreasoningistocalculatetheconditionalprobabilityp=P(two-headedcoinwaschosenjheadscameup)=P(two-headedcoinwaschosenandheadscameup)P(headscameup):WehaveP(two-headedcoinwaschosenandheadscameup)=13;P(headscameup)=12;sobytakingtheratiooftheabovetwoprobabil