商务与经济统计习题答案（第8版，中文版）SBE8-SM04

4-1Chapter4IntroductiontoProbabilityLearningObjectives1.Obtainanappreciationoftheroleprobabilityinformationplaysinthedecisionmakingprocess.2.Understandprobabilityasanumericalmeasureofthelikelihoodofoccurrence.3.Knowthethreemethodscommonlyusedforassigningprobabilitiesandunderstandwhentheyshouldbeused.4.Knowhowtousethelawsthatareavailableforcomputingtheprobabilitiesofevents.5.Understandhownewinformationcanbeusedtoreviseinitial(prior)probabilityestimatesusingBayes’theorem.Chapter44-2Solutions:1.NumberofexperimentalOutcomes=(3)(2)(4)=242.636!3365432132132120FHGIKJ!!()()ABCACEBCDBEFABDACFBCECDEABEADEBCFCDFABFADFBDECEFACDAEFBDFDEF3.P366!63654120()!()()()BDFBFDDBFDFBFBDFDB4.a.HTHTHTHTHTHTHT(H,H,H)(H,H,T)(H,T,H)(H,T,T)(T,H,H)(T,H,T)(T,T,H)(T,T,T)1stToss2ndToss3rdTossb.Let:HbeheadandTbetail(H,H,H)(T,H,H)(H,H,T)(T,H,T)(H,T,H)(T,T,H)(H,T,T)(T,T,T)c.Theoutcomesareequallylikely,sotheprobabilityofeachoutcomesis1/8.5.P(Ei)=1/5fori=1,2,3,4,5P(Ei)0fori=1,2,3,4,5P(E1)+P(E2)+P(E3)+P(E4)+P(E5)=1/5+1/5+1/5+1/5+1/5=1Theclassicalmethodwasused.IntroductiontoProbability4-36.P(E1)=.40,P(E2)=.26,P(E3)=.34Therelativefrequencymethodwasused.7.No.Requirement(4.3)isnotsatisfied;theprobabilitiesdonotsumto1.P(E1)+P(E2)+P(E3)+P(E4)=.10+.15+.40+.20=.858.a.Therearefouroutcomespossibleforthis2-stepexperiment;planningcommissionpositive-councilapproves;planningcommissionpositive-councildisapproves;planningcommissionnegative-councilapproves;planningcommissionnegative-councildisapproves.b.Letp=positive,n=negative,a=approves,andd=disapproves.PlanningCommissionCouncilpnadad(p,a)(p,d)(n,a)(n,d)9.50450!446!504948474321230300FHGIKJ!,10.a.Usetherelativefrequencyapproach:P(California)=1,434/2,374=.60b.Numbernotfrom4states=2,374-1,434-390-217-112=221P(Notfrom4States)=221/2,374=.09c.P(NotinEarlyStages)=1-.22=.78d.EstimateofnumberofMassachusettscompaniesinearlystageofdevelopment-(.22)39086Chapter44-4e.Ifweassumethesizeoftheawardsdidnotdifferbystates,wecanmultiplytheprobabilityanawardwenttoColoradobythetotalventurefundsdisbursedtogetanestimate.EstimateofColoradofunds=(112/2374)($32.4)=$1.53billionAuthors'Note:TheactualamountgoingtoColoradowas$1.74billion.11.a.No,theprobabilitiesdonotsumtoone.Theysumto.85.b.Ownermustrevisetheprobabilitiessotheysumto1.00.12.a.Usethecountingruleforcombinations:495495444948474645543211906884FHGIKJ!!!()()()()()()()()()(),,b.Verysmall:1/1,906,884=0.0000005c.Multiplytheanswertopart(a)by42togetthenumberofchoicesforthesixnumbers.No.ofChoices=(1,906,884)(42)=80,089,128ProbabilityofWinning=1/80,089,128=0.000000012513.Initiallyaprobabilityof.20wouldbeassignedifselectionisequallylikely.Datadoesnotappeartoconfirmthebeliefofequalconsumerpreference.Forexampleusingtherelativefrequencymethodwewouldassignaprobabilityof5/100=.05tothedesign1outcome,.15todesign2,.30todesign3,.40todesign4,and.10todesign5.14.a.P(E2)=1/4b.P(any2outcomes)=1/4+1/4=1/2c.P(any3outcomes)=1/4+1/4+1/4=3/415.a.S={aceofclubs,aceofdiamonds,aceofhearts,aceofspades}b.S={2ofclubs,3ofclubs,...,10ofclubs,Jofclubs,Qofclubs,Kofclubs,Aofclubs}c.Thereare12;jack,queen,orkingineachofthefoursuits.d.Fora:4/52=1/13=.08Forb:13/52=1/4=.25Forc:12/52=.23IntroductiontoProbability4-516.a.(6)(6)=36samplepointsb..123456123456234567345678456789567891011109876789101112Die1TotalforBothDie2c.6/36=1/6d.10/36=5/18e.No.P(odd)=18/36=P(even)=18/36or1/2forboth.f.Classical.Aprobabilityof1/36isassignedtoeachexperimentaloutcome.17.a.(4,6),(4,7),(4,8)b..05+.10+.15=.30c.(2,8),(3,8),(4,8)d..05+.05+.15=.25e..1518.a.0;probabilityis.05b.4,5;probabilityis.10+.10=.20c.0,1,2;probabilityis.05+.15+.35=.5519.a.Yes,theprobabilitiesareallgreaterthanorequaltozeroandtheysumtoone.b.P(A)=P(0)+P(1)+P(2)=.08+.18+.32=.58Chapter44-6c.P(B)=P(4)=.1220.a.P(N)=56/500=.112b.P(T)=43/500=.086c.Totalin6states=56+53+43+37+28+28=245P(B)=245/500=.49AlmosthalftheFortune500companiesareheadquarteredinthesestates.21.a.P(A)=P(1)+P(2)+P(3)+P(4)+P(5)=20501250650350150=.40+.24+.12+.06+.02=.84b.P(B)=P(3)+P(4)+P(5)=.12+.06+.02=.20c.P(2)=12/50=.2422.a.P(A)=.40,P(B)=.40,P(C)=.60b.P(AB)=P(E1,E2,E3,E4)=.80.YesP(AB)=P(A)+P(B).c.Ac={E3,E4,E5}Cc={E1,E4}P(Ac)=.60P(Cc)=.40d.ABc={E1,E2,E5}P(ABc)=.60e.P(BC)=P(E2,E3,E4,E5)=.8023.a.P(A)=P(E1)+P(E4)+P(E6)=.05+.25+.10=.40P(B)=P(E2)+P(E4)+P(E7)=.20+.25+.05=.50P(C)=P(E2)+P(E3)+P(E5)+P(E7)=.20+.20+.15+.05=.60b.AB={E1,E2,E4,E6,E7}P(AB)=P(E1)+P(E2)+P(E4)+P(E6)+P(E7)=.05+.20+.25+.10+.05=.65c.AB={E4}P(AB)=P(E4)=.25IntroductiontoProbability4-7d.Yes,theyaremutuallyexclusive.e.Bc={E1,E3,E5,E6};P(Bc)=P(E1)+P(E3)+P(E5)+P(E6)=.05+.20+.15+.10=.5024.P(CrashNotLikely)=1-.14-.43=.4325.LetY=highone-yearreturnM=highfive-yearreturna.P(Y)=15/30=.50P(M)=12/30=.40P(YM)=6/30=.20b.P(YM)=P(Y)+P(M)-P(YM)=.50+.40-.20=.70c.1-P(YM)=1-.70=.3026.LetY=highone-yearreturnM=highfive-yearreturna.P(Y)=9/30=.30P(M)=7/30=.23b.P(YM)=5/30=.17c.P(YM)=.30+.23-.17=.36P(Neither)=1-.36=.6427.Let:D=consumesorservesdomesticwineI=consumesorservesimportedwineWearegivenP(D)=0.57,P(I)=0.33,P(DI)=0.63P(DI)=P(D)+P(I)-P(D