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1、Chapter3:DiscreteRandomVariablesandTheirProbabilityDistributions3.1P(Y=0)=P(noimpurities)=.2,P(Y=1)=P(exactlyoneimpurity)=.7,P(Y=2)=.1.3.2WeknowthatP(HH)=P(TT)=P(HT)=P(TH)=0.25.So,P(Y=-1)=.5,P(Y=1)=.25=P(Y=2).3.3p(2)=P(DD)=1/6,p(3)=P(DGD)+P(GDD)=2(2/4)(2/3)(1/2)=2/6,p(4)=P(GGDD)+P(DG
2、GD)+P(GDGD)=3(2/4)(1/3)(2/2)=1/2.3.4Definetheevents:A:value1failsB:valve2failsC:valve3fails3P(Y=2)=P(A∩B∩C)=.8=0.5122P(Y=0)=P(A∩(B∪C))=P(A)P(B∪C)=.2(.2+.2-.2)=0.072.Thus,P(Y=1)=1-.512-.072=0.416.3.5Thereare3!=6possiblewaystoassignthewordstothepictures.Ofthese,oneisaperfectmatch,three
3、haveonematch,andtwohavezeromatches.Thus,p(0)=2/6,p(1)=3/6,p(3)=1/6.⎛5⎞3.6Thereare⎜⎜⎟⎟=10samplepoints,andallareequallylikely:(1,2),(1,3),(1,4),(1,5),⎝2⎠(2,3),(2,4),(2,5),(3,4),(3,5),(4,5).a.p(2)=.1,p(3)=.2,p(4)=.3,p(5)=.4.b.p(3)=.1,p(4)=.1,p(5)=.2,p(6)=.2,p(7)=.2,p(8)=.1,p(9)=.1.33.7T
4、hereare3=27waystoplacethethreeballsintothethreebowls.LetY=#ofemptybowls.Then:3!6p(0)=P(nobowlsareempty)==2727p(2)=P(2bowlsareempty)=3276318p(1)=P(1bowlisempty)=1−−=.2727273.8Notethatthenumberofcellscannotbeodd.p(0)=P(nocellsinthenextgeneration)=P(thefirstcelldiesorthefirstcellsplitsa
5、ndbothdie)=.1+.9(.1)(.1)=0.109p(4)=P(fourcellsinthenextgeneration)=P(thefirstcellsplitsandbothcreatedcellssplit)=.9(.9)(.9)=0.729.p(2)=1–.109–.729=0.162.3.9TherandomvariableYtakesonvales0,1,2,and3.a.LetEdenoteanerroronasingleentryandletNdenotenoerror.Thereare8samplepoints:EEE,EEN,ENE
6、,NEE,ENN,NEN,NNE,NNN.WithP(E)=.05andP(N)=.95andassumingindependence:3P(Y=3)=(.05)=0.000125P(Y=2)=3(.05)2(.95)=0.00712523P(Y=1)=3(.05)(.95)=0.135375P(Y=0)=(.95)=0.857375.Chapter3:DiscreteRandomVariablesandTheirProbabilityDistributions32Instructor’sSolutionsManualb.Thegraphisomitted.c.
7、P(Y>1)=P(Y=2)+P(Y=3)=0.00725.3.10DenoteRastheeventarentaloccursonagivendayandNdenotesnorental.Thus,thesequenceofinterestisRR,RNR,RNNR,RNNNR,….ConsiderthepositionimmediatelyfollowingthefirstR:itisfilledbyanRwithprobability.2andbyanNwithprobability.8.Thus,P(Y=0)=.2,P(Y=1)=.8(.2)=.16,P(
8、Y=2)=.128,….
