Introduction To Probability Solutions to the Odd Numbered Exercises.pdf

Introduction To Probability Solutions to the Odd Numbered Exercises.pdf

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页数:46页

时间:2019-03-12

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1、DEPOSITBONUSESTOGETSTARTEDFullTiltPoker(100%DepositBonusupto$600)PlayOnlinePokerCakePoker(110%DepositBonusupto$600)WilliamHill(100%DepositBonus&$120,000RakeRace)PrestigeCasino($1,500inFREESpinsNoDepositNeeded)PokerStrategy($50usdStartingCapitalFREENoDepositNeeded)WilliamHillBingo($40usd

2、FREEToStartPlayingNoDepositRequired)CharlesM.GrinsteadandJ.LaurieSnell:INTRODUCTIONtoPROBABILITYPublishedbyAMSSolutionstotheoddnumberedexercisesSECTION1.11.Asnincreases,theproportionofheadsgetscloserto1/2,butthedi®erencebetweenthenumberofheadsandhalfthenumberof°ipstendstoincrease(althou

3、ghitwilloccasionallybe0).3.(b)Ifonesimulatesasu±cientlylargenumberofrolls,oneshouldbeabletoconcludethatthegamblerswerecorrect.5.Thesmallestnshouldbeabout150.7.Thegraphofwinningsforbettingonacolorismuchsmoother(i.e.hassmaller°uctuations)thanthegraphforbettingonanumber.9.Eachtimeyouwin,yo

4、ueitherwinanamountthatyouhavealreadylostoroneoftheoriginalnumbers1,2,3,4,andhenceyournetwinningisjustthesumofthesefournumbers.Thisisnotafoolproofsystem,sinceyoumayreachapointwhereyouhavetobetmoremoneythanyouhave.Ifyouandthebankhadunlimitedresourcesitwouldbefoolproof.11.Fortwotosses,thep

5、robabilitiesthatPeterwins0and2are1/2and1/4,respectively.Forfourtosses,theprobabilitiesthatPeterwins0,2,and4are3/8,1/4,and1/16,respectively.13.Yoursimulationshouldresultinabout25daysinayearhavingmorethan60percentboysinthelargehospitalandabout55daysinayearhavingmorethan60percentboysinthes

6、mallhospital.15.Inabout25percentofthegamestheplayerwillhaveastreakof¯ve.SECTION1.21.P(fa;b;cg)=1P(fag)=1=2P(fa;bg)=5=6P(fbg)=1=3P(fb;cg)=1=2P(fcg)=1=6P(fa;cg)=2=3P(Á)=03.(b),(d)5.(a)1/2(b)1/4(c)3/8(d)7/87.11/129.3=4;111.1:12;1:3;1:3513.11:415.Letthesamplespacebe:!1=fA;Ag!4=fB;Ag!7=fC;Ag

7、1!2=fA;Bg!5=fB;Bg!8=fC;Bg!3=fA;Cg!6=fB;Cg!9=fC;Cgwherethe¯rstgradeisJohn'sandthesecondisMary's.YouaregiventhatP(!4)+P(!5)+P(!6)=:3;P(!2)+P(!5)+P(!8)=:4;P(!5)+P(!6)+P(!8)=:1:Addingthe¯rsttwoequationsandsubtractingthethird,weobtainthedesiredprobabilityasP(!2)+P(!4)+P(!5)=:6:17.Th

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