零度小说网

Chapter 2 The workings of probability（第2页）

Assumingtwofactorstobei，whe，isoistakesmadeinassessingprobabilities。Supposethat，iy’sgraduateschool，halfthestudentsarefemale，andoudyengineeriudentatrandom:theprobabilitythisstudentisfemalebetakeheprobabilitythestudentstudiesengineerih。However，youwillfindthattheprobabilitythestudentisafemaleengineerismuchlessthantheirprodue-tenth。

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&ionLatofindtheceofatleastowoevents，providedtheyaredisjoint。Whatiftheyarenotdisjoint?Forexample，indrawirandom，whatistheceitiseitheraSpadeoraheAceofSpadesfallsinbothcategories，soifwejustaddedtherespectiveprobabilities，wewouldtthatcardtwice。Tocorrectfortheoutesthatwouldbedouble-dfindtheprobabilitythat　atleastowoeventsoccurs，

addtheirindividualsubtractthecebothoccur。Ofcourse，ifthetwoeve，itisimpossiblethatbothhappen，sothisextratermhasvaluezero，ainalAdditionLaw。

&hisnotioninathetwoearlierexamples。WiththedRedBlackcard，thecewegetatleastothearithmetic12+12-14，whichis34。Iherexample，thelyberedballiseitherLreenis12+12-0。4=0。6。

AndtheceeitheraSpadeoranAcearisesas1352+452-152=1652，edbynotily16ofthe52cardssatisfythis。

Thislastcalshainstmakihmeticsimplifis。Yes，1352isthesameas14，and452isthesameas113，buttoadd14and113，youarebetteroffwiththeirinalfraditisseldomhelpfultore-writeafriendlyfralike513asitsuglydecimalapproximation0。38461538。。。

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Aasisdescribedasiwheneverkherorhemoccurmakesheprobabilitiesofahers。Inthiscase，theMultiplieansthat，whateverseleofeventswemakefromthis，theprobabilitythatallofthemoccurisjusttheproductoftheirindividualprobabilities。

Buthowmightwefindtheprobabilitythatthreeormoreeventsallotheyare　?Forexample，inwhiste，thecardsarerandomlyshuffledandsharedequallyamongfourplayers。HowlikelyisitthattheyallreceiveexaeAce?

siderthefourseparateevesexaeAgetsoneAhasosoneAly，thesefourevei，asifahemhappeheris。Wewillfindtheprobabilitytheyalloccurviaa　three-stageprocess。

First，wefindtheprobabilitythatAlyoneAce。Assumingallpossiblewaysofdealingthecardsareequallylikely，wehaveaing:tthetotalnumberofpossibledeals，ainhowmasexaeAce。Believeme，theceworksoutasjustunder44%。

AssumeAoneAdhewelveleavesthreeAdthirty-sixheotherplayers，andBriaeenofthem，atrandom。AsimilargexerallerpackshowsthatthecehewouldgetexaeAceisjustover46%。TheMultipliLawthehattheceofbothevebothAnneandBrialyoheproductofthesetwovalues，justover20%。

SonowassumeAnneandBrianeachhaveexaeAreceivesthirteencardsatrandomfromthetwoAdtwenty-fourremaihathegetsexaeAdtobe52%。

&epistousetheMultipliLawoobiwocals:theneandBrian，andthenalso，allhaveexaeAceisalittleover10%。Ifthishappens，DebbyiablyhasthefinalAce，sowehavefoundtheanswerweseek。

Thisaselfisofnorealce–althoughthedealistotallyraequitableoutefortheAcesisratheruthemethodusedisuofihateveryeventiionsdowniheeevent;then，assumiodtheceofaseingbothoftheseodtheceofathird;thenassumihreeeveheceofafourth–andsoon。Finally，multiplyallthesequaher。

Whereelsemightwehavetofolloosemyjourhreestages，andIassesstheirseparatecesofhavingnodelay:however，allthestageswillbeaffectedbytheweather，anddelayoagewillgetheceofdelayelsewhere。Inmanufadustry，thesafetyofapieentwillrelyonseveralposwhiotoperateily–somemayusethesamely，othersmayhavebeeestedbythesameunreliableemployee。Withamedicalprocedure，whetherorhatgareiofeachothermakeahugediffereheoverallcethatallturnsoutwell。

&sareiheheyalloccurisjusttheproductoftheirindividualces。Butweareseldomluoughforthistohold:astage-by-stageassessment，withprobabilitiesgasthewresses，isthenorm。

Whataboutthecethat　atleastohreeormoreeventsoccurs?TheAdditioendtothiscase，butastheexpressionise，Iwilldown。ItsrecipefolloathasdescribedwhenusiipliLawforthecethatallofmasoccur:takeitoime。

Usingtheword　iwhen　disjoi，andviceversa，areoheexampleofgorandomhelpyouseehowtoavoidit。Here，theeveaSpade’aaClub’aredisjoint，butfarfromi，asifeitheroccurs，theotherot，sothecebothoccuriszero!Also‘GetaSpade’aanAdepeplainlynotdisjoint。

&heAdditionLawisusedtofindtheceofatleastoheMultipliLawisusedtogivethecetheyalloccur。

Itissometimessaidthatgreallygoesowo，infinity。Thisaphorismcarriesthetruththatifwemakethestepfromdealingwithoodealingwithtwocases，theepstothree，four，five，etc。casesaretrivialinparison。ThissurelyappliestoboththeAdditionandMultipliLaws。

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&herhappens，oritdoesalprobabilityissplitbetweehappening，anditfailingtohappen。Soifwedtheevehappen，wededucetheceitdoesoccurbysubtra100%。

Toillustrate，let’sfindtheceofatleastoneSixwhenanordinaryfairdieisrolledtwieisapairthescoresofthefirstroll，thenthesed，e。g。（5，2）or（4，4），aakeallsuesasequallylikely。Eachrollhassixpossibleresults，leadingto6x6=36outesaltogether。OurevehappeherdieshowsaSix，forwhichthereare5x5=25outes。TheoSixesis2536，sotheceofatleastoneSixis1136，abitlessthahree。

ThisleadsontoajuniorversionofagamblingproblemsolvedbyBlaisePasdPierredeFermatin1654。Howoftenmustwerolladietomakeit　morelikelythannotthatwegetatleastoneSix，i。e。thattheceofgettingaSixismorethanonehalf?Wehavejustseenthattworollsarenotenough。

&rarollihenumberofpossibleoutesbyafactorofsix，whiletheeswithoutaSixgetsmultipliedbyfive。Seesinall，and125ofthem–overhalf–oSix;threerollsareher。However，five1，296outes，andonly625ofthemhavenoSixes–fewerthanhalf。ThatleavesmoreoutesthatihahnoSixes，sonowaSixismorelikelythannot。Fourrollssuffice。

&ualgameanalysedbyPasdFermatinvtwodicetogether;andaskinghowofteobedomorelikelythannotthata　Double-Sixturnsupatleastohodofsolutionisthesame，buttherawarithmetiidable。Today，wequicklyreaswerwithamiputerorapocketcalculator，whilelogarithmsandslideruleshadtlyjustbeeavailableihtury。Withupto24rolls，itismorelikelythannotthatnoDouble-Sixappears，buta25throlltipsthebalaher

Mostproblemsoftheform‘Findtheceofatleastoshappesolvedinthismanner:workouttheohemarises，araunity。