概率統(tǒng)計(jì)中文概統(tǒng)課件lecture2

Lecture2FiniteSampleSpaceTheProbabilityofaUnionofEvents1.6FiniteSampleSpaceAprobabilitydistributiononSisspecifiedbyassigningaprobabilitypitoeachpointsi.Fori=1,...,n,Also,TheprobabilityofanyeventAcanbefoundbyaddingtheprobabilitiespiofallessithatbelongtoA.ThesamplespaceScontainsonlyafinitenumberofpoints(es)s1,...,sn.Example:ConsumerComplaintsAmanufacturerofanelectromechanicalkitchenutensilconductedananalysisofalargenumberofconsumercomplaintsandfoundthattheyfellintothethreecategoriesshowninthetablebelow.Whatistheprobabilitythatthereasonforacomplaintiseitherelectricalormechanical?SimpleSampleSpacesAsamplespaceScontainingness1,...,sniscalledasimplesamplespaceif，i=1,...,n.IfaneventAcontainsexactlymes,thenExample:CokevsPepsi?Supposethatinsomepopulation,theprobabilityofpreferringcokeoverPepsiis50%.Threepeoplearesurveyed.WhatistheprobabilitythatexactlytwopeopleperferCoke?Simplesamplespace!Acontains3es,soPr(A)=3/8Thesamplespacecontains8es.S1:CCCS2:PCCS3:CPCS4:CCPS5:CPPS6:PCPS7:PPCS8:PPP InasimplesamplespaceS,howtodeterminethenumberoftotalesinthespaceSandinvariouseventsinSwithoutcompilingalistofthees?Permutation

(排列)n

個(gè)相異物體，有放回的抽出k個(gè)物體的排列總數(shù)為：

nkn

個(gè)相異物體，無放回（一次性）抽出k個(gè)物體的排列總數(shù)為：Pn,k=n(n-1)...(n-k+1)n

個(gè)相異物體的排列總數(shù)為:Pn,n=n(n-1)...1=n!(nfactorial)Define0!=1.Eachdifferentarrangementiscalledapermutation. Note:(a,b)and(b,a)aredifferentpermutations.Pn,k=n(n-1)...(n-k+1)iscalledthenumberofpermutationsofnelementstakenkatatime.Example1.7.1ChoosingOfficersApresidentandasecretaryaretobechosenfrom25members.Whatisthetotalnumberofwaysinwhichthesetwopositionscanbefilled?

P25,2=(25)(24)=600TheBirthdayProblemWhatistheprobabilitythatatleasttwopeopleinagroupofkpeople()willhavethesamebirthday?

Assumethebirthdaysofthekpeopleareunrelated.Alsoassumethateachofthe365daysisequallylikelytobethebirthdayofanypersoninthegroup.

Theprobabilitythatallkpeoplewillhavedifferentbirthdayis

TheprobabilitythatatleasttwopeoplewillhavethesamebirthdayisSimplesampleSpace!樣本空間中的樣本點(diǎn)個(gè)數(shù)?事件中的樣本點(diǎn)個(gè)數(shù)? k=10,p=0.1169482 k=20,p=0.4114384 k=30,p=0.7063162 k=40,p=0.8912318 k=50,p=0.9703736Combination（組合）Asubsetofkelementsistobeselectedfromasetofndistinctelements.Thearrangementoftheelementsinasubsetisirrelevant.Eachsubsetiscalledacombination.Thesubsets{a,b}and{b,a}areidentical.LetCn,k

denotethenumberofcombinationsofnelementstakenkatatime.HowtocalculateCn,k?WecanconstructalistofallPn,kpermutationswithkelmementsoutofnelementsasfollows:First,aparticularcombinationofkelmentsisselected.Thesekelementscanbepermutatedink!ways.Thenumberofpermutationscanbecalculatedby

k!Cn,kSoExample1.8.1SelectingACommitteeAcommitteeof8peopleistobeselectedfromagroupof20people.ThenumberofdifferentwaysofselectingthecommitteeisTheProbabilityofAUnionofEventsThesamplespaceSmaycontaineitherafinitenumberofesoraninfinitenumber.Wewillstudytheprobabilityoftheunionofnevents.Iftheeventsaredisjoint,ForanytwoeventsA1andA2,Theorem1.ForanythreeeventsA1,A2andA3,ProofTheorem.ForanyneventsA1,...,

An,TheCollector’sProblemTheCollector’sProblem Supposethateachpackageofbubblegumcontainsthepictureofabasketballplayer;thatthepicturesofrdifferentplayersareused;thatthepictureofeachplayerisequallylikelytobeplacedinanygivenpackageofgum;andthatpicturesareplacedindifferentpackagesindependentlyofeachother.Whatistheprobabilitypthatapersonwhobuysnpackagesofgum()willobtainacompletesetofrdifferentpictures.Thecomplementaryeventisthatthepictureofatleastoneplayerismissing.LetAidenotetheeventthatthepictureofplaye