Probability - Study Notes | Washington State University

University:
Washington State University
Course:
MATH 456 | Introduction to Statistical Theory
Academic year:

2023
Views:

46

Pages:

6
Author:

NeatGold

classmatenDatePageProbabilityAnexperimenthasai,e.,thesamplespace,setofallpossibleactionsEventsareoutcomesofanexperimentMutuallyExclusiveEventsIfA&BaremutuallyexclusiveAnB=4A,B&CarethennC=0,butAnBBnclCAAmay/maynotbe$.ExhaustiveEventsIfAB&,CareexhaustiveeventsAUBUC=SProbabilityofoccurrenceofaneventEisP(E) favourabletotal outcomesTheoddsinfavour =fav unfav, whiletheoddsagainstunfau/taxnoteHT&TH}areconsidereddifferentevenifthecoinsareidenticaland tossedsimultaneously. Independent EventsA&Bindependentevents itareofAdoesn'tthe occurrenceaffectoccurrenceof B &thevice versa .ifA&R are independent -A&BarealsoA&Baretoo.A & R as wellP(A/B) = P(A/B) = P(A)P (B/A) = P(B/A) = P(B)P(AB) = P(AnB) = P(A) . P(B/A) = P(A) 1/8HA,B & C are independent events - -P(A n B) = P(A)P(B)P(B n C) = P(B) P(C)P(CNA) = P(c) P(A)P(AnBnc) = P(A) P(B)P(C)TocheckiteventsarecheckitP(AnB) =independent,P(A)P(B)NotethatifA&BareitdoesNOTindependentimplyexclusivetheyaremutually DatePageConditional ProbabilityTheprobabilityofaneventAgiventhathappeningevenBhasoccurredisP(AB) =P(AnB)n(AnB) uts)P(B)Forexample,itadiceisvolledonebyone event A = sum is 7 andB = 2nd outcome is 5 Find P(A/B) & P(B/A) .P(A/B) = P(A n B) =1/36=IP(B)6/366P(B/A) = 1/6A = h.(1,6) (2,5) (3,4),(4,3),(5,2); (64,113B= ({(iis), (2,5) (3,5),(4,5), (5,5). (6,5)3)multiplicationtheoremstatesthatnoteTheP(AnB)=P(A)P(B/A)P(SAnB)(derived from the conditional prob. eqn) Binomial TrialsLet Kas an experiment havenwhereeachindependent trials,trialhas two outcomes - successorfailure For example,A diceisthrownrepeatedlyTheexperimentisasuccessofthe number 6 appears and afailureifitdoesn't(E)(E)ThetreediagramcanthusbedrawnlikeSO:(1/6)E(1/6)E(6/6)Etillnhere, 1/6 is the(5/6) 3 EP(E) & 5/6 is P(E)Ingeneral,letP(success) = P andP(failure) = qandptq = 1.Letarandomvariablexbethenumberofsuccesses.ThenP (exactly r success)choose placement +xsuccesses, since=FSSFS is diffCrPqP (atleastfrom S5SFFsuccess)rw=nciin-iq 2 success & 3failure, (n=5) classmatechoose posDatefor success,Pageeg. mc2 for2summationP(atmostrsuccess)of casesr>xr=P ( first.h.cn-LCiopsuccessqinrthtrial)=all failuresqPfirstsuccessbefore thatThetreediagramwillbe-EPEfor exactly 3 successes,EalaP(E)=xp(p)4q EPEfor exactlysuccess,outof2,qE-rP(E) = pxqPq9ETotalProbabilityTheoremLetE1 , E,N,/Eubemutuallyexclusive&exhaustiveeventsLetAbeaneventThenP(A)=P(EnAl+P(E2nA)+P(En n A)P(E,) P (A/E,) + P(E, ) P (A/E2)+InverseProbabilityTheorem(EL/A) = P(ExnA) =P(EunA)P(A)P(E,nA)+ P(E,nA)+ForexampleA ,B&Caremachinesthatproduceglassbottlesinafactory.TheymayproducedefectivebottlessometimesThetreediagramlooks classrateDatesample spacePagefor2nd Qlikethis :St.DP(D)= P(AnD) + P(BnD) + P(C2A)25%AIS=25.5 + 35.4 + 40.347.D.35%B= 3.45 %40%31-00Pc5(prob that a glass bottleproduced isdefective)Giventhataglassbottleisdefective. whatistheprobabilitythatitcamefrommachine B)(B/D) = P(BnD) =P(B n D)P(D)P(AnD)+ P(BnD) + P(CnD)=35 x4=40.5%25*5 + 35x4 + 40x3

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