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Differential Equation l (MATH 2230)

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CHAPTER 4

Conditionalprobability

Denition,Bayes'Ruleandexamples

Supp osethere are 200 men,of which 100 aresmokers,and 100 women,of which 20 are smokers. Whatistheprobabilitythatap ersonchosenatrandomwillb easmoker? The answeris 120 / 300. Now,letusask,whatistheprobabilitythatap ersonchosenatrandom isasmokergiventhatthep ersonisawomen? Onewouldexp ecttheanswertob e 20 / 100 anditis.

Whatwehavecomputedis

numb erofwomensmokers numb erofwomen

=

numb erofwomensmokers/ 300 numb erofwomen/ 300

,

whichisthesame asthe probabilitythat ap ersonchosenat randomisawomananda smokerdividedbytheprobabilitythatap ersonchosenatrandomisawoman.

Withthisinmind,wegivethefollowingdenition.

Denition4(Conditionalprobability)

IfP(F)> 0 ,wedenetheprobabilityof EgivenFas

P(E|F) :=

P(E∩F)

P(F)

.

NoteP(E∩F) =P(E|F)P(F).

Example4. Supp oseyourolltwodice. Whatistheprobabilitythesumis8?

Solution:therearevewaysthiscanhapp en{(2,6),(3,5),(4,4),(5,3),(6,2)},sotheprob- abilityis 5 / 36. LetuscallthiseventA. Whatistheprobabilitythatthesumis 8 given thattherstdieshowsa3? LetBb etheeventthattherstdieshowsa3(A∩B) istheprobabilitythattherstdieshowsa 3 andthesumis8,or 1 / 36. P(B) = 1/ 6 , so

P(A|B) = 11 // 366 = 1/ 6.

Example4. Supp oseab oxhas 3 redmarblesand 2 blackones. Weselect 2 marbles. Whatistheprobabilitythatsecondmarbleisredgiventhattherstoneisred?

42 4

Solution: LetAb etheeventthesecondmarbleisred,andBtheeventthattherstone isred. P(B) = 3/ 5 ,whileP(A∩B)istheprobabilityb otharered,oristheprobability thatwe chose 2 redoutof 3 and 0 blackoutof2. ThenP(A∩B) =

( 3

)( 2

)

/

( 5

)

, andso

P(A|B) = 33 // 105 = 1/ 2.

Example4. Afamilyhas 2 children. Giventhatoneofthechildrenisab oy,whatis theprobabilitythattheotherchildisalsoab oy?

Solution:LetBb etheeventthatonechildisab oy,andAtheeventthatb othchildrenare b oys. Thep ossibilitiesarebb, bg, gb, gg,eachwithprobability 1 / 4 .P(A∩B) =P(bb) = 1/ 4

andP(B) =P(bb, bg, gb) = 3/ 4 .Sotheansweris 13 // 44 = 1/ 3.

Example4. Supp osethetestforHIVis99%accurateinb othdirectionsand0%ofthe p opulationisHIVp ositive. Ifsomeonetestsp ositive,whatistheprobabilitytheyactually areHIVp ositive?

Solution:LetDistheeventthatap ersonisHIVp ositive,andTistheeventthatthep erson testsp ositive.

P(D|T) =

P(D∩T)

P(T)

=

(0)(0)

(0)(0) + (0)(0)

≈23%.

Ashortreasonwhythissurprisingresultholdsisthattheerrorinthetestismuchgreater thanthep ercentageofp eoplewithHIV osethatwehave 1000 p eople. Onaverage, 3 ofthemwillb eHIVp ositiveand 10 willtestp ositive. Sothe chancesthatsomeonehasHIVgiventhatthep ersontestsp ositiveisapproximately 3 / 10. Thereasonthatitisnotexactly 0. 3 isthatthereissomechancesomeonewhoisp ositive willtestnegative.

Supp oseyouknowP(E|F)andyouwanttondP(F|E).Recallthat

P(E∩F) =P(E|F)P(F),

andso

P(F|E) =

P(F∩E)

P(E)

=

P(E|F)P(F)

P(E)

Example4. Supp ose36%offamiliesownadog, 30%offamiliesownacat,and22% ofthefamiliesthathaveadogalsohaveacat haveacat?

Solution:LetDb ethefamiliesthatownadog,andCthefamiliesthatownacat givenP(D) = 0. 36 ,P(C) = 0. 30 ,P(C |D) = 0. 22 WewanttoknowP(D|C). Weknow

44 4

Hereisanotherexamplerelatedtoconditionalprobability,althoughthisisnotanexample ofBayes'rule. ThisisknownastheMontyHal lproblemafterthehostoftheTVshowin the60scalledLet'sMakeaDeal.

Example 4. Therearethreedo ors, b ehindonea nicecar,b ehindeach of theother twoagoateatingabaleofstraw. Youcho oseado or. ThenMontyHallop ensoneofthe otherdo ors,whichshowsabaleofstraw. Hegivesyoutheopp ortunityofswitchingtothe remainingdo or?

Solution: Let'ssupp oseyoucho osedo or1,sincethesameanalysisapplieswhicheverdo or youchose. Strategyoneistostickwithdo or1. With probability 1 / 3 youchosethecar. MontyHallshowsyouoneoftheotherdo ors,butthatdo esn'tchangeyourprobabilityof winning.

Strategy 2 istochange. Let'ssaythecarisb ehinddo or1,whichhapp enswithprobability 1/3 ors,saydo or2 eagoat,soyou switchtodo or3,andlose. Thesameargumentappliesifheshowsyoudo or3. Supp ose thecarisb ehinddo or2. Hewillshowyoudo or3,sincehedo esn'twanttogiveawaythe car or 2 andwin. Thishapp enswithprobability 1 / 3. Thesameargument appliesifthecarisb ehinddo or3/3andlosewithprobability 1/3 2 ismuchsup erior.

Aproblemthatcomesupinactuarialsciencefrequentlyisgambler'sruin.

Example4 (Gambler'sruin). Supp oseyouplaythegamebytossingafaircoinre- p eatedlyandindep endently,youwinadollar,andifitcomesuptails, youloseadollar. Supp oseyoustartwith$50'stheprobabilityyouwillgetto$ withoutrstgettingruined(runningoutofmoney)?

Solution:itiseasiertosolveaslightlyharderproblem edescrib edashaving probability 1 / 2 ofwinning 1 dollarandaprobability 1 / 2 oflosing 1 dollar egins withagivennumb erof dollars,andintendstoplay thegamerep eatedly untiltheplayer eithergo esbrokeorincreaseshisholdingstoNdollars.

Foranygivenamountnofcurrentholdings,theconditionalprobabilityofreachingNdollars b eforegoing brokeisindep endent ofhow weacquiredthendollars, sothereisaunique probabilityP(N|n)ofreachingN ontheconditionthatwecurrentlyholdndollars. Of course,foranyniteNweseethatP(N|n) = 0andP(N|N) = 1. Theproblemisto determinethevaluesofP(N|n)fornb etween 0 andN.

WeareconsideringthissettingforN= 200,andwewouldliketondP(200|50). Denote y(n) :=P(200|n), whichistheprobabilityyougetto 200 withoutrstgettingruinedif youstartwithndollars. Wesawthaty(0) = 0andy(200) = 1. Supp osetheplayerhas ndollarsatthemoment,the nextroundwill leavetheplayerwitheithern+ 1orn− 1 dollars,b othwithprobability 1 / 2 .Thusthecurrentprobabilityofwinningisthesameasa weightedaverageoftheprobabilitiesofwinninginplayer'stwop ossiblenext states

4.1,BAYES'RULEANDEXAMPLES 45

have y(n) = 12 y(n+ 1) + 12 y(n−1).

Multiplyingby2,andsubtractingy(n) +y(n−1)fromeachside,wehave

y(n+ 1)−y(n) =y(n)−y(n−1).

Thissaysthatslop esofthegraphofy(n)ontheadjacentintervalsareconstant(rememb er thatxmustb eaninteger).Inotherwords,thegraphofy(n)mustb ealine(0) = 0 andy(200) = 1,wehavey(n) =n/ 200 ,andthereforey(50) = 1/ 4.

Anotherwaytoseewhatthefunctiony(n)istousethetelescopingsumasfollows

(4.1) y(n) =y(n)−y(0) = (y(n)−y(n−1)) +...+ (y(1)−y(0))

=n(y(1)−y(0)) =ny(1).

sincetheallthesedierencesarethesame,andy(0) = 0. Tondy(1)wecanusethefact thaty(200) = 1,soy(1) = 1/ 200 ,andthereforey(n) =n/ 200 andy(50) = 1/ 4.

Example4. Supp oseweareinthesamesituation,butyouareallowedtogoarbitrarily farindebt. Letz(n)b etheprobabilityyouevergetto$200ifyoustartwithndollars. Whatisaformulaforz(n)?

Solution:Justasab ove,weseethatzsatisestherecursiveequation

z(n) = 12 z(n+ 1) + 12 z(n−1).

Whatweneedtodeterminenowareb oundaryconditions. Nowthatthe gamblercango todebt,thecondition thatifwestartwith 0 wenevergetto$200,thatis,probabilityof getting$200is 0 ,isnottrue.FollowingEquation4.1(0) 6 = 0weseethat

z(n)−z(0) = (z(n)−z(n−1)) +...+ (z(1)−z(0)) =n(z(1)−z(0)),

therefore z(n) =n(z(1)−z(0)) +z(0).

Ifwedenotea:=z(1)−z(0)andb:=z(0)weseethatasafunctionofnwehave

z(n) =an+b.

Wewouldliketondaandbnow,soforanynwe have 06 z(n) 61 .Thisisp ossibleonlyifa= 0,thatis,

z(1) =z(0),

z(n) =z(0)

foranyn(200) = 1,therefore

4.2 47

Furtherexamplesandapplications

4.2. Examples,basicprop erties,multiplicationrule,lawoftotalprobability.

Example 4. Landonis 80% surehe forgot his textb o okeither at the Unionor in Monteith%surethattheb o okisattheunion,and40%surethatitisinMonteith. GiventhatLandonalreadywenttoMonteithandnoticedhistextb o okisnotthere,whatis theprobabilitythatitisattheUnion?

Solution:CallingU=textbookisattheUnion,andU=textbookisinMonteith,noticethat U⊆McandhenceU∩Mc=U. Thus,

P(U|Mc) =

P(U∩Mc) P(Mc)

=

P(U)

1 −P(M)

=

4 / 10

6 / 10

=

2

3 .

Example 4. SarahandBobdraw 13 cardseachfromastandarddeckof 52. Given thatSarahhasexactlytwoaces,whatistheprobabilitythatBobhasexactlyoneace?

Solution: LetA=Sarahhastwoaces,andletB=Bobhasexactlyoneace. Inorderto computeP(B|A),weneedtocalculateP(A)andP(A∩B). Ontheonehand,Sarahcould haveanyof

( 52

)

p ossiblehands,

( 4

)

· ( 48

)

willhaveexactlytwoacessothat

P(A) =

( 4

)

· ( 48

)

( 52

).

Ontheotherhand,thenumb erofwaysinwhichSarahcanpickahandandBobanother (dierent)is

( 52

)

· ( 39

)

.Thethenumb erofwaysinwhichAandBcansimultaneouslyo ccur is

( 4

)

· ( 48

)

· ( 2

)

· ( 37

)

andhence

P(A∩B) =

( 4

)

· ( 48

)

· ( 2

)

· ( 37

)

( 52

)

· ( 39

).

Applyingthedenitionofconditionalprobabilitywenallyget

P(B|A) =

P(A∩B)

P(A)

=

( 4

)

· ( 48

)

· ( 2

)

· ( 37

)/( 52

)

· ( 39

)

( 4

)

· ( 48

)/( 52

) =

2 ·

( 37

)

( 39

)

Example 4. Atotalof 500 marriedcouplesarep olledab outtheirsalarieswiththe followingresults

husbandmakeslessthan$25K husbandmakesmorethan$25K wifemakeslessthan$25K 212 198 wifemakesmorethan$25K 36 54

(a)Findtheprobabilitythatahusbandearnslessthan$25K. Solution:

P(husbandmakes <$25K) =

212

500 +

36

500 =

248

500 = 0. 496.

48 4

(b)Findtheprobabilitythatawifeearnsmorethan$25K,giventhatthehusbandearns asthatmuchaswell. Solution:

P(wifemakes >$25K|husbandmakes >$25K) =

54 / 500

(198 + 54)/ 500

=

54

252 = 0. 214

(c)Findtheprobabilitythatawifeearnsmorethan$25K,giventhatthehusbandmakes lessthan$25K. Solution:

P(wifemakes >$25K|husbandmakes <$25K) =

36 / 500

248 / 500

= 0. 145.

Fromthedenitionofconditionalprobabilitywecandeducesomeusefulrelations.

Prop osition4.

Proof(i)whichisarewritingofthedenitionofconditionalprobability

P(F|E) =P(PE(∩EF)). Letusprove(ii): wecanwriteEastheunionofthepairwisedisjoint

setsE∩FandE∩Fc. Using(i)wehave

P(E) =P(E∩F) +P(E∩Fc) =P(E|F)P(F) +P(E|Fc)P(Fc).

Finally,writingF=EinthepreviousequationandsinceP(E|Ec) = 0,weobtain(iii).

Example4. PhanwantstotakeeitheraBiologycourseoraChemistrycourse. His adviserestimatesthattheprobabilityofscoringanAinBiologyis 45 ,whiletheprobability ofscoringanAinChemistryis 17. IfPhandecidesrandomly,byacointoss,whichcourse totake,whatishisprobabilityofscoringanAinChemistry?

Solution: denotebyBtheeventthatPhan takesBiology,andbyC theeventthat Phan takesChemistry,andbyA=theeventthatthescoreisanA,sinceP(B) =P(C) = 12 wehave

P(A∩C) =P(C)P(A|C) =

1

2

·

1

7 =

1

14 .

TheidentityP(E∩F) =P(E)P(F|E)fromProp osition4(i)canb egeneralizedtoany numb erofeventsinwhatissometimescalledthemultiplicationrule.

50 4

Solution:

(a)denotebyA 1 =theeventthatapolicyholderwil lhaveanaccidentwithinayear,and denotebyA =the eventthata policyholder is accidentprone. ApplyingProp osi- tion4(ii)wehave P(A 1 ) =P(A 1 |A)P(A) +P(A 1 |Ac) (1−P(A)) = 0. 4 · 0 .3 + 0(1− 0 .3) = 0. 26 (b)UseBayes'formulatoseethat

P(A|A 1 ) =

P(A∩A 1 )

P(A 1 )

=

P(A)P(A 1 |A)

0. 26

=

0. 3 · 0. 4

0. 26

=

6

14 .

UsingthelawoftotalprobabilityfromProp osition4'rule,which app earedinProp osition4.

Prop osition4(GeneralizedBayes'rule)

LetF 1 ,... , Fn⊆ S b emutuallyexclusiveandexhaustiveevents,i. S =

⋃n i=1Fi. Then,foranyeventE⊆Sandanyj= 1,... , nitholdsthat

P(Fj|E) =

P(E|Fj)P(Fj) ∑n i=1P(E|Fi)P(Fi)

Example4. Supp oseafactoryhasmachinesI,I I,andI I Ithatpro duceiSung phones. Thefactory'srecordshowsthat

MachinesI,I IandI I Ipro duce,resp ectively,2%,1%,and3%defectiveiSungs. Outofthetotalpro duction,machinesI,I I,andI I Ipro duce,resp ectively,35%,25%and 40%ofalliSungs.

AniSung isselectedatrandomfromthefactory.

(a)WhatisprobabilitythattheiSungselectedisdefective? Solution:Bythelawoftotalprobability, P(D) = P(I)P(D|I) +P(II)P(D|II) +P(III)P(D|III)

= 0. 35 · 0 .02 + 0. 25 · 0 .01 + 0. 4 · 0 .03 =

215 10 , 000

.

(b)GiventhattheiSungisdefective,whatistheconditionalprobabilitythatitwaspro- ducedbymachineI I I? Solution:ApplyingBayes'rule,

P(III|D) =

P(III)P(D|III)

P(D)

=

0. 4 · 0. 03

215 / 10 , 000

=

120

215 .

Example4. Inamultiplechoicetest,astudenteitherknowstheanswertoaquestion orshe/hewillrandomlyguessit. Ifeachquestionhasmp ossibleanswersandthestudent

4.2 51

knowstheanswertoaquestionwithprobabilityp,whatistheprobabilitythatthestudent actuallyknowstheanswertoaquestion,giventhathe/sheanswerscorrectly?

Solution:denotebyKtheeventthatastudentknowstheanswer,andbyC theeventthat astudentanswerscorrectly. ApplyingBayes'rulewehave

P(K|C) =

P(C|K)P(K)

P(C|K)P(K) +P(C|Kc)P(Kc)

=

1 ·p 1 ·p+m 1 (1−p)

=

mp 1 + (m−1)p

.

4.3 53

Exercise4. Amultiplechoiceexamhas 4 choicesforeachquestion. Thestudenthas studiedenoughsothattheprobabilitytheywillknowtheanswertoaquestionis0,the probabilitythat the studentwill b eable toeliminate onechoiceis0, otherwise all 4 choicesseemequallyplausible. Iftheyknowtheanswertheywillgetthequestioncorrect. Ifnottheyhavetoguessfromthe 3 or 4 choices measurewhatthestudentknows,andnothowwelltheycanguess. Ifthestudentanswers aquestioncorrectlywhatistheprobabilitythattheyactuallyknowtheanswer?

Exercise4. Ablo o dtestindicatesthepresenceofAmyotrophiclateralsclerosis(ALS) 95%ofthetimewhenALSisactuallypresent 0%ofthetimewhenthediseaseisnotactuallypresent. Onep ercentofthep opulation actuallyhasALS ersonactuallyhasALSgiventhatthe testindicatesthepresenceofALS.

Exercise 4. Asurveyconductedinacollegefoundthat 40%of thestudentswatch showAand17%ofthestudentswhofollowshowA,alsowatchshowB,20% ofthestudentswatchshowB.

(1)Whatistheprobabilitythatarandomlychosenstudentfollowsb othshows? (2)Whatis theconditional probabilitythat the studentfollowsshow Agiven that she/hefollowsshowB?

Exercise4. UseBayes'formulatosolvethefollowingproblem orthasproblems withbirds. Iftheweatherissunny, theprobabilitythattherearebirdson therunwayis 1 / 2 ;ifitiscloudy,butdry,theprobabilityis 1 / 3 ;andifitisraining,thentheprobability is 1 / 4. Theprobabilityofeachtyp eoftheweatheris 1 / 3. Giventhatthebirdsareonthe runaway,whatistheprobability

(1)thattheweatherissunny? (2)thattheweatheriscloudy(dryorrainy)?

Exercise4. Supp oseyoutossafaircoinrep eatedlyandindep endently. Ifitcomesup heads,youwinadollar,andifitcomesuptails,youloseadollar. Supp oseyoustartwith $20$150b eforeyougobroke? (SeeExample4 asolution).

Exercise∗4. Supp oseweplaygambler'sruingameinExample4, butratherinsuchawaythatyouwinadollarwithprobabilityp,andyouloseadollarwith probability 1 −p, 0 < p < 1. FindtheprobabilityofreachingN dollarsb eforegoingbroke ifwestartwithndollars.

Exercise∗ 4. Supp oseF isanevent, anddenePF(E) :=P(E|F). Showthat the conditionalprobabilityPF isaprobabilityfunction,thatis,itsatisestheaxiomsofproba- bility.

54 4

Exercise∗ 4. ShowdirectlythatProp osition2 conditionalprobability PF. Inparticular,foranyeventsEandF

E(Ec|F) = 1−E(E|F).

56 4

SolutiontoExercise4: denotebyM theeventanAmericanisaman,byCtheevent anAmericaniscolorblind

P(M|C) =

P(C|M)P(M)

P(C|M)P(M) +P(C|Mc)P(Mc)

(0) (0)

(0) (0) + (0) (0)

≈ 0. 9505.

SolutiontoExercise4(A):letHb etheeventabulbworksover 5000 hours,Xb ethe eventthatabulbcomesfromfactoryX,andY b etheeventabulbcomesfromfactoryY. Thenbythelawoftotalprobability

P(H) =P(H|X)P(X) +P(H|Y)P(Y) = (0) (0) + (0) (0) = 0. 974.

SolutiontoExercise4(B):ByPart(a)wehave

P(Y |H) =

P(H|Y)P(Y)

P(H)

=

(0) (0)

0. 974

≈ 0. 39.

SolutiontoExercise4(C):WeagainusetheresultfromPart(a)

P(X|Hc) =

P(Hc|X)P(X) P(Hc)

=

P(Hc|X)P(X) 1 −P(H)

(1− 0 .99) (0)

1 − 0. 974

=

(0) (0)

0. 026

≈ 0. 23

SolutiontoExercise4:denotebyDtheeventthataboltisdefective,Atheeventthat aboltisfrommachineA,byBtheeventthataboltisfrommachineC. ThenbyBayes' theorem

P(A|D) =

P(D|A)P(A)

P(D|A)P(A) +P(D|B)P(B) +P(D|C)P(C)

=

(0) (0)

(0) (0) + (0) (0) + (0) (0)

= 0. 362.

SolutiontoExercise4: LetC b etheeventastudentgivesthecorrectanswer,Kb e theeventastudentknows thecorrectanswer,Eb etheeventastudentcaneliminateone incorrect answer, andG b e theevent astudent havetoguessananswer. UsingBayes'

4.4 57

theoremwehave

P(K|C) =

P(C|K)P(K)

P(C)

=

P(C|K)P(K)

P(C|K)P(K) +P(C|E)P(E) +P(C|G)P(G)

=

1 · 12

1 · 12 + 13 · 14 + 14 · 14

=

24

31 ≈. 774 ,

thatis,approximately 77 .4%ofthestudentsknowtheansweriftheygivethecorrectanswer.

SolutiontoExercise4:Let+denotetheeventthatatestresultispositive,andbyD theeventthatthediseaseispresent. Then

P(D|+) =

P(+|D)P(D)

P(+|D)P(D) +P(+|Dc)P(Dc)

(0) (0)

(0) (0) + (0) (0)

= 0. 657.

SolutiontoExercise4:itisclearthatPF(E) =P(E|F)isb etween 0 and 1 sincethe right-handsideoftheidentitydeningPF is. Toseethesecondaxiom,observethat

PF(S) =P(S|F) =

P(S∩F)

P(F)

=

P(F)

= 1.

Nowtake{Ei}∞i=1, Ei∈ Ftob epairwisedisjoint,then

PF

(∞

⋃

)

=P

(∞

⋃

Ei|F

)

=

P((

⋃∞

i=1Ei)∩F) P(F)

P(

⋃∞

i=1(Ei∩F)) P(F)

=

∑∞

i=1P(Ei∩F) P(F)

∑∞

P(Ei∩F) P(F)

=

∑∞

PF(Ei).

Inthisweusedthedistributionlawforsets(E∪F)∩G= (E∩G)∪(F∩G)andthefact that{Ei∩F}∞i=1arepairwisedisjointaswell.

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Prob3160ch4 - math

Course: Differential Equation l (MATH 2230)

12 Documents

Students shared 12 documents in this course

University: Central Luzon State University

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CHAPTER 4

Conditional probability

4.1. Denition, Bayes' Rule and examples

Suppose there are 200 men, of which 100 are smokers, and 100 women, of which 20 are

smokers. What is the probability that a person chosen at random will be a smoker? The

answer is

120/300

. Now, let us ask, what is the probability that a person chosen at random

is a smoker given that the person is a women? One would expect the answer to be

20/100

and it is.

What we have computed is

number of women smokers

number of women

number of women smokers

/300

number of women

/300 ,

which is the same as the probability that a person chosen at random is a woman and a

smoker divided by the probability that a person chosen at random is a woman.

With this in mind, we give the following denition.

Denition 4.1 (Conditional probability)

P(F)>0

, we dene

the probability of

given

P(E|F) := P(E∩F)

P(F).

Note

P(E∩F) = P(E|F)P(F)

Example 4.1.

Suppose you roll two dice. What is the probability the sum is 8?

Solution

: there are ve ways this can happen

{(2,6),(3,5),(4,4),(5,3),(6,2)}

, so the prob-

ability is

5/36

. Let us call this event

. What is the probability that the sum is 8 given

that the rst die shows a 3? Let

be the event that the rst die shows a 3. Then

P(A∩B)

is the probability that the rst die shows a 3 and the sum is 8, or

1/36

P(B) = 1/6

, so

P(A|B) = 1/36

1/6= 1/6

Example 4.2.

Suppose a box has 3 red marbles and 2 black ones. We select 2 marbles.

What is the probability that second marble is red given that the rst one is red?

Prob3160ch4 - math

Differential Equation l (MATH 2230)

Central Luzon State University

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CHAPTER 4

Conditionalprobability

=

,

P(E∩F)

P(F)

.

( 3

)( 2

)

/

( 5

)

P(D|T) =

P(D∩T)

P(T)

=

(0)(0)

(0)(0) + (0)(0)

≈23%.

P(E∩F) =P(E|F)P(F),

P(F|E) =

P(F∩E)

P(E)

=

P(E|F)P(F)

P(E)

=

P(U)

1 −P(M)

=

4 / 10

6 / 10

=

2

3

.

( 52

)

( 4

)

·

( 48

)

P(A) =

( 4

)

·

( 48

)

( 52

).

( 52

)

·

( 39

)

( 4

)

·

( 48

)

·

( 2

)

·

( 37

)

P(A∩B) =

( 4

)

·

( 48