ProblemSolutions–Chapter1
Problem1.1.1Solution
BasedontheVenndiagramontheright,thecompleteGerlandas pizzamenuis
• Regularwithouttoppings
• Regularwithmushrooms
• Regularwithonions
• Regularwithmushroomsandonions
• Tuscanwithouttoppings
• Tuscanwithmushrooms
Problem1.1.2Solution
BasedontheVenndiagramontheright,theanswersaremostly fairlystraightforward.Theonlytrickinessisthatapizzaiseither Tuscan(T )orNeapolitan(N )so {N,T } isapartitionbutthey arenotdepictedasapartition.Specifically,theevent N isthe regionoftheVenndiagramoutsideofthe“squareblock”ofevent T .Ifthisisclear,thequestionsareeasy.
(a) Since N = T c , N ∩ M = φ.Thus N and M arenotmutuallyexclusive.
(b) EverypizzaiseitherNeapolitan(N ),orTuscan(T ).Hence N ∪ T = S so that N and T arecollectivelyexhaustive.Thusitsalso(trivially)truethat N ∪ T ∪ M = S.Thatis, R, T and M arealsocollectivelyexhaustive.
(c) FromtheVenndiagram, T and O aremutuallyexclusive.Inwords,this meansthatTuscanpizzasneverhaveonionsorpizzaswithonionsarenever Tuscan.Asanaside,“Tuscan”isafakepizzadesignation;oneshouldn’t concludethatpeoplefromTuscanyactuallydislikeonions.
(d) FromtheVenndiagram, M ∩ T and O aremutuallyexclusive.ThusGerlanda’sdoesn’tmakeTuscanpizzawithmushroomsandonions.
(e) Yes.IntermsoftheVenndiagram,thesepizzasareintheset(T ∪ M ∪ O)c .
M O T
M O T
2
AtRicardo’s,thepizzacrustiseitherRoman(R)orNeapolitan (N ).TodrawtheVenndiagramontheright,wemakethefollowingobservations:
R N M O W
• Theset {R,N } isapartitionsowecandrawtheVenndiagramwiththis partition.
• OnlyRomanpizzascanbewhite.Hence W ⊂ R.
• OnlyaNeapolitanpizzacanhaveonions.Hence O ⊂ N .
• BothNeapolitanandRomanpizzascanhavemushroomssothatevent M straddlesthe {R,N } partition.
• TheNeapolitanpizzacanhavebothmushroomsandonionsso M ∩ O cannot beempty.
• Theproblemstatementdoesnotprecludeputtingmushroomsonawhite Romanpizza.Hencetheintersection W ∩ M shouldnotbeempty.
Problem1.2.1Solution
(a) Anoutcomespecifieswhethertheconnectionspeedishigh(h),medium(m), orlow(l)speed,andwhetherthesignalisamouseclick(c)oratweet(t).
Thesamplespaceis
S = {ht,hc,mt,mc,lt,lc} .
(b) Theeventthatthewi-ficonnectionismediumspeedis A1 = {mt,mc}.
(c) Theeventthatasignalisamouseclickis A2 = {hc,mc,lc}.
(d) Theeventthataconnectioniseitherhighspeedorlowspeedis A3 = {ht,hc,lt,lc}.
Problem1.1.3Solution
(1)
3
(e) Since A1 ∩ A2 = {mc} andisnotempty, A1, A2,and A3 arenotmutually exclusive.
(f) Since
A1, A2, A3 iscollectivelyexhaustive.
Problem1.2.2Solution
(a) Thesamplespaceoftheexperimentis
(b) Theeventthatthecircuitfrom Z failsis
(c) Since ZF ∩ XA = {aaf,aff } = φ, ZF and XA arenotmutuallyexclusive.
(d) Since ZF ∪ XA = {aaa,aaf,afa,aff,faf,fff } = S, ZF and XA arenot collectivelyexhaustive.
(e) Theeventthatmorethanonecircuitisacceptableis
= {
(f) Inspectionshowsthat C ∩ D = φ so C and D aremutuallyexclusive.
(g) Since C ∪ D = S, C and D arecollectivelyexhaustive.
A1 ∪ A2 ∪ A3
ht,hc,mt,mc,lt,lc} = S, (2)
= {
thecollection
S = {aaa,aaf,afa,faa,ffa,faf,aff,fff } . (1)
ZF = {aaf,aff,faf,fff } . (2) Theeventthatthecircuitfrom
XA =
aaa,aaf,afa,aff
. (3)
X isacceptableis
{
}
C
aaa,aaf,afa,faa
(4) Theeventthatatleasttwocircuitsfailis D
ffa,faf,aff,fff
(5)
} .
= {
} .
4
Problem1.2.3Solution
Thesamplespaceis
Theevent H thatthefirstcardisaheartistheset
Theevent H has13outcomes,correspondingtothe13heartsinadeck.
Problem1.2.4Solution
Thesamplespaceis
Theevent H definedbytheeventofaJulybirthdayisgivenbythefollowinget with31sampleoutcomes:
Problem1.2.5Solution
Ofcourse,therearemanyanswerstothisproblem.Herearefourpartitions.
1. Wecandividestudentsintoengineersornon-engineers.Let A1 equalthe setofengineeringstudentsand A2 thenon-engineers.Thepair {A1,A2} isa partition.
2. ToseparatestudentsbyGPA,let Bi denotethesubsetofstudentswithGPAs G satisfying i 1 ≤ G<i.AtRutgers, {B1,B2,...,B5} isapartition.Note that B5 isthesetofallstudentswithperfect4.0GPAs.Ofcourse,other schoolsusedifferentscalesforGPA.
3. Wecanalsodividethestudentsbyage.Let Ci denotethesubsetofstudents ofage i inyears.Atmostuniversities, {C10,C11,...,C100} wouldbeanevent space.Sinceauniversitymayhaveprodigieseitherunder10orover100,we notethat {C0,C1,...} isalwaysapartition.
S = {A♣,...,K♣,A♦,...,K♦,A♥,...,K♥,A♠,...,K♠} . (1)
H = {A♥,...,K♥} . (2)
S = 1/1 ... 1/31, 2/1 ... 2/29, 3/1 ... 3/31, 4/1 ... 4/30, 5/1 ... 5/31, 6/1 ... 6/30, 7/1 ... 7/31, 8/1 ... 8/31, 9/1 ... 9/31, 10/1 ... 10/31, 11/1 ... 11/30, 12/1 ... 12/31 . (1)
H = {7/1, 7/2,..., 7/31} . (2)
5
4. Lastly,wecancategorizestudentsbyattendance.Let D0 denotethenumber ofstudentswhohavemissedzerolecturesandlet D1 denoteallotherstudents. Althoughitislikelythat D0 isanemptyset, {D0,D1} isawelldefined partition.
Problem1.2.6Solution
Let R1 and R2 denotethemeasuredresistances.Thepair(R1,R2)isanoutcome oftheexperiment.Somepartitionsinclude
1. Ifweneedtocheckthatneitherresistanceistoohigh,apartitionis
2. Ifweneedtocheckwhetherthefirstresistanceexceedsthesecondresistance, apartitionis
3. Ifweneedtocheckwhethereachresistancedoesn’tfallbelowaminimum value(inthiscase50ohmsfor R1 and100ohmsfor R2),apartitionis
4. Ifwewanttocheckwhethertheresistorsinparallelarewithinanacceptable rangeof90to110ohms,apartitionis
A1 = {R1 < 100,R2 < 100} ,A2 = {R1 ≥ 100}∪{R2 ≥ 100} . (1)
B1 = {R1 >R2} B2 = {R1 ≤ R2} . (2)
C1,C2,C3,C4 where C1 = {R1 < 50,R2 < 100} ,C2 = {R1 < 50,R2 ≥ 100} , (3) C3 = {R1 ≥ 50,R2 < 100} ,C4 = {R1 ≥ 50,R2 ≥ 100} . (4)
D1 = (1/R1 +1/R2) 1 < 90 , (5) D2 = 90 ≤ (1/R1 +1/R2) 1 ≤ 110 , (6) D2 = 110 < (1/R1 +1/R2) 1 . (7) 6
(a) A and B mutuallyexclusiveandcollectivelyexhaustiveimplyP[A]+P[B]=1.
SinceP[A]=3P[B],wehaveP[B]=1/4.
(b) SinceP[A ∪ B]=P[A],weseethat B ⊆ A.ThisimpliesP[A ∩ B]=P[B]. SinceP[A ∩ B]=0,thenP[B]=0.
(c) Sinceit’salwaystruethatP[A
B] P[AB],wehavethat
Thisimplies2P[B]=P[AB].However,since AB ⊂ B,wecanconcludethat 2P[B]=P[AB] ≤ P[B].ThisimpliesP[B]=0.
Problem1.3.2Solution
Therolloftheredandwhitedicecanbeassumedtobeindependent.Foreach die,allrollsin {1, 2,..., 6} haveprobability1/6.
(a) Thus
(b) Infact,eachpairofpossiblerolls RiWj hasprobability1/36.TofindP[S5], weadduptheprobabilitiesofallpairsthatsumto5:
Problem1.3.3Solution
Anoutcomeisapair(i,j)where i isthevalueofthefirstdieand j isthevalueof theseconddie.Thesamplespaceistheset
with36outcomes,eachwithprobability1/36Notethattheeventthattheabsolute valueofthedifferenceofthetworollsequals3is
Problem1.3.1Solution
∪ B]=P[A
P[A]+P[B] P[AB]=P[A] P[B]
(1)
]+P[
.
P[R3W2]=P[R3]P[W2]= 1 36
.
P[S5]=P[R1W4]+P[R2W3]+P[R3W2]+P[R4W1]=4/36=1/9.
S = {(1, 1), (1, 2),..., (6, 5), (6, 6)} . (1)
D3 = {(1, 4), (2, 5), (3, 6), (4, 1), (5, 2), (6, 3)} . (2) Sincethereare6outcomesin D3,P[D3]=6/36=1/6. 7
Problem1.3.4Solution
FALSE.SinceP[A]=1 P[Ac]=2P[Ac]impliesP[Ac]=1/3. (b) FALSE.Suppose A = B andP[A]=1/2.Inthatcase, P[AB]=P[A]=1/2 > 1/4=P[A]P[B] . (1) (c) TRUE.Since AB ⊆ A,P[AB] ≤ P[A],Thisimplies P[AB] ≤ P[A] < P[B] . (2) (d) FALSE:Foracounterexample,let A = φ andP[B] > 0sothat A = A∩B = φ andP[A]=P[A ∩ B]=0but0=P[A] < P[B]. Problem1.3.5Solution Thesamplespaceoftheexperimentis S = {LF,BF,LW,BW } . (1) Fromtheproblemstatement,weknowthatP[LF ]=0.5,P[BF ]=0.2and P[BW ]=0.2.ThisimpliesP[LW ]=1 0.5 0.2 0.2=0.1.Thequestions canbeansweredusingTheorem1.5. (a) Theprobabilitythataprogramisslowis P[W ]=P[LW ]+P[BW ]=0.1+0.2=0.3. (2) (b) Theprobabilitythataprogramisbigis P[B]=P[BF ]+P[BW ]=0.2+0.2=0.4. (3) (c) Theprobabilitythataprogramissloworbigis P[W ∪ B]=P[W ]+P[B] P[BW ]=0.3+0.4 0.2=0.5. (4) 8
(a)
Problem1.3.6Solution
Asampleoutcomeindicateswhetherthecellphoneishandheld(H)ormobile(M ) andwhetherthespeedisfast(F )orslow(W ).Thesamplespaceis
Thisimplies
Also,sincetheprobabilitiesmustsumto1,
Nowthatwehavefoundtheprobabilitiesoftheoutcomes,findinganyotherprobabilityiseasy.
Problem1.3.7Solution
Areasonableprobabilitymodelthatisconsistentwiththenotionofashuffleddeck isthateachcardinthedeckisequallylikelytobethefirstcard.Let Hi denote theeventthatthefirstcarddrawnisthe ithheartwherethefirstheartistheace,
S = {HF,HW,MF,MW } . (1) TheproblemstatementtellsusthatP[HF ]=0.2,P[MW ]=0.1andP[F ]=0.5. Wecanusethesefactstofindtheprobabilitiesoftheotheroutcomes.Inparticular, P[F ]=P[HF ]+P[MF ] . (2)
P[MF ]=P[F ] P[HF ]=0.5 0.2=0.3. (3)
P[HW ]=1 P[HF ] P[MF ] P[MW ] =1 0.2 0.3 0.1=0.4. (4)
(a) Theprobabilityacellphoneisslowis P[W ]=P[HW ]+P[MW ]=0.4+0.1=0.5. (5) (b) TheprobabilitythatacellhponeismobileandfastisP[MF ]=0.3. (c) Theprobabilitythatacellphoneishandheldis P[H]=P[HF ]+P[HW ]=0.2+0.4=0.6. (6)
9
Thisistheansweryouwouldexpectsince13outof52cardsarehearts.Thepoint tokeepinmindisthatthisisnotjustthecommonsenseanswerbutistheresult ofaprobabilitymodelforashuffleddeckandtheaxiomsofprobability.
Problem1.3.8Solution
Let si denotetheoutcomethatthedownfacehas i dots.Thesamplespaceis S
Problem1.3.9Solution
Let si equaltheoutcomeofthestudent’squiz.Thesamplespaceisthencomposed ofallthepossiblegradesthatshecanreceive.
Sinceeachofthe11possibleoutcomesisequallylikely,theprobabilityofreceiving agradeof
/11.Theprobabilitythatthe studentgetsanAistheprobabilitythatshegetsascoreof9orhigher.Thatis
Theprobabilityoffailingrequiresthestudenttogetagradelessthan4.
thesecondheartisthedeuceandsoon.Inthatcase,P[Hi]=1/52for1 ≤ i ≤ 13. Theevent H
H = H1 ∪ H2 ∪···∪ H13. (1) UsingTheorem1.1,wehave P[H]= 13 i=1 P[Hi]=13/52. (2)
thatthefirstcardisaheartcanbewrittenasthemutuallyexclusive union
{s1,...,s
si]=1/
thattherollisevenis P[E]=P[s2]+P[s4]+P[s6]=3/6. (1)
=
6}.TheprobabilityofeachsampleoutcomeisP[
6.From Theorem1.1,theprobabilityoftheevent E
S = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10} . (1)
1
10isP[si]=1
P[GradeofA]=P[9]+P[10]=1/11+1/11=2/11. (2)
i,foreach i =0,
,...,
P[Failing]=P[3]+P[2]+P[1]+P[0] =1/11+1/11+1/11+1/11=4/11. (3) 10
Eachstatementisaconsequenceofpart4ofTheorem1.4.
(a) Since
(b) Since
Specifically,wewilluseTheorem1.4(c)whichstatesthatforanyevents
Problem1.3.10Solution
A ⊂ A ∪ B,P[A] ≤ P[A ∪ B].
B ⊂ A ∪ B,P[B] ≤ P[A ∪ B].
A ∩ B ⊂ A,P[A ∩ B] ≤ P[A].
A ∩ B ⊂ B,P[A ∩ B] ≤ P[B]. Problem1.3.11Solution
(c) Since
(d) Since
A and B, P[A ∪ B]=P[A]+P[B] P[A ∩ B] . (1)
P[A1 ∪ A2]=P[A1]+P[A2] P[A1 ∩ A2] . (2) Bythefirstaxiomofprobability,P[A1 ∩ A2] ≥ 0.Thus, P[A1 ∪ A2] ≤ P[A1]+P[A2] . (3) whichprovestheunionboundforthecase n =2.Nowwemakeourinduction
n 1subsets.Inthis case,givensubsets A1,...,An,wedefine A = A1 ∪ A2 ∪···∪ An 1,B = An. (4) Byourinductionhypothesis, P[A]=P[A1 ∪ A2 ∪···∪ An 1] ≤ P[A1]+ ··· +P[An 1] . (5) Thispermitsustowrite P[A1 ∪···∪ An]=P[A ∪ B] ≤ P[A]+P[B](bytheunionboundfor n =2) =P[A1 ∪···∪ An 1]+P[An] ≤ P[A1]+ ··· P[An 1]+P[An](6) whichcompletestheinductiveproof. 11
Toprovetheunionboundbyinduction,wefirstprovethetheoremforthecaseof n =2events.Inthiscase,byTheorem1.4(c),
hypothesisthattheunion-boundholdsforanycollectionof
Itistemptingtousethefollowingproof:
Problem1.3.12Solution
S = S ∪ φ, 1=P[S ∪ φ]=P[S]+P[φ] . SinceP[S]=1,wemusthaveP[φ]=0.
A1 and A2, P[A1 ∪ A2]=P[A1]+P[A2] . (1) Theproblemisthatthispropertyisaconsequenceofthethreeaxioms,andthus mustbeproven.Foraproofthatusesjustthethreeaxioms,let A1 beanarbitrary setandfor n =2, 3,...,let An = φ.Since A1 = ∪∞ i=1Ai,wecanuseAxiom3to write P[A1]=P[∪∞ i=1Ai]=P[A1]+P[A2]+ ∞ i=3 P[Ai] . (2) BysubtractingP[A1]frombothsides,thefactthat A2 = φ permitsustowrite P[φ]+ ∞ n=3 P[Ai]=0. (3) ByAxiom1,P[Ai] ≥ 0forall i.Thus, ∞ n=3 P[Ai] ≥ 0.ThisimpliesP[φ] ≤ 0. SinceAxiom1requiresP[φ] ≥ 0,wemusthaveP[φ]=0. Problem1.3.13Solution Followingthehint,wedefinethesetofevents {Ai|i =1, 2,...} suchthat i = 1,...,m, Ai = Bi andfor i>m, Ai = φ.Byconstruction, ∪m i=1Bi = ∪∞ i=1Ai. Axiom3thenimplies P[∪m i=1Bi]=P[∪∞ i=1Ai]= ∞ i=1 P[Ai] . (1) 12
Since S and φ aremutuallyexclusive,andsince
Theabove“proof”usedthepropertythatformutuallyexclusivesets
EachclaiminTheorem1.4requiresaprooffromwhichwecancheckwhichaxioms areused.However,theproblemissomewhathardbecausetheremaystillbe asimplerproofthatusesfeweraxioms.Still,theproofofeachpartwillneed Theorem1.3whichwenowprove.
For i>m,P[Ai]=P[φ]=0,yieldingtheclaimP[∪m i=1Bi]= m i=1 P[Ai]= m i=1 P[Bi]. NotethatthefactthatP[φ]=0followsfromAxioms1and2.Thisproblemis morechallengingifyoujustuseAxiom3.Westartbyobserving P[∪m i=1Bi]= m 1 i=1 P[Bi]+ ∞ i=m P[Ai] . (2) Now,weuseAxiom3againonthecountablyinfinitesequence Am,Am+1,... to write ∞ i=m P[Ai]=P[Am ∪ Am+1 ∪···]=P[Bm] . (3) Thus,wehaveusedjustAxiom3toproveTheorem1.3: P[∪m i=1Bi]= m i=1 P[Bi] . (4)
Problem1.3.14Solution
Forthemutuallyexclusiveevents B1,...,Bm,let Ai = Bi for i =1,...,m and let Ai = φ for i>m.Inthatcase,byAxiom3, P[B1 ∪ B2 ∪···∪ Bm]=P[A1 ∪ A2 ∪···] = m 1 i=1 P[Ai]+ ∞ i=m P[Ai] = m 1 i=1 P[Bi]+ ∞ i=m P[Ai] . (1) 13
Thus,wehaveusedjustAxiom3toproveTheorem1.3:
(a) ToshowP[φ]=0,let B1 = S andlet B2 = φ.ThusbyTheorem1.3,
Thus,P[φ]=0.NotethatthisproofusesonlyTheorem1.3whichusesonly Axiom3.
(b) UsingTheorem1.3with B1 = A and B2 = Ac,wehave
Since,Axiom2saysP[S]=1,P[
c]=1 P[A].ThisproofusesAxioms2 and3.
(c) ByTheorem1.8,wecanwriteboth A and B asunionsofmutuallyexclusive events:
NowweapplyTheorem1.3towrite
NotethatsofarwehaveusedonlyAxiom3.Finally,weobservethat A ∪ B canbewrittenastheunionofmutuallyexclusiveevents
Now,weuseAxiom3againon Am,Am+1,... towrite ∞ i=m P[Ai]=P[Am ∪ Am+1 ∪···]=P[Bm] . (2)
P[B1 ∪ B2 ∪···∪ Bm]= m i=1 P[Bi] . (3)
P[S]=P[B1 ∪ B2]=P[B1]+P[B2]=P[S]+P[φ] . (4)
P[S]=P[A ∪ Ac]=P[A]+P[Ac] . (5)
A
A =(AB) ∪ (ABc),B =(AB) ∪ (AcB). (6)
P[A]=P[AB]+P[ABc] , P[B]=P[AB]+P[AcB] . (7) Wecanrewritethesefactsas P[ABc]=P[A] P[AB], P[AcB]=P[B] P[AB]. (8)
A ∪ B =(AB) ∪ (ABc) ∪ (AcB). (9) 14
Onceagain,usingTheorem1.3,wehave
SubstitutingtheresultsofEquation(8)intoEquation(10)yields
whichcompletestheproof.NotethatthisclaimrequiredonlyAxiom3.
(d) Observethatsince A ⊂ B,wecanwrite B asthemutuallyexclusiveunion B = A ∪ (AcB).ByTheorem1.3(whichusesAxiom3),
and3.
Problem1.4.1Solution
Eachquestionrequestsaconditionalprobability.
(a) Notethattheprobabilityacallisbriefis
P[A ∪ B]=P[AB]+P[ABc]+P[AcB](10)
P[A ∪ B]=P[AB]+P[A] P[AB]+P[B] P[AB]
(11)
,
P[B]=P[A]+P[AcB] . (12) ByAxiom1,P[
≥ 0,hichimpliesP[A] ≤ P[B].ThisproofusesAxioms1
AcB]
P[B]=P[H0B]+P[H1B]+P[H2B]=0.6. (1) Theprobabilityabriefcallwillhavenohandoffsis P[H0|B]= P[H0B] P[B] = 0.4 0.6 = 2 3 . (2) (b) TheprobabilityofonehandoffisP[H1]=P[H1B]+P[H1L]=0.2.The probabilitythatacallwithonehandoffwillbelongis P[L|H1]= P[H1L] P[H1] = 0.1 0.2 = 1 2 . (3) (c) TheprobabilityacallislongisP[L]=1 P[B]=0.4.Theprobabilitythat alongcallwillhaveoneormorehandoffsis P[H1 ∪ H2|L]= P[H1L ∪ H2L] P[L] = P[H1L]+P[H2L] P[L] = 0.1+0.2 0.4 = 3 4 . (4) 15
Let si denotetheoutcomethattherollis i.So,for1 ≤ i ≤ 6, Ri = {si}.Similarly, Gj = {sj+1,...,s6}. (a) Since G1 = {s2,s3,s4,s5,s6} andalloutcomeshaveprobability1/6,P[G1]= 5/6.Theevent R3G1 = {s3} andP[R3G1]=1/6sothat P[R3|G1]= P[R3G1] P[G1] = 1 5 . (1) (b) Theconditionalprobabilitythat6isrolledgiventhattherollisgreaterthan 3is P[R6|G3]= P[R6G3] P[G3] = P[s6] P[s4,s5,s6] = 1/6 3/6 . (2) (c) Theevent E thattherollisevenis E = {s2,s4,s6} andhasprobability3/6. Thejointprobabilityof G3 and E is P[G3E]=P[s4,s6]=1/3. (3) Theconditionalprobabilitiesof G3 given E is P[G3|E]= P[G3E] P[E] = 1/3 1/2 = 2 3 . (4)
Theconditionalprobabilitythattherollisevengiventhatit’sgreaterthan 3is P[E|G3]= P[EG3] P[G3] = 1/3 1/2 = 2 3 . (5) Problem1.4.3Solution Sincethe2ofclubsisanevennumberedcard, C2 ⊂ E sothatP[C2E]=P[C2]= 1/3.SinceP[E]=2/3, P[C2|E]= P[C2E] P[E] = 1/3 2/3 =1/2. (1) Theprobabilitythatanevennumberedcardispickedgiventhatthe2ispickedis P[E|C2]= P[C2E] P[C2] = 1/3 1/3 =1. (2) 16
Problem1.4.2Solution
(d)
Let Ai and Bi denotetheeventsthatthe ithphonesoldisanApricotoraBanana respectively.OurgoalistofindP[B1B2],butsinceitisnotclearwheretostart, weshouldplanonfillinginthetable
Westillneedthreemoreequationstosolveforthefourunknowns.From“salesof ApricotsandBananasareequallylikely,”weknowthatP[
Thefinalequationcomesfrom“giventhatthefirstphonesoldisaBanana,thesecondphoneistwiceaslikelytobeaBanana,”whichimpliesP[
Problem1.4.4Solution
A2 B2 A1 B1 Thistablehasfourunknowns:P[A1A2],P[A1B2],P[B1A2],andP[B1B2].Westart
P[A1A2]+P[A1B2]+P[B1A2]+P[B1B2]=1. (1)
knowingthat
Ai]=P[Bi]=1/2for
, 2.Thisimplies P[A1]=P[A1A2]+P[A1B2]=1/2, (2) P[A2]=P[A1A2]+P[B1A2]=1/2. (3)
i =1
B2|B1]=2P[A2|B1].
P[B1B2] P[B1] =2 P[B1A2] P[B1] =⇒ P[B1A2]= 1 2 P[B1B2] . (4) ReplacingP[B1A2]withP[B1B2]/2inthethefirstthreeequationsyields P[A1A2]+P[A1B2]+ 3 2 P[B1B2]=1, (5) P[A1A2]+P[A1B2]=1/2, (6) P[A1A2]+ 1 2 P[B1B2]=1/2. (7) Subtracting(6)from(5)yields(3/2)P[B1B2]=1/2,orP[B1B2]=1/3,whichis theanswerwearelookingfor. 17
UsingBayes’theorem,wehave
Atthispoint,ifyouarecurious,wecansolvefortherestoftheprobabilitytable.
1A2]=1/3.Itthen followsfrom(6)thatP[A1B2]=1/6.Theprobabilitytableis
From(4),wehaveP[B1A2]=1/6andfrom(7)weobtainP[
Problem1.4.5Solution
Thefirstgenerationconsistsoftwoplantseachwithgenotype yg or gy.Theyare crossedtoproducethefollowingsecondgenerationgenotypes, S = {yy,yg,gy,gg}.
Eachgenotypeisjustaslikelyasanyothersotheprobabilityofeachgenotypeis consequently1/4.Apeaplanthasyellowseedsifitpossessesatleastonedominant y gene.Thesetofpeaplantswithyellowseedsis
Problem1.4.6Solution
Define D astheeventthatapeaplanthastwodominant y genes.Tofindthe conditionalprobabilityof D giventheevent Y ,correspondingtoaplanthaving yellowseeds,welooktoevaluate
NotethatP[DY ]isjusttheprobabilityofthegenotype yy.FromProblem1.4.5, wefoundthatwithrespecttothecolorofthepeas,thegenotypes yy, yg, gy,and gg wereallequallylikely.Thisimplies
Thus,theconditionalprobabilitycanbeexpressedas
A2 B2 A1 1/31/6 B1 1/61/3 .
A
Y = {yy,yg,gy} . (1) Sotheprobabilityofapeaplantwithyellowseedsis P[Y ]=P[yy]+P[yg]+P[gy]=3/4. (2)
P[D|Y ]= P[DY ] P[Y ] . (1)
P[DY ]=P[yy]=1/4P[Y ]=P[yy,gy,yg]=3/4. (2)
P[D|Y ]= P[DY ] P[Y ] = 1/4 3/4 =1/3. (3) 18
