Chapter 1 m
Principles of Probability
m m
1. Combiningm independentm probabilities.
Youmhavemappliedmtomthreemmedicalmschools:m UniversitymofmCaliforniamatmSanmFranciscom(UCSF)
,
DuluthmSchoolmofmMinesm(DSM),mandmHarvardm(H).mYoumguessmthatmthemprobabilitiesmyo
u’llmbemacceptedmare:m p(UCSF)m=m0.10,mp(DSM)m=m0.30,mandmp(H)m=m0.50.m Assumemthat
mthemacceptancemeventsmaremindependent.
(a) Whatm ism them probabilitym thatm youm getm inm somewherem (atm leastm onem acceptance)?
(b) Whatm ism them probabilitym thatm youm willm bem acceptedm bym bothm Harvardm andm Duluth?
(a) Themsimplestmwaymtomsolvemthismproblemmismtomrecallmthatmwhenmprobabilitiesmaremind
ependent,mandmyoumwantmthemprobabilitymofmeventsmAmANDmB,myoumcanmmultiplym them
.mWhenmeventsmaremmutuallymexclusivem andmyoumwantmthemprobabilitymofmeventsmAmO
RmB,myoumcanmaddm themprobabilities.m Thereforemwemtrymtomstructuremthemproblemmint
omanmANDmandm ORm problem.m Wem wantm them probabilitym ofm gettingm intom Hm ORm DSMm o
rm UCSF.m Butmthismdoesn’tmhelpmbecausemthesemeventsmaremnotmmutuallymexclusivem(mu
tuallymexclusivemmeansm thatm ifm onem happens,m them otherm cannotm happen).m Som wem trym a
gain.m Them probabilitymofmacceptancem somewhere,mPm(a),mismPm(a)m=m1m−mPm(r),mwheremPm
(r)mismthemprobabilitymthatmyou’rem rejectedm everywhere.m (You’rem eitherm acceptedm somew
herem orm you’rem not.)m Butm thismprobabilitymcanm bemputminmthemabovemterms.m Pm(r)m=mth
emprobabilitymthatmyou’remrejectedmatmHmANDmatmDSMmANDmatmUCSF.mThesemeventsmare
mindependent,msomwemhavemthemanswer.mThem probabilitym ofm rejectionm atm Hm ism p(rH)m
=m1m−m0.5m=m0.5.m Rejectionm atm DSMm ismp(rDSM)m =m 1m−m0.3m =m 0.7.m Rejectionm atm U
CSFm ism p(rUCSF)m =m 1m−m0.1m =m 0.9.m ThereforemPm(r)m=m(0.5)(0.7)(0.9)m=m0.315.m There
foremthemprobabilitymofmatmleastmonemacceptance
=m Pm(a)m =m 1m−mPm(r)m=m 0.685.
1
,(b) Themsimplem answerm ism thatmthism ismthem intersectionm ofm twom independentm events:
p(aH)p(aDSM)m =m (0.50)(0.30)
=m m 0.15.
Am morem mechanicalm approachm tomeithermpartm (a)mormthism partm ism tomwritem outm allmthem
possiblemcircumstances.m RejectionmandmacceptancematmHmaremmutuallymexclusive.m Theirm
probabilitiesm addmtomone.m Themsamemformthemothermtwomschools.m Thereforemallmpossible
mcircumstancesm aremtakenmintomaccountmbymaddingm themmutuallym exclusivemeventsmtoget
her,mandmmultiplyingm independentmevents:
[p(aH)m +mp(rH)][p(aDSM)m+m p(rDSM)][p(aUCSF)m +mp(rUCSF)]m =m 1,
orm equivalently,
=m p(aH)p(aDSM)p(aUCSF)m+mp(aH)p(aDSM)p(rUCSF)
+p(aH)p(rDSM)p(aUCSF)m+m ·m·m·
wheremthemfirstmtermmismthemprobabilitymofmacceptancematmallm3,mthemsecondmtermmrepres
entsmacceptancem atmHmandmDSMmbutmrejectionm atmUCSF,mthemthirdmtermmrepresentsm acc
eptancem atm Hm andm UCSFm butm rejectionm atm DSM,m etc.m Eachm ofm thesem eventsm ism mutu
allym exclusivemwithmrespectmtomeachmother;mthereforem theymaremallmadded.m Eachmindivid
ualmtermm representsmindependentmeventsmof,mformexample,maHmandm aDSMmandm aUCSF.m
Thereforem itm ismsimplemtomreadmoffmthemanswerminmthismproblem:m wemwantmaHmandm aDS
M,mbutmnoticemwemdon’tm carem aboutm UCSF.m Thism probabilitym is
p(aH)p(aDSM)m m =m m p(aH)p(aDSM)[p(aUCSF)m+mp(rUCSF)]
=m m (0.50)(0.30)
=m m 0.15.
Notemthatmwemcouldmhavemsolvedmpartm(a)mthemsamemway;mitmwouldmhavemrequiredmadd
ingmupmallmthemappropriatempossiblemmutuallymexclusivemevents.m Youmcanmcheckmthatmit
mgivesmthemsamemanswer masmabovem(butmnoticemhowmmuchmmoremtediousmitmis).
2
,2. Probabilitiesm ofm sequences.
AssumemthatmthemfourmbasesmA,mC,mT,mandmGmoccurmwithmequalmlikelihoodminmamDNAmsequ
ence
ofm ninem monomers.
(a) Whatm ism them probabilitym ofm findingm them sequencem AAATCGAGTm throughm rando
mmchance?
(b) Whatm ism them probabilitym ofm findingm them sequencem AAAAAAAAAm throughm rando
mmchance?
(c) Whatm ism them probabilitym ofm findingm anym sequencem thatm hasm fourm A’s,m twom T’s,m two
m G’s,mandmonemC,msuchmasmthatminm(a)?
(a) Eachmbasemoccursmwithmprobabilitym1/4.m ThemprobabilitymofmanmAminmpositionm1mism1/
4,mofmAminmpositionm2mism1/4,mofmAminmpositionm3mism1/4,mofmTm inmpositionm4mism1/4,man
dmsomon.mTheremarem9mbases.m Themprobabilitymofmthismspecificmsequencemism(1/4)9m=m
3.8m×m10−6.
(b) Samem answerm asm (a)m above.
(c) Eachmspecificmsequencemhasmthemprobabilitymgivenmabove,mbutminmthismcasemtheremaremma
nympossiblem sequencesm whichm satisfym them requirementm thatmwemhavem 4mA’s,m2mTm’s,m2mG’
s,mandm1
C.m Howmmanymaremthere?m Wemstartmasmwemhavemdonembefore,mbymassumingmallmninemobj
ectsmaremdistinguishable.m Theremarem9!m arrangementsmofmninemdistinguishablem objectsmi
nmam linearmsequence.m (Themfirstmonemcanmbeminmanymofmninemplaces,mthemsecondminmanym
ofmthemremainingmeightmplaces,mandmsomon.)m Butmwemcan’tmdistinguishmthemfourmA’s,msom
wemhavemovercountedm bym am factorm ofm 4!,m andm mustm dividem thism out.m Wem can’tm disting
uishm them twom Tm’s,msomwemhavemovercountedmbym2!,mandmmustmalsomdividemthismout.m A
ndmsomon.m Somthemprobabilitymofmhavingmthismcompositionmis
"m #m m mm 9
9! 1
=m 0.014
4!2!2!1! 4 .
3
, 3. Them probabilitym ofm am sequencem (givenm am composition).
Amscientistmhasmconstructedmamsecretmpeptidemtomcarrymammessage.m Youmknowmonlymthemcomp
osi-
tionmofmthempeptide,mwhichmismsixmaminomacidsmlong.m ItmcontainsmonemserinemS,monemthreoni
nemT,monemcysteinemC,monemargininemR,mandmtwomglutamatesmE.mWhatmismthemprobabilit
ymthatmthemsequencemSECRETmwillmoccurmbymchance?
ThemSm couldmbeminmanymonemofmthem6mpositionsmwithmequalmlikelihood.m Themprobabilitymtha
tmitmisminmpositionm1mism(1/6).mGivenmthatmSmisminmthemfirstmposition,mwemhavem2mE ' s mwhichm
couldmoccurminmanymofmthemremainingm5mpositions.m Themprobabilitymthatmonemofmthemmismin
mpositionm2mism(2/5).m Givenmthosemtwomlettersminmposition,mthemprobabilitymthatmthem1mCm ismi
nmthemnextmofmthem4mremainingmpositionsmism(1/4).mThemprobabilitymformthemRmism(1/3).m Fo
rmthemremainingmEmism(1/2),mandmformthemlastmTm ism(1/1),msomthemprobabilitymis
"m #−1
6!
(1/6)(2/5)(1/4)(1/3)(1/2)m =m 1/360m = .
1!2!1!1!
4. Combiningm independentm probabilities.
Youm havem am fairm six-
sidedm die.m Youm wantm tom rollm itm enoughm timesm tom ensurem thatm am 2m occursm at
leastmonce.m Whatmnumbermofmrollsmkmismrequiredmtomensuremthatmthemprobabilitymismatmleastm2
/3mthatmatmleastmonem2mwillmappear?
5
q = m =mprobabilitymthatmam2mdoesmnotmappearmonm thatmroll.
6
qk =m probabilitymthatmam2mdoesmnotmappearmonmkmINDEPENDENTm rolls.
Pm(k)m =m 1m−mqkm=mprobabilitymthatmatmleastmonem2mappearsmonmkmrolls.
For
m m m
2m 2m 1m 1
Pm(k)m m ≥ ,m1m−mqkm≥m =⇒m qkm ≤m =⇒m kmlnmqm≤mln
3 3 3 3
ln(1/3)
=⇒m k ≥ =m 6.03
ln(5/6)
Approximatelym sixmormmoremrollsmwillmensuremwithmprobabilitymPm ≥m2/3mthatmam2mwillmappear.
4
Principles of Probability
m m
1. Combiningm independentm probabilities.
Youmhavemappliedmtomthreemmedicalmschools:m UniversitymofmCaliforniamatmSanmFranciscom(UCSF)
,
DuluthmSchoolmofmMinesm(DSM),mandmHarvardm(H).mYoumguessmthatmthemprobabilitiesmyo
u’llmbemacceptedmare:m p(UCSF)m=m0.10,mp(DSM)m=m0.30,mandmp(H)m=m0.50.m Assumemthat
mthemacceptancemeventsmaremindependent.
(a) Whatm ism them probabilitym thatm youm getm inm somewherem (atm leastm onem acceptance)?
(b) Whatm ism them probabilitym thatm youm willm bem acceptedm bym bothm Harvardm andm Duluth?
(a) Themsimplestmwaymtomsolvemthismproblemmismtomrecallmthatmwhenmprobabilitiesmaremind
ependent,mandmyoumwantmthemprobabilitymofmeventsmAmANDmB,myoumcanmmultiplym them
.mWhenmeventsmaremmutuallymexclusivem andmyoumwantmthemprobabilitymofmeventsmAmO
RmB,myoumcanmaddm themprobabilities.m Thereforemwemtrymtomstructuremthemproblemmint
omanmANDmandm ORm problem.m Wem wantm them probabilitym ofm gettingm intom Hm ORm DSMm o
rm UCSF.m Butmthismdoesn’tmhelpmbecausemthesemeventsmaremnotmmutuallymexclusivem(mu
tuallymexclusivemmeansm thatm ifm onem happens,m them otherm cannotm happen).m Som wem trym a
gain.m Them probabilitymofmacceptancem somewhere,mPm(a),mismPm(a)m=m1m−mPm(r),mwheremPm
(r)mismthemprobabilitymthatmyou’rem rejectedm everywhere.m (You’rem eitherm acceptedm somew
herem orm you’rem not.)m Butm thismprobabilitymcanm bemputminmthemabovemterms.m Pm(r)m=mth
emprobabilitymthatmyou’remrejectedmatmHmANDmatmDSMmANDmatmUCSF.mThesemeventsmare
mindependent,msomwemhavemthemanswer.mThem probabilitym ofm rejectionm atm Hm ism p(rH)m
=m1m−m0.5m=m0.5.m Rejectionm atm DSMm ismp(rDSM)m =m 1m−m0.3m =m 0.7.m Rejectionm atm U
CSFm ism p(rUCSF)m =m 1m−m0.1m =m 0.9.m ThereforemPm(r)m=m(0.5)(0.7)(0.9)m=m0.315.m There
foremthemprobabilitymofmatmleastmonemacceptance
=m Pm(a)m =m 1m−mPm(r)m=m 0.685.
1
,(b) Themsimplem answerm ism thatmthism ismthem intersectionm ofm twom independentm events:
p(aH)p(aDSM)m =m (0.50)(0.30)
=m m 0.15.
Am morem mechanicalm approachm tomeithermpartm (a)mormthism partm ism tomwritem outm allmthem
possiblemcircumstances.m RejectionmandmacceptancematmHmaremmutuallymexclusive.m Theirm
probabilitiesm addmtomone.m Themsamemformthemothermtwomschools.m Thereforemallmpossible
mcircumstancesm aremtakenmintomaccountmbymaddingm themmutuallym exclusivemeventsmtoget
her,mandmmultiplyingm independentmevents:
[p(aH)m +mp(rH)][p(aDSM)m+m p(rDSM)][p(aUCSF)m +mp(rUCSF)]m =m 1,
orm equivalently,
=m p(aH)p(aDSM)p(aUCSF)m+mp(aH)p(aDSM)p(rUCSF)
+p(aH)p(rDSM)p(aUCSF)m+m ·m·m·
wheremthemfirstmtermmismthemprobabilitymofmacceptancematmallm3,mthemsecondmtermmrepres
entsmacceptancem atmHmandmDSMmbutmrejectionm atmUCSF,mthemthirdmtermmrepresentsm acc
eptancem atm Hm andm UCSFm butm rejectionm atm DSM,m etc.m Eachm ofm thesem eventsm ism mutu
allym exclusivemwithmrespectmtomeachmother;mthereforem theymaremallmadded.m Eachmindivid
ualmtermm representsmindependentmeventsmof,mformexample,maHmandm aDSMmandm aUCSF.m
Thereforem itm ismsimplemtomreadmoffmthemanswerminmthismproblem:m wemwantmaHmandm aDS
M,mbutmnoticemwemdon’tm carem aboutm UCSF.m Thism probabilitym is
p(aH)p(aDSM)m m =m m p(aH)p(aDSM)[p(aUCSF)m+mp(rUCSF)]
=m m (0.50)(0.30)
=m m 0.15.
Notemthatmwemcouldmhavemsolvedmpartm(a)mthemsamemway;mitmwouldmhavemrequiredmadd
ingmupmallmthemappropriatempossiblemmutuallymexclusivemevents.m Youmcanmcheckmthatmit
mgivesmthemsamemanswer masmabovem(butmnoticemhowmmuchmmoremtediousmitmis).
2
,2. Probabilitiesm ofm sequences.
AssumemthatmthemfourmbasesmA,mC,mT,mandmGmoccurmwithmequalmlikelihoodminmamDNAmsequ
ence
ofm ninem monomers.
(a) Whatm ism them probabilitym ofm findingm them sequencem AAATCGAGTm throughm rando
mmchance?
(b) Whatm ism them probabilitym ofm findingm them sequencem AAAAAAAAAm throughm rando
mmchance?
(c) Whatm ism them probabilitym ofm findingm anym sequencem thatm hasm fourm A’s,m twom T’s,m two
m G’s,mandmonemC,msuchmasmthatminm(a)?
(a) Eachmbasemoccursmwithmprobabilitym1/4.m ThemprobabilitymofmanmAminmpositionm1mism1/
4,mofmAminmpositionm2mism1/4,mofmAminmpositionm3mism1/4,mofmTm inmpositionm4mism1/4,man
dmsomon.mTheremarem9mbases.m Themprobabilitymofmthismspecificmsequencemism(1/4)9m=m
3.8m×m10−6.
(b) Samem answerm asm (a)m above.
(c) Eachmspecificmsequencemhasmthemprobabilitymgivenmabove,mbutminmthismcasemtheremaremma
nympossiblem sequencesm whichm satisfym them requirementm thatmwemhavem 4mA’s,m2mTm’s,m2mG’
s,mandm1
C.m Howmmanymaremthere?m Wemstartmasmwemhavemdonembefore,mbymassumingmallmninemobj
ectsmaremdistinguishable.m Theremarem9!m arrangementsmofmninemdistinguishablem objectsmi
nmam linearmsequence.m (Themfirstmonemcanmbeminmanymofmninemplaces,mthemsecondminmanym
ofmthemremainingmeightmplaces,mandmsomon.)m Butmwemcan’tmdistinguishmthemfourmA’s,msom
wemhavemovercountedm bym am factorm ofm 4!,m andm mustm dividem thism out.m Wem can’tm disting
uishm them twom Tm’s,msomwemhavemovercountedmbym2!,mandmmustmalsomdividemthismout.m A
ndmsomon.m Somthemprobabilitymofmhavingmthismcompositionmis
"m #m m mm 9
9! 1
=m 0.014
4!2!2!1! 4 .
3
, 3. Them probabilitym ofm am sequencem (givenm am composition).
Amscientistmhasmconstructedmamsecretmpeptidemtomcarrymammessage.m Youmknowmonlymthemcomp
osi-
tionmofmthempeptide,mwhichmismsixmaminomacidsmlong.m ItmcontainsmonemserinemS,monemthreoni
nemT,monemcysteinemC,monemargininemR,mandmtwomglutamatesmE.mWhatmismthemprobabilit
ymthatmthemsequencemSECRETmwillmoccurmbymchance?
ThemSm couldmbeminmanymonemofmthem6mpositionsmwithmequalmlikelihood.m Themprobabilitymtha
tmitmisminmpositionm1mism(1/6).mGivenmthatmSmisminmthemfirstmposition,mwemhavem2mE ' s mwhichm
couldmoccurminmanymofmthemremainingm5mpositions.m Themprobabilitymthatmonemofmthemmismin
mpositionm2mism(2/5).m Givenmthosemtwomlettersminmposition,mthemprobabilitymthatmthem1mCm ismi
nmthemnextmofmthem4mremainingmpositionsmism(1/4).mThemprobabilitymformthemRmism(1/3).m Fo
rmthemremainingmEmism(1/2),mandmformthemlastmTm ism(1/1),msomthemprobabilitymis
"m #−1
6!
(1/6)(2/5)(1/4)(1/3)(1/2)m =m 1/360m = .
1!2!1!1!
4. Combiningm independentm probabilities.
Youm havem am fairm six-
sidedm die.m Youm wantm tom rollm itm enoughm timesm tom ensurem thatm am 2m occursm at
leastmonce.m Whatmnumbermofmrollsmkmismrequiredmtomensuremthatmthemprobabilitymismatmleastm2
/3mthatmatmleastmonem2mwillmappear?
5
q = m =mprobabilitymthatmam2mdoesmnotmappearmonm thatmroll.
6
qk =m probabilitymthatmam2mdoesmnotmappearmonmkmINDEPENDENTm rolls.
Pm(k)m =m 1m−mqkm=mprobabilitymthatmatmleastmonem2mappearsmonmkmrolls.
For
m m m
2m 2m 1m 1
Pm(k)m m ≥ ,m1m−mqkm≥m =⇒m qkm ≤m =⇒m kmlnmqm≤mln
3 3 3 3
ln(1/3)
=⇒m k ≥ =m 6.03
ln(5/6)
Approximatelym sixmormmoremrollsmwillmensuremwithmprobabilitymPm ≥m2/3mthatmam2mwillmappear.
4
