15/06/2023o01:49 Solutionomanualoforointroductionotooprobabilit
y
Introduction to Probability 2n
o o o
d Editiono
Problem Solutions o
(lastoupdated:o 9/26/17)
co DimitrioP.oBertsekasoandoJohnoN.oTsitsiklis
Massachusettso Instituteo ofo Technology
WWW ositeoforobookoinformationoandoorder
so http://www.athenasc.com
Athenao Scientific,o Belmont,o Massachusetts
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CoHoAoPoToEoRo 1
Solutiono too Problemo 1.1.o Weo have
Ao=o{2,o4,o6}, Bo=o{4,o5,o6},
sooAo∪oBo=o{2,o4,o5,o6},
(Ao∪oB)co =o{1,o3}.
oando Onotheootheroha
nd,
Aco∩oBco=o{1,o3,o5}o∩o{1,o2,o3}o=o{1,o3}.
Similarly,oweohaveoAo∩oBo=o{4,o6},oand
(Ao∩oB)co=o{1,o2,o3,o5}.
Ono theo othero hand,
Aco∪oBco=o{1,o3,o5}o∪o{1,o2,o3}o=o{1,o2,o3,o5}.
SolutionotooProblemo1.2.o (a)oByousingoaoVennodiagramoitocanobeoseenotha
toforoanyo setsoSoandoTo,oweohave
So=o(So∩oTo)o∪o(So∩oTc).
(Alternatively,oargueothatoanyoxomustobelongotooeitheroTo orotooTc,osooxobel
ongsotooSo ifoandoonlyoifoitobelongsotooSo∩oTo orotooSo∩oTc.)o Applyothisoe
qualityowithoSo=oA coando To =oB,otooobtainotheofirstorelation
Aco=o(Aco∩oB)o∪o(Aco∩oBc).
InterchangeotheorolesoofoAoandoBotooobtainotheosecondorelation.
(b) Byo Deo Morgan’so law,o weo have
(Ao∩oB)co=oA co∪oBc,
ando byo usingo theo equalitieso ofo parto (a),o weo obtain
o
(A∩B)co =o (A c∩B)∪(A c∩B c) ∪ (A∩B c)∪(A c∩B c) =o (Ac∩B)∪(Ac∩Bc)∪(A∩Bc).
(c) Weo haveo Ao=o{1,o3,o5}oando Bo =o{1,o2,o3},o soo Ao∩oBo =o{1,o3}.o Therefore,
(Ao∩oB)co=o{2,o4,o5,o6},
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and
Aco∩oBo=o{2}, Aco∩oBco=o{4,o6}, Ao∩oBco=o{5}.
Thus,o theo equalityo ofo parto (b)o iso verified.
Solutiono too Problemo 1.5.o Leto Go ando Co beo theo eventso thato theo choseno st
udento iso ao geniuso ando ao chocolateo lover,o respectively.o Weo haveo P(G)o =o 0.
6,o P(C)o =o 0.7,o and
P(Go∩oC)o=o0.4.o WeoareointerestedoinoP(Gco∩oCc),owhichoisoobtainedowithotheofollowing
calculation:
o
P(Gco∩C c)o =o 1−P(G∪C)o =o 1− P(G)+P(C)−P(G∩C) =o 1−(0.6+0.7−0.4)o =o 0.1.
SolutionotooProblemo1.6.o Weofirstodetermineotheoprobabilitiesoofotheosixop
ossibleo outcomes.o Leto ao =o P({1})o =o P({3})o =o P({5})o ando bo =o P({2})o =
o P({4})o =o P({6}).
Weoareogivenothatobo=o2a.oByotheoadditivityoandonormalizationoaxioms,o1o=
o3ao+o3bo=o 3ao+o6ao=o9a.o Thus,o ao=o1/9,o bo=o2/9,o ando P({1,o2,o3})o=o4/
9.
SolutionotooProblemo1.7.o Theooutcomeoofothisoexperimentocanobeoanyofiniteos
equenceo ofotheoformo(a 1,oa2,o.o.o.o,oa n),owhereonoisoanoarbitraryopositiveoint
eger,oa 1,oa2,o.o.o.o,oa n—
1o belongotoo{1,o3},oandoa no belongsotoo{2,o4}.o Inoaddition,othereoareopossi
bleooutcomes
ino whicho ano eveno numbero iso nevero obtained.o Sucho outcomeso areo infiniteo sequences
(a1,oa2 ,o.o.o.),owithoeachoelementoinotheosequenceobelongingotoo{1,o3}.o Theosam
pleospaceo consistsoofoallopossibleooutcomesoofotheoaboveotwootypes.
SolutionotooProblemo1.8.o Letopio beotheoprobabilityoofowinningoagainstotheoop
ponento playedoinotheoithoturn.oThen,oyouowillowinotheotournamentoifoyouowi
noagainstotheo2ndo playero(probabilityop2 )oandoalsooyouowinoagainstoatoleasto
oneoofotheotwoootheroplayers
[probabilityop1 o+o(1o−op1 )p3 o =op1 o+op3 o−op1 p3 ].o Thus,otheoprobabilityoofowin
ningotheo tournamentois
p2(p1 o+op3 o−op1p3).
Theo ordero (1,o2,o3)o iso optimalo ifo ando onlyo ifo theo aboveo probabilityo iso noo less
o thano theo probabilitiesocorrespondingotootheotwooalternativeoorders,oi.e.,
p2 (p1 o +op 3 o−op1p3)o≥op1(p 2o +op 3o−op
2p3 ),o p2 (p 1 o +op 3o −op1p3 )o≥op3(p 2 o +op
1o −op2p1).
Itocanobeoseenothatotheofirstoinequalityoaboveoisoequivalentotoop2o ≥op1,owhileothe
osecondo inequalityoaboveoisoequivalentotoop2o ≥op3.
i=1 ∪ o Si,o weo have
Solutiono too Problemo 1.9.o (a)o Sinceo Ωo=o n
n
[
Ao=o o (Ao∩oSi),
i=1
whileotheosetsoAo∩oSio areodisjoint.oTheoresultofollowsobyousingotheoadditivityoaxiom.
(b)oTheoeventsoBo∩oCc,oBco∩oC,oBo∩oC,oandoBco∩oCcoformoaopartitionoofo
Ω,osoobyoparto (a),oweohave
P(A)o=oP(Ao∩oBo∩oCc)o+oP(Ao∩oBco∩oC)o+oP(Ao∩oBo∩oC)o+oP(Ao∩oBco∩oCc).o o (1)
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TheoeventoAo∩oBocanobeowrittenoasotheounionoofotwoodisjointoeventsoasofollows:
Ao∩oBo=o(Ao∩oBo∩oC)o∪o(Ao∩oBo∩oCc),
soothato
P(Ao∩oB)o=oP(Ao∩oBo∩oC)o+oP(Ao∩oBo∩oCc). (2)
Similarly,
P(Ao∩oC)o=oP(Ao∩oBo∩oC)o+oP(Ao∩oBco∩oC). (3)
Combiningo Eqs.o (1)-(3),o weo obtaino theo desiredo result.
Solutiono too Problemo 1.10.o SinceotheoeventsoAo∩oBco andoA co∩oBo areodisj
oint,oweo haveousingotheoadditivityoaxiomorepeatedly,
o
P (A∩Bc)∪(A co∩B) =o P(A∩Bc)+P(A co∩B)o =o P(A)−P(A∩B)o+P(B)−P(A∩B).
SolutionotooProblemo1.14.o (a)oEachopossibleooutcomeohasoprobabilityo1/36.
oThereo areo6opossibleooutcomesothatoareodoubles,osootheoprobabilityoofodoub
lesoiso6/36o=o1/6.
(b) Theoconditioningo evento (sumo iso 4o oro less)o consistso ofo theo 6o outcomes
(1,o1),o(1,o2),o(1,o3),o(2,o1),o(2,o2),o(3,o1)o ,
2oofowhichoareodoubles,osootheoconditionaloprobabilityoofodoublesoiso2/6o=o1/3.
(c) Thereoareo11opossibleooutcomesowithoatoleastooneo6,onamely,o(6,o6),o(6,oi),o
ando(i,o6),o foro io=o1,o2,o.o.o.o,o5.o Thus,o theo probabilityo thato ato leasto oneo die
o iso ao 6o iso 11/36.
(d) Thereoareo30opossibleooutcomesowhereotheodiceolandoonodifferentonumbers.
oOutoofo these,othereoareo10ooutcomesoinowhichoatoleastooneoofotheorollsoisoao
6.oThus,otheodesiredo conditionaloprobabilityoiso10/30o=o1/3.
Solutiono too Problemo 1.15.o Leto Ao beo theo evento thato theo firsto tosso iso ao
heado ando letoBobeotheoeventothatotheosecondotossoisoaohead.o Weomustocom
pareotheoconditionalo probabilitieso P(Ao∩oBo|oA)o ando P(Ao∩oBo|oAo∪oB).o W
eo have
o
P (Ao∩oB)o∩oA P(Ao∩oB)o
P(Ao∩oBo|oA)o=o =o ,
P(A) P(A)
and o
P (Ao∩oB)o∩o(Ao∪oB) P(Ao∩oB)o
P(Ao∩oBo|oAo∪oB)o=o =o .
P(Ao∪oB P(Ao∪oB
) )
SinceoP(Ao∪oB)o≥oP(A),otheofirstoconditionaloprobabilityoaboveoisoatoleastoasol
arge,osoo Aliceoisoright,oregardlessoofowhetherotheocoinoisofairooronot.o Inotheo
caseowhereotheocoino iso fair,o thato is,o ifo allo fouro outcomeso HH,o HTo,o TH,o TT
o areo equallyo likely,o weo have
P(Ao∩oB)o 1/4o 1o P(Ao∩oB)o 1/4o 1o
=o =o , =o =o .
P(A) 1/2o P(Ao∪oB 3/4o o 3
o 2 )
Ao generalizationo ofo Alice’so reasoningo iso thato ifo A′,o B′ ,o ando C ′ oareo eve
ntso sucho thato B′ o ⊂o C′o ando A′ o∩oB′o =o A′ o∩oC′o (foro exampleo ifo A′ o⊂o B′ o
⊂o C′),o theno theo event
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y
Introduction to Probability 2n
o o o
d Editiono
Problem Solutions o
(lastoupdated:o 9/26/17)
co DimitrioP.oBertsekasoandoJohnoN.oTsitsiklis
Massachusettso Instituteo ofo Technology
WWW ositeoforobookoinformationoandoorder
so http://www.athenasc.com
Athenao Scientific,o Belmont,o Massachusetts
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CoHoAoPoToEoRo 1
Solutiono too Problemo 1.1.o Weo have
Ao=o{2,o4,o6}, Bo=o{4,o5,o6},
sooAo∪oBo=o{2,o4,o5,o6},
(Ao∪oB)co =o{1,o3}.
oando Onotheootheroha
nd,
Aco∩oBco=o{1,o3,o5}o∩o{1,o2,o3}o=o{1,o3}.
Similarly,oweohaveoAo∩oBo=o{4,o6},oand
(Ao∩oB)co=o{1,o2,o3,o5}.
Ono theo othero hand,
Aco∪oBco=o{1,o3,o5}o∪o{1,o2,o3}o=o{1,o2,o3,o5}.
SolutionotooProblemo1.2.o (a)oByousingoaoVennodiagramoitocanobeoseenotha
toforoanyo setsoSoandoTo,oweohave
So=o(So∩oTo)o∪o(So∩oTc).
(Alternatively,oargueothatoanyoxomustobelongotooeitheroTo orotooTc,osooxobel
ongsotooSo ifoandoonlyoifoitobelongsotooSo∩oTo orotooSo∩oTc.)o Applyothisoe
qualityowithoSo=oA coando To =oB,otooobtainotheofirstorelation
Aco=o(Aco∩oB)o∪o(Aco∩oBc).
InterchangeotheorolesoofoAoandoBotooobtainotheosecondorelation.
(b) Byo Deo Morgan’so law,o weo have
(Ao∩oB)co=oA co∪oBc,
ando byo usingo theo equalitieso ofo parto (a),o weo obtain
o
(A∩B)co =o (A c∩B)∪(A c∩B c) ∪ (A∩B c)∪(A c∩B c) =o (Ac∩B)∪(Ac∩Bc)∪(A∩Bc).
(c) Weo haveo Ao=o{1,o3,o5}oando Bo =o{1,o2,o3},o soo Ao∩oBo =o{1,o3}.o Therefore,
(Ao∩oB)co=o{2,o4,o5,o6},
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and
Aco∩oBo=o{2}, Aco∩oBco=o{4,o6}, Ao∩oBco=o{5}.
Thus,o theo equalityo ofo parto (b)o iso verified.
Solutiono too Problemo 1.5.o Leto Go ando Co beo theo eventso thato theo choseno st
udento iso ao geniuso ando ao chocolateo lover,o respectively.o Weo haveo P(G)o =o 0.
6,o P(C)o =o 0.7,o and
P(Go∩oC)o=o0.4.o WeoareointerestedoinoP(Gco∩oCc),owhichoisoobtainedowithotheofollowing
calculation:
o
P(Gco∩C c)o =o 1−P(G∪C)o =o 1− P(G)+P(C)−P(G∩C) =o 1−(0.6+0.7−0.4)o =o 0.1.
SolutionotooProblemo1.6.o Weofirstodetermineotheoprobabilitiesoofotheosixop
ossibleo outcomes.o Leto ao =o P({1})o =o P({3})o =o P({5})o ando bo =o P({2})o =
o P({4})o =o P({6}).
Weoareogivenothatobo=o2a.oByotheoadditivityoandonormalizationoaxioms,o1o=
o3ao+o3bo=o 3ao+o6ao=o9a.o Thus,o ao=o1/9,o bo=o2/9,o ando P({1,o2,o3})o=o4/
9.
SolutionotooProblemo1.7.o Theooutcomeoofothisoexperimentocanobeoanyofiniteos
equenceo ofotheoformo(a 1,oa2,o.o.o.o,oa n),owhereonoisoanoarbitraryopositiveoint
eger,oa 1,oa2,o.o.o.o,oa n—
1o belongotoo{1,o3},oandoa no belongsotoo{2,o4}.o Inoaddition,othereoareopossi
bleooutcomes
ino whicho ano eveno numbero iso nevero obtained.o Sucho outcomeso areo infiniteo sequences
(a1,oa2 ,o.o.o.),owithoeachoelementoinotheosequenceobelongingotoo{1,o3}.o Theosam
pleospaceo consistsoofoallopossibleooutcomesoofotheoaboveotwootypes.
SolutionotooProblemo1.8.o Letopio beotheoprobabilityoofowinningoagainstotheoop
ponento playedoinotheoithoturn.oThen,oyouowillowinotheotournamentoifoyouowi
noagainstotheo2ndo playero(probabilityop2 )oandoalsooyouowinoagainstoatoleasto
oneoofotheotwoootheroplayers
[probabilityop1 o+o(1o−op1 )p3 o =op1 o+op3 o−op1 p3 ].o Thus,otheoprobabilityoofowin
ningotheo tournamentois
p2(p1 o+op3 o−op1p3).
Theo ordero (1,o2,o3)o iso optimalo ifo ando onlyo ifo theo aboveo probabilityo iso noo less
o thano theo probabilitiesocorrespondingotootheotwooalternativeoorders,oi.e.,
p2 (p1 o +op 3 o−op1p3)o≥op1(p 2o +op 3o−op
2p3 ),o p2 (p 1 o +op 3o −op1p3 )o≥op3(p 2 o +op
1o −op2p1).
Itocanobeoseenothatotheofirstoinequalityoaboveoisoequivalentotoop2o ≥op1,owhileothe
osecondo inequalityoaboveoisoequivalentotoop2o ≥op3.
i=1 ∪ o Si,o weo have
Solutiono too Problemo 1.9.o (a)o Sinceo Ωo=o n
n
[
Ao=o o (Ao∩oSi),
i=1
whileotheosetsoAo∩oSio areodisjoint.oTheoresultofollowsobyousingotheoadditivityoaxiom.
(b)oTheoeventsoBo∩oCc,oBco∩oC,oBo∩oC,oandoBco∩oCcoformoaopartitionoofo
Ω,osoobyoparto (a),oweohave
P(A)o=oP(Ao∩oBo∩oCc)o+oP(Ao∩oBco∩oC)o+oP(Ao∩oBo∩oC)o+oP(Ao∩oBco∩oCc).o o (1)
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TheoeventoAo∩oBocanobeowrittenoasotheounionoofotwoodisjointoeventsoasofollows:
Ao∩oBo=o(Ao∩oBo∩oC)o∪o(Ao∩oBo∩oCc),
soothato
P(Ao∩oB)o=oP(Ao∩oBo∩oC)o+oP(Ao∩oBo∩oCc). (2)
Similarly,
P(Ao∩oC)o=oP(Ao∩oBo∩oC)o+oP(Ao∩oBco∩oC). (3)
Combiningo Eqs.o (1)-(3),o weo obtaino theo desiredo result.
Solutiono too Problemo 1.10.o SinceotheoeventsoAo∩oBco andoA co∩oBo areodisj
oint,oweo haveousingotheoadditivityoaxiomorepeatedly,
o
P (A∩Bc)∪(A co∩B) =o P(A∩Bc)+P(A co∩B)o =o P(A)−P(A∩B)o+P(B)−P(A∩B).
SolutionotooProblemo1.14.o (a)oEachopossibleooutcomeohasoprobabilityo1/36.
oThereo areo6opossibleooutcomesothatoareodoubles,osootheoprobabilityoofodoub
lesoiso6/36o=o1/6.
(b) Theoconditioningo evento (sumo iso 4o oro less)o consistso ofo theo 6o outcomes
(1,o1),o(1,o2),o(1,o3),o(2,o1),o(2,o2),o(3,o1)o ,
2oofowhichoareodoubles,osootheoconditionaloprobabilityoofodoublesoiso2/6o=o1/3.
(c) Thereoareo11opossibleooutcomesowithoatoleastooneo6,onamely,o(6,o6),o(6,oi),o
ando(i,o6),o foro io=o1,o2,o.o.o.o,o5.o Thus,o theo probabilityo thato ato leasto oneo die
o iso ao 6o iso 11/36.
(d) Thereoareo30opossibleooutcomesowhereotheodiceolandoonodifferentonumbers.
oOutoofo these,othereoareo10ooutcomesoinowhichoatoleastooneoofotheorollsoisoao
6.oThus,otheodesiredo conditionaloprobabilityoiso10/30o=o1/3.
Solutiono too Problemo 1.15.o Leto Ao beo theo evento thato theo firsto tosso iso ao
heado ando letoBobeotheoeventothatotheosecondotossoisoaohead.o Weomustocom
pareotheoconditionalo probabilitieso P(Ao∩oBo|oA)o ando P(Ao∩oBo|oAo∪oB).o W
eo have
o
P (Ao∩oB)o∩oA P(Ao∩oB)o
P(Ao∩oBo|oA)o=o =o ,
P(A) P(A)
and o
P (Ao∩oB)o∩o(Ao∪oB) P(Ao∩oB)o
P(Ao∩oBo|oAo∪oB)o=o =o .
P(Ao∪oB P(Ao∪oB
) )
SinceoP(Ao∪oB)o≥oP(A),otheofirstoconditionaloprobabilityoaboveoisoatoleastoasol
arge,osoo Aliceoisoright,oregardlessoofowhetherotheocoinoisofairooronot.o Inotheo
caseowhereotheocoino iso fair,o thato is,o ifo allo fouro outcomeso HH,o HTo,o TH,o TT
o areo equallyo likely,o weo have
P(Ao∩oB)o 1/4o 1o P(Ao∩oB)o 1/4o 1o
=o =o , =o =o .
P(A) 1/2o P(Ao∪oB 3/4o o 3
o 2 )
Ao generalizationo ofo Alice’so reasoningo iso thato ifo A′,o B′ ,o ando C ′ oareo eve
ntso sucho thato B′ o ⊂o C′o ando A′ o∩oB′o =o A′ o∩oC′o (foro exampleo ifo A′ o⊂o B′ o
⊂o C′),o theno theo event
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