2/25/24,s8:02sPM Solutionsmanualsforsintroductionstosprobability
Introduction to Probability 2n
s s s
d Editions
Problem Solutions s
(lastsupdated:s 9/26/17)
cs DimitrisP.sBertsekassandsJohnsN.sTsitsiklis
Massachusettss Institutes ofs Technology
WWW ssitesforsbooksinformationsandsorderss http://www.athena
sc.com
Athenas Scientific,s Belmont,s Massachusetts
1
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CsHsAsPsTsEsRs 1
Solutions tos Problems 1.1.s Wes have
As=s{2,s4,s6}, Bs=s{4,s5,s6},
sosAs∪sBs=s{2,s4,s5,s6},s
(As∪sB)cs=s{1,s3}.
ands Onsthesothershand,
Acs∩sBcs=s{1,s3,s5}s∩s{1,s2,s3}s=s{1,s3}.
Similarly,sweshavesAs∩sBs=s{4,s6},sand
(As∩sB)cs =s{1,s2,s3,s5}.
Ons thes others hand,
Acs∪sBcs=s{1,s3,s5}s∪s{1,s2,s3}s=s{1,s2,s3,s5}.
Solutions tos Problems 1.2.s (a)s Bys usings as Venns diagrams its cans bes seens thats fo
rs anys setssSs andsTs,sweshave
Ss=s(Ss∩sTs)s∪s(Ss∩sTsc).
(Alternatively,sargues thats anys xs musts belongs tos eithersTs ors tos Tsc,s sos xs belon
gss tosSs ifsands onlysifsitsbelongsstosSs∩sTs orstosSs∩sTsc.)s Applysthissequality
swithsSs =sAcs ands Ts =sB,stosobtainsthesfirstsrelation
Acs =s(Acs∩sB)s∪s(Acs∩sBc).
InterchangesthesrolessofsAsandsBstosobtainsthessecondsrelation.
(b) Bys Des Morgan’ss law,s wes have
(As∩sB)cs =sAcs∪sBc,
ands bys usings thes equalitiess ofs parts (a),s wes obtain
s
(A∩B)cs =s (Ac∩B)∪(Ac∩Bc) ∪ (A∩Bc)∪(Ac∩Bc) =s (Ac∩B)∪(Ac∩Bc)∪(A∩Bc).
(c) Wes haves As=s{1,s3,s5}sands Bs =s{1,s2,s3},s sos As∩sBs =s{1,s3}.s Therefore,
(As∩sB)cs =s{2,s4,s5,s6},
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and
Acs∩sBs=s{2}, Acs∩sBcs=s{4,s6}, As∩sBcs=s{5}.
Thus,s thes equalitys ofs parts (b)s iss verified.
Solutions tos Problems 1.5.s Lets Gs ands Cs bes thes eventss thats thes chosens stud
ents iss as geniuss ands as chocolates lover,s respectively.s Wes haves P(G)s =s 0.6,s P(
C)s =s 0.7,s and
P(Gs∩sC)s=s0.4.s WesaresinterestedsinsP(Gcs∩sCc),swhichsissobtainedswithsthesfollowing
calculation:
s
P(Gcs∩Cc)s =s 1−P(G∪C)s =s 1− P(G)+P(C)−P(G∩C) =s 1−(0.6+0.7−0.4)s =s 0.1.
SolutionstosProblems1.6.s Wesfirstsdeterminesthesprobabilitiessofsthessixspos
sibles outcomes.s Lets as =s P({1})s =s P({3})s =s P({5})s ands bs =s P({2})s =s P({4
})s =s P({6}).
Wesaresgivensthatsbs=s2a.s Bysthesadditivitysandsnormalizationsaxioms,s1s=s3as+
s3bs=s 3as+s6as=s9a.s Thus,s as=s1/9,s bs=s2/9,s ands P({1,s2,s3})s=s4/9.
SolutionstosProblems1.7.s Thesoutcomesofsthissexperimentscansbesanysfinitessequ
ences ofsthes forms (a1,sa2,s.s.s.s,sa n),s wheres ns iss ans arbitrarys positives integer,s a1
,sa2,s.s.s.s,san−1
belongs tos {1,s3},s ands a ns belongss tos {2,s4}.s Ins addition,s theres ares possibles outcomes
ins whichs ans evens numbers iss nevers obtained.s Suchs outcomess ares infinites seque
ncess (a1,sa2,s.s.s.),swithseachselementsinsthessequencesbelongingstos{1,s3}.s Thessam
plesspaces consistssofsallspossiblesoutcomessofsthesabovestwostypes.
SolutionstosProblems1.8.s Letspis besthesprobabilitysofswinningsagainststhesoppo
nents playedsinsthesithsturn.sThen,syouswillswinsthestournamentsifsyouswinsagai
nststhes2nds players(probabilitysp2 )sandsalsosyouswinsagainstsatsleastsonesofsth
estwos othersplayers
[probabilitysp1 s+s(1s−sp1)p3s =sp1s+sp3 s−sp1p3 ].s Thus,sthesprobabilitysofswinnin
gsthes tournamentsis
p2 (p1 s +sp3s−sp1p3).
Thes orders (1,s2,s3)s iss optimals ifs ands onlys ifs thes aboves probabilitys iss nos lesss th
ans thes probabilitiesscorrespondingstosthestwosalternativesorders,si.e.,
p2 (p1 s+sp3 s−sp1p3 )s≥sp1(p2 s+sp3s−sp2p3
),s p2 (p1 s +sp3s−sp1p3 )s≥sp3(p2 s +sp1 s−sp
2p1).
Itscansbesseensthatsthesfirstsinequalitysabovesissequivalentstosp2s ≥sp1,swhilesthessec
onds inequalitysabovesissequivalentstosp2s ≥sp3.
Solutions tos Problems1.9.s (a)sSincesΩs=s∪
i=1 s Si ,sweshave
n
n
[
As=s s (As∩sSi),
i=1
whilesthessetssAs∩sSis aresdisjoint.s Thesresultsfollowssbysusingsthesadditivitysaxiom.
(b)s Thes eventss Bs∩sC c,s Bcs∩sC,s Bs∩sC,s ands Bcs∩sCcs forms as partitions ofs Ω,s so
s bys parts (a),sweshave
P(A)s=sP(As∩sBs∩sCc)s+sP(As∩sBcs∩sC)s+sP(As∩sBs∩sC)s+sP(As∩sBcs∩sCc).s s (1)
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Thes events As∩sBs cans bes writtens ass thes unions ofs twos disjoints eventss ass follows:
As∩sBs =s(As∩sBs∩sC)s∪s(As∩sBs∩sCc),
sosthats
P(As∩sB)s=sP(As∩sBs∩sC)s+sP(As∩sBs∩sCc). (2)
Similarly,
P(As∩sC)s=sP(As∩sBs∩sC)s+sP(As∩sBcs∩sC). (3)
Combinings Eqs.s (1)-(3),s wes obtains thes desireds result.
SolutionstosProblems1.10.s SincestheseventssAs∩sBcs andsAcs∩sBsaresdisjoint,
swes havesusingsthesadditivitysaxiomsrepeatedly,
s
P (A∩Bc )∪(A c s∩B) =s P(A∩Bc )+P(A c s∩B)s =s P(A)−P(A∩B)s+P(B)−P(A∩B).
Solutions tos Problems 1.14.s (a)sEachspossiblesoutcomeshassprobabilitys1/36.s The
res ares6spossiblesoutcomessthatsaresdoubles,ssosthesprobabilitysofsdoublessiss6/
36s=s1/6.
(b) Thesconditionings events(sums iss4s ors less)s consistssofsthes 6s outcomes
(1,s1),s(1,s2),s(1,s3),s(2,s1),s(2,s2),s(3,s1) ,
2s ofs whichs ares doubles,s sos thes conditionals probabilitys ofs doubless iss 2/6s=s1/3.
(c) Theresares11spossiblesoutcomesswithsatsleastsones6,snamely,s(6,s6),s(6,si),sand
s(i,s6),s fors is=s1,s2,s.s.s.s,s5.s Thus,s thes probabilitys thats ats leasts ones dies iss as 6
s iss 11/36.
(d) Theresares30spossiblesoutcomesswheresthesdiceslandsonsdifferentsnumbers.s Out
sofs these,stheresares10soutcomessinswhichsatsleastsonesofsthesrollssissas6.s Thus,sth
esdesireds conditionalsprobabilitysiss10/30s=s1/3.
Solutions tos Problems 1.15.s Lets As bes thes events thats thes firsts tosss iss as hea
ds ands letsBsbestheseventsthatsthessecondstosssissashead.s Wesmustscomparesth
esconditionals probabilitiess P(As∩sBs|sA)s ands P(As∩sBs|sAs∪sB).s Wes have
s
P (As∩sB)s∩sA P(As∩sB)s
P(As∩sBs|sA)s=s =s ,
P(A) P(A)
and s
P (As∩sB)s∩s(As∪sB) P(As∩sB)s
P(As∩sBs|sAs∪sB)s=s =s .
P(As∪sB) P(As∪sB)
Sinces P(As∪sB)s ≥sP(A),s thes firsts conditionals probabilitys aboves iss ats leasts ass lar
ge,s sos Alices iss right,s regardlesss ofs whethers thes coins iss fairs ors not.s Ins thes cases
wheres thes coin
iss fair,s thats is,s ifs alls fours outcomess HH,s HTs,s TsH,s TsTs ares equallys likely,s wes have
P(As∩sB)s =s 1/4s 1s P(As∩sB)s =s 1/4s 1s
=s , =s .
P(A) 1/2s s P(As∪sB 3/4s s 3
2 )
As generalizations ofs Alice’ss reasonings iss thats ifs A′,s B′,s ands C′sares events
s suchs thats B′s ⊂s C′s ands A′s∩sB′s =s A′s∩sC′s (fors examples ifs A′s ⊂s B′s ⊂s C′),
s thens thes event
4
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Introduction to Probability 2n
s s s
d Editions
Problem Solutions s
(lastsupdated:s 9/26/17)
cs DimitrisP.sBertsekassandsJohnsN.sTsitsiklis
Massachusettss Institutes ofs Technology
WWW ssitesforsbooksinformationsandsorderss http://www.athena
sc.com
Athenas Scientific,s Belmont,s Massachusetts
1
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CsHsAsPsTsEsRs 1
Solutions tos Problems 1.1.s Wes have
As=s{2,s4,s6}, Bs=s{4,s5,s6},
sosAs∪sBs=s{2,s4,s5,s6},s
(As∪sB)cs=s{1,s3}.
ands Onsthesothershand,
Acs∩sBcs=s{1,s3,s5}s∩s{1,s2,s3}s=s{1,s3}.
Similarly,sweshavesAs∩sBs=s{4,s6},sand
(As∩sB)cs =s{1,s2,s3,s5}.
Ons thes others hand,
Acs∪sBcs=s{1,s3,s5}s∪s{1,s2,s3}s=s{1,s2,s3,s5}.
Solutions tos Problems 1.2.s (a)s Bys usings as Venns diagrams its cans bes seens thats fo
rs anys setssSs andsTs,sweshave
Ss=s(Ss∩sTs)s∪s(Ss∩sTsc).
(Alternatively,sargues thats anys xs musts belongs tos eithersTs ors tos Tsc,s sos xs belon
gss tosSs ifsands onlysifsitsbelongsstosSs∩sTs orstosSs∩sTsc.)s Applysthissequality
swithsSs =sAcs ands Ts =sB,stosobtainsthesfirstsrelation
Acs =s(Acs∩sB)s∪s(Acs∩sBc).
InterchangesthesrolessofsAsandsBstosobtainsthessecondsrelation.
(b) Bys Des Morgan’ss law,s wes have
(As∩sB)cs =sAcs∪sBc,
ands bys usings thes equalitiess ofs parts (a),s wes obtain
s
(A∩B)cs =s (Ac∩B)∪(Ac∩Bc) ∪ (A∩Bc)∪(Ac∩Bc) =s (Ac∩B)∪(Ac∩Bc)∪(A∩Bc).
(c) Wes haves As=s{1,s3,s5}sands Bs =s{1,s2,s3},s sos As∩sBs =s{1,s3}.s Therefore,
(As∩sB)cs =s{2,s4,s5,s6},
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and
Acs∩sBs=s{2}, Acs∩sBcs=s{4,s6}, As∩sBcs=s{5}.
Thus,s thes equalitys ofs parts (b)s iss verified.
Solutions tos Problems 1.5.s Lets Gs ands Cs bes thes eventss thats thes chosens stud
ents iss as geniuss ands as chocolates lover,s respectively.s Wes haves P(G)s =s 0.6,s P(
C)s =s 0.7,s and
P(Gs∩sC)s=s0.4.s WesaresinterestedsinsP(Gcs∩sCc),swhichsissobtainedswithsthesfollowing
calculation:
s
P(Gcs∩Cc)s =s 1−P(G∪C)s =s 1− P(G)+P(C)−P(G∩C) =s 1−(0.6+0.7−0.4)s =s 0.1.
SolutionstosProblems1.6.s Wesfirstsdeterminesthesprobabilitiessofsthessixspos
sibles outcomes.s Lets as =s P({1})s =s P({3})s =s P({5})s ands bs =s P({2})s =s P({4
})s =s P({6}).
Wesaresgivensthatsbs=s2a.s Bysthesadditivitysandsnormalizationsaxioms,s1s=s3as+
s3bs=s 3as+s6as=s9a.s Thus,s as=s1/9,s bs=s2/9,s ands P({1,s2,s3})s=s4/9.
SolutionstosProblems1.7.s Thesoutcomesofsthissexperimentscansbesanysfinitessequ
ences ofsthes forms (a1,sa2,s.s.s.s,sa n),s wheres ns iss ans arbitrarys positives integer,s a1
,sa2,s.s.s.s,san−1
belongs tos {1,s3},s ands a ns belongss tos {2,s4}.s Ins addition,s theres ares possibles outcomes
ins whichs ans evens numbers iss nevers obtained.s Suchs outcomess ares infinites seque
ncess (a1,sa2,s.s.s.),swithseachselementsinsthessequencesbelongingstos{1,s3}.s Thessam
plesspaces consistssofsallspossiblesoutcomessofsthesabovestwostypes.
SolutionstosProblems1.8.s Letspis besthesprobabilitysofswinningsagainststhesoppo
nents playedsinsthesithsturn.sThen,syouswillswinsthestournamentsifsyouswinsagai
nststhes2nds players(probabilitysp2 )sandsalsosyouswinsagainstsatsleastsonesofsth
estwos othersplayers
[probabilitysp1 s+s(1s−sp1)p3s =sp1s+sp3 s−sp1p3 ].s Thus,sthesprobabilitysofswinnin
gsthes tournamentsis
p2 (p1 s +sp3s−sp1p3).
Thes orders (1,s2,s3)s iss optimals ifs ands onlys ifs thes aboves probabilitys iss nos lesss th
ans thes probabilitiesscorrespondingstosthestwosalternativesorders,si.e.,
p2 (p1 s+sp3 s−sp1p3 )s≥sp1(p2 s+sp3s−sp2p3
),s p2 (p1 s +sp3s−sp1p3 )s≥sp3(p2 s +sp1 s−sp
2p1).
Itscansbesseensthatsthesfirstsinequalitysabovesissequivalentstosp2s ≥sp1,swhilesthessec
onds inequalitysabovesissequivalentstosp2s ≥sp3.
Solutions tos Problems1.9.s (a)sSincesΩs=s∪
i=1 s Si ,sweshave
n
n
[
As=s s (As∩sSi),
i=1
whilesthessetssAs∩sSis aresdisjoint.s Thesresultsfollowssbysusingsthesadditivitysaxiom.
(b)s Thes eventss Bs∩sC c,s Bcs∩sC,s Bs∩sC,s ands Bcs∩sCcs forms as partitions ofs Ω,s so
s bys parts (a),sweshave
P(A)s=sP(As∩sBs∩sCc)s+sP(As∩sBcs∩sC)s+sP(As∩sBs∩sC)s+sP(As∩sBcs∩sCc).s s (1)
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Thes events As∩sBs cans bes writtens ass thes unions ofs twos disjoints eventss ass follows:
As∩sBs =s(As∩sBs∩sC)s∪s(As∩sBs∩sCc),
sosthats
P(As∩sB)s=sP(As∩sBs∩sC)s+sP(As∩sBs∩sCc). (2)
Similarly,
P(As∩sC)s=sP(As∩sBs∩sC)s+sP(As∩sBcs∩sC). (3)
Combinings Eqs.s (1)-(3),s wes obtains thes desireds result.
SolutionstosProblems1.10.s SincestheseventssAs∩sBcs andsAcs∩sBsaresdisjoint,
swes havesusingsthesadditivitysaxiomsrepeatedly,
s
P (A∩Bc )∪(A c s∩B) =s P(A∩Bc )+P(A c s∩B)s =s P(A)−P(A∩B)s+P(B)−P(A∩B).
Solutions tos Problems 1.14.s (a)sEachspossiblesoutcomeshassprobabilitys1/36.s The
res ares6spossiblesoutcomessthatsaresdoubles,ssosthesprobabilitysofsdoublessiss6/
36s=s1/6.
(b) Thesconditionings events(sums iss4s ors less)s consistssofsthes 6s outcomes
(1,s1),s(1,s2),s(1,s3),s(2,s1),s(2,s2),s(3,s1) ,
2s ofs whichs ares doubles,s sos thes conditionals probabilitys ofs doubless iss 2/6s=s1/3.
(c) Theresares11spossiblesoutcomesswithsatsleastsones6,snamely,s(6,s6),s(6,si),sand
s(i,s6),s fors is=s1,s2,s.s.s.s,s5.s Thus,s thes probabilitys thats ats leasts ones dies iss as 6
s iss 11/36.
(d) Theresares30spossiblesoutcomesswheresthesdiceslandsonsdifferentsnumbers.s Out
sofs these,stheresares10soutcomessinswhichsatsleastsonesofsthesrollssissas6.s Thus,sth
esdesireds conditionalsprobabilitysiss10/30s=s1/3.
Solutions tos Problems 1.15.s Lets As bes thes events thats thes firsts tosss iss as hea
ds ands letsBsbestheseventsthatsthessecondstosssissashead.s Wesmustscomparesth
esconditionals probabilitiess P(As∩sBs|sA)s ands P(As∩sBs|sAs∪sB).s Wes have
s
P (As∩sB)s∩sA P(As∩sB)s
P(As∩sBs|sA)s=s =s ,
P(A) P(A)
and s
P (As∩sB)s∩s(As∪sB) P(As∩sB)s
P(As∩sBs|sAs∪sB)s=s =s .
P(As∪sB) P(As∪sB)
Sinces P(As∪sB)s ≥sP(A),s thes firsts conditionals probabilitys aboves iss ats leasts ass lar
ge,s sos Alices iss right,s regardlesss ofs whethers thes coins iss fairs ors not.s Ins thes cases
wheres thes coin
iss fair,s thats is,s ifs alls fours outcomess HH,s HTs,s TsH,s TsTs ares equallys likely,s wes have
P(As∩sB)s =s 1/4s 1s P(As∩sB)s =s 1/4s 1s
=s , =s .
P(A) 1/2s s P(As∪sB 3/4s s 3
2 )
As generalizations ofs Alice’ss reasonings iss thats ifs A′,s B′,s ands C′sares events
s suchs thats B′s ⊂s C′s ands A′s∩sB′s =s A′s∩sC′s (fors examples ifs A′s ⊂s B′s ⊂s C′),
s thens thes event
4
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