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1、Chapter2:Probability2.1A={FF},B={MM},C={MF,FM,MM}.Then,A∩B=nullset,B∩C={MM},C∩B={MF,FM},A∪B={FF,MM},A∪C=S,B∪C=C.2.2a.A∩Bb.A∪Bc.A∪Bd.(A∩B)∪(A∩B)2.32.4a.b.Chapter2:Probability9Instructor’sSolutionsManual2.5a.(.A∩B)∪(A∩B)=A∩(B∪B)=A∩S=Ab.B∪(.A∩B)=(B∩A)∪(B∩B)=(B∩A)=Ac.(A
2、∩B)∩(A∩B)=A∩(B∩B)=nullset.Theresultfollowsfromparta.d.B∩(A∩B)=A∩(B∩B)=nullset.Theresultfollowsfrompartb.2.6a.36+6=42b.33c.182.7A={twomales}={M1,M2),(M1,M3),(M2,M3)}B={atleastonefemale}={(M1,W1),(M2,W1),(M3,W1),(M1,W2),(M2,W2),(M3,W2),{W1,W2)}B={nofemales}=AA∪B=SA∩B=
3、nullA∩B=A2.8A={(1,2),(2,2),(3,2),(4,2),(5,2),(6,2),(1,4),(2,4),(3,4),(4,4),(5,4),(6,4),(1,6),(2,6),(3,6),(4,6),(5,6),(6,6)}C={(2,2),(2,4),(2,6),(4,2),(4,4),(4,6),(6,2),(6,4),(6,6)}A∩B={(2,2),(4,2),(6,2),(2,4),(4,4),(6,4),(2,6),(4,6),(6,6)}A∩B={(1,2),(3,2),(5,2),(1,4
4、),(3,4),(5,4),(1,6),(3,6),(5,6)}A∪B=everythingbut{(1,2),(1,4),(1,6),(3,2),(3,4),(3,6),(5,2),(5,4),(5,6)}A∩C=A2.9S={A+,B+,AB+,O+,A-,B-,AB-,O-}2.10a.S={A,B,AB,O}b.P({A})=0.41,P({B})=0.10,P({AB})=0.04,P({O})=0.45.c.P({A}or{B})=P({A})+P({B})=0.51,sincetheeventsaremutual
5、lyexclusive.2.11a.SinceP(S)=P(E1)+…+P(E5)=1,1=.15+.15+.40+3P(E5).So,P(E5)=.10andP(E4)=.20.b.Obviously,P(E3)+P(E4)+P(E5)=.6.Thus,theyareallequalto.22.12a.LetL={lefttern},R={rightturn},C={continuesstraight}.b.P(vehicleturns)=P(L)+P(R)=1/3+1/3=2/3.2.13a.Denotetheevents
6、asVL,SL,U,O.b.Notequallylikely:P(VL)=.24,P(SL)=.24,P(U)=.40,P(O)=.12.c.P(atleastSL)=P(SL)+P(VL)=.48.2.14a.P(needsglasses)=.44+.14=.48b.P(needsglassesbutdoesn’tusethem)=.14c.P(usesglasses)=.44+.02=.462.15a.SincetheeventsareM.E.,P(S)=P(E1)+…+P(E4)=1.So,P(E2)=1–.01–.09
7、–.81=.09.b.P(atleastonehit)=P(E1)+P(E2)+P(E3)=.19.Chapter2:Probability10Instructor’sSolutionsManual2.16a.1/3b.1/3+1/15=6/15c.1/3+1/16=19/48d.49/2402.17LetB=bushingdefect,SH=shaftdefect.a.P(B)=.06+.02=.08b.P(BorSH)=.06+.08+.02=.16c.P(exactlyonedefect)=.06+.08=.14d.P(
8、neitherdefect)=1–P(BorSH)=1-.16=.842.18a.S={HH,TH,HT,TT}b.ifthecoinisfair,alleventshaveprobability.25.c.A={HT,TH},B={HT,TH,HH}d.P(A)=.5,P(
