Solutions for Probability And Statistics for Engineers And Scientists 9th Us Edition by Walpole by 6alsm

Chapter1 IntroductiontoStatisticsandData

Analysis

1.1(a)15.

(b)¯ x = 1 15 (3.4+2.5+4.8+ +4.8)=3.787.

(c)Samplemedianisthe8thvalue,afterthedataissortedfromsmallesttolargest:3.6.

(d)Adotplotisshownbelow.

2.53.03.54.04.55.05.5

(e)Aftertrimmingtotal40%ofthedata(20%highestand20%lowest),thedatabecomes:

2.93.03.33.43.6 3.74.04.44.8

So.thetrimmedmeanis xtr20 = 1 9 (2 9+3 0+ +4 8)=3 678

(f)Theyareaboutthesame.

1.2(a)Mean=20.7675andMedian=20.610.

(b)¯ xtr10 =20.743.

(c)Adotplotisshownbelow. 181920212223

(d)No.Theyareallclosetoeachother.

1.3(a)Adotplotisshownbelow.

200205210215220225230

Intheﬁgure,“×”representsthe“Noaging”groupand“◦”representsthe“Aging” group.

(b)Yes;tensilestrengthisgreatlyreducedduetotheagingprocess.

(c)MeanAging =209.90,andMeanNoaging =222.10.

(d)MedianAging =210 00,andMedianNoaging =221 50.Themeansandmediansforeach grouparesimilartoeachother.

1.4(a) XA =7.950and XA =8.250; XB =10.260and XB =10.150.

(b)Adotplotisshownbelow.

6.57.58.59.510.511.5

Intheﬁgure,“×”representscompany A and“◦”representscompany B .Thesteelrods madebycompany B showmoreﬂexibility.

1.5(a)Adotplotisshownbelow.

−10010203040

Intheﬁgure,“×”representsthecontrolgroupand“◦”representsthetreatmentgroup.

(b) XControl =5 60, XControl =5 00,and Xtr(10);Control =5 13; XTreatment =7.60, XTreatment =4.50,and Xtr(10);Treatment =5.63.

(c)Thediﬀerenceofthemeansis2.0andthediﬀerencesofthemediansandthetrimmed meansare0.5,whicharemuchsmaller.Thepossiblecauseofthismightbeduetothe extremevalues(outliers)inthesamples,especiallythevalueof37.

1.6(a)Adotplotisshownbelow.

1.952.052.152.252.352.452.55

Intheﬁgure,“×”representsthe20◦ Cgroupand“◦”representsthe45◦ Cgroup.

(b) X20◦ C =2.1075,and X45◦ C =2.2350.

(c)Basedontheplot,itseemsthathightemperatureyieldsmorehighvaluesoftensile strength,alongwithafewlowvaluesoftensilestrength.Overall,thetemperaturedoes haveaninﬂuenceonthetensilestrength.

(d)Italsoseemsthatthevariationofthetensilestrengthgetslargerwhenthecuretemperatureisincreased.

1.7 s2 = 1 15 1 [(3 4 3 787)2 +(2 5 3 787)2 +(4 8 3 787)2 + ··· +(4 8 3 787)2 ]=0 94284; s = √s2 = √0.9428=0.971.

1.8 s2 = 1 20 1 [(18 71 20 7675)2 +(21 41 20 7675)2 + +(21 12 20 7675)2 ]=2 5329; s = √2.5345=1.5915.

1.9(a) s2 NoAging = 1 10 1 [(227 222 10)2 +(222 222 10)2 + ··· +(221 222 10)2 ]=23 66;

sNoAging = √23 62=4 86.

s2 Aging = 1 10 1 [(219 209 90)2 +(214 209 90)2 + +(205 209 90)2 ]=42 10;

sAging = √42 12=6 49.

(b)Basedonthenumbersin(a),thevariationin“Aging”issmallerthatthevariationin “NoAging”althoughthediﬀerenceisnotsoapparentintheplot.

1.10Forcompany A: s2 A =1 2078and sA = √1 2072=1 099. Forcompany B : s2 B =0 3249and sB = √0 3249=0 570.

1.11Forthecontrolgroup: s2 Control =69.38and sControl =8.33.

Forthetreatmentgroup: s2 Treatment =128.04and sTreatment =11.32.

1.12Forthecuretemperatureat20◦ C: s2 20◦ C =0.005and s20◦ C =0.071.

Forthecuretemperatureat45◦ C: s2 45◦ C =0.0413and s45◦ C =0.2032. Thevariationofthetensilestrengthisinﬂuencedbytheincreaseofcuretemperature.

1.13(a)Mean= X =124.3andmedian= X =120; (b)175isanextremeobservation.

1.14(a)Mean= X =570 5andmedian= X =571; (b)Variance= s2 =10;standarddeviation= s =3.162;range=10; (c)Variationofthediametersseemstoobigsothequalityisquestionable.

1.15Yes.Thevalue0.03125isactuallya P -valueandasmallvalueofthisquantitymeansthat theoutcome(i.e., HHHHH )isveryunlikelytohappenwithafaircoin.

1.16Thetermontheleftsidecanbemanipulatedto

=0, whichisthetermontherightside.

1.17(a) Xsmokers =43.70and Xnonsmokers =30.32; (b) ssmokers =16 93and snonsmokers =7 13; (c)Adotplotisshownbelow.

10203040506070

Intheﬁgure,“×”representsthenonsmokergroupand“◦”representsthesmokergroup. (d)Smokersappeartotakelongertimetofallasleepandthetimetofallasleepforsmoker groupismorevariable.

1.18(a)Astem-and-leafplotisshownbelow.

Chapter1 IntroductiontoStatisticsandDataAnalysis

StemLeafFrequency 10573 2352 32463 411384 5224575 60012344577911 70124445667889914 80001122344558914 902584

(b)Thefollowingistherelativefrequencydistributiontable.

RelativeFrequencyDistributionofGrades ClassInterval

(c)Ahistogramplotisgivenbelow.

Relative Frequency

14.524.534.544.554.564.574.584.594.5 FinalExamGrades

Thedistributionskewstotheleft.

(d) X =65 48, X =71 50and s =21 13. 1.19(a)Astem-and-leafplotisshownbelow.

StemLeafFrequency 0222334578 10235586 20353 3032 40573 505694 600054

(b)Thefollowingistherelativefrequencydistributiontable.

RelativeFrequencyDistributionofYears ClassInterval ClassMidpoint Frequency, f RelativeFrequency

1.20(a)Astem-and-leafplotisshownnext.

StemLeafFrequency 0*342 05666777777788999917 1*000000122333334416 1556678889910 2*0343 271 3*21

(b)Therelativefrequencydistributiontableisshownnext.

RelativeFrequencyDistributionofFruitFlyLives

(c)Ahistogramplotisshownnext. 271217222732 Fruitflylives(seconds)

Relative Frequency

(d) X =10.50. Copyright c 2012PearsonEducation,Inc.PublishingasPrenticeHall.

1.21(a) X =74 02and ˜ X =78; (b) s =39 26.

1.22(a) X =6.7261and X =0.0536.

(b)Ahistogramplotisshownnext.

6.626.666.76.746.786.82

RelativeFrequencyHistogramforDiameter

(c)Thedataappeartobeskewedtotheleft.

1.23(a)Adotplotisshownnext.

(b) X1980 =395.1and X1990 =160.2.

(c)Thesamplemeanfor1980isovertwiceaslargeasthatof1990.Thevariabilityfor 1990decreasedalsoasseenbylookingatthepicturein(a).Thegaprepresentsan increaseofover400ppm.Itappearsfromthedatathathydrocarbonemissionsdecreased considerablybetween1980and1990andthattheextremelargeemission(over500ppm) werenolongerinevidence.

1.24(a) X =2.8973and s =0.5415.

(b)Ahistogramplotisshownnext.

Relative Frequency

(c)Usethedouble-stem-and-leafplot,wehavethefollowing.

StemLeafFrequency 1(84)1

2*(05)(10)(14)(37)(44)(45)6

2(52)(52)(67)(68)(71)(75)(77)(83)(89)(91)(99)11

3*(10)(13)(14)(22)(36)(37)6

3(51)(54)(57)(71)(79)(85)6

1.25(a) X =33 31;

(b) ˜ X =26 35;

(c)Ahistogramplotisshownnext.

Relative Frequency

102030405060708090 Percentageofthefamilies

(d) Xtr(10) =30.97.Thistrimmedmeanisinthemiddleofthemeanandmedianusingthe fullamountofdata.Duetotheskewnessofthedatatotheright(seeplotin(c)),itis commontousetrimmeddatatohaveamorerobustresult.

1.26Ifamodelusingthefunctionofpercentoffamiliestopredictstaﬀsalaries,itislikelythatthe modelwouldbewrongduetoseveralextremevaluesofthedata.Actuallyifascatterplot ofthesetwodatasetsismade,itiseasytoseethatsomeoutlierwouldinﬂuencethetrend.

1.27(a)Theaveragesofthewearareplottedhere. 7008009001000110012001300

(b)Whentheloadvalueincreases,thewearvaluealsoincreases.Itdoesshowcertain relationship.

(c)Aplotofwearsisshownnext.

7008009001000110012001300

(d)Therelationshipbetweenloadandwearin(c)isnotasstrongasthecasein(a),especially fortheloadat1300.Onereasonisthatthereisanextremevalue(750)whichinﬂuence themeanvalueattheload1300.

1.28(a)Adotplotisshownnext.

HighLow

71.4571.6571.8572.0572.2572.4572.6572.8573.05

Intheﬁgure,“×”representsthelow-injection-velocitygroupand“◦”representsthe high-injection-velocitygroup.

(b)Itappearsthatshrinkagevaluesforthelow-injection-velocitygroupishigherthanthose forthehigh-injection-velocitygroup.Also,thevariationoftheshrinkageisalittlelarger forthelowinjectionvelocitythanthatforthehighinjectionvelocity.

1.29Aboxplotisshownnext.

1.30Aboxplotplotisshownnext.

1.31(a)Adotplotisshownnext.

LowHigh

76798285889194

Intheﬁgure,“×”representsthelow-injection-velocitygroupand“◦”representsthe high-injection-velocitygroup.

(b)Inthistime,theshrinkagevaluesaremuchhigherforthehigh-injection-velocitygroup thanthoseforthelow-injection-velocitygroup.Also,thevariationfortheformergroup ismuchhigheraswell.

(c)Sincetheshrinkageeffectschangeindifferentdirectionbetweenlowmodetemperature andhighmoldtemperature,theapparentinteractionsbetweenthemoldtemperature andinjectionvelocityaresignificant.

1.32Aninteractionplotisshownnext.

Itisquiteobvioustoﬁndtheinteractionbetweenthetwovariables.Sinceinthisexperimental data,thosetwovariablescanbecontrolledeachattwolevels,theinteractioncanbeinvesCopyright c 2012PearsonEducation,Inc.PublishingasPrenticeHall.

Chapter1 IntroductiontoStatisticsandDataAnalysis tigated.However,ifthedataarefromanobservationalstudies,inwhichthevariablevalues cannotbecontrolled,itwouldbediﬃculttostudytheinteractionsamongthesevariables.

Chapter2

Probability

2.1(a) S = {8, 16, 24, 32, 40, 48}

(b)For x2 +4x 5=(x +5)(x 1)=0,theonlysolutionsare x = 5and x =1. S = {−5, 1}

(d) S = {N.America, S.America, Europe, Asia, Africa, Australia, Antarctica}

(e)Solving2x 4 ≥ 0gives x ≥ 2.Sincewemustalsohave x< 1,itfollowsthat S = φ.

2.2 S = {(x,y ) | x2 + y 2 < 9; x ≥ 0,y ≥ 0}

2.3(a) A = {1, 3}.

(b) B = {1, 2, 3, 4, 5, 6}

(d) D = {0, 1, 2, 3, 4, 5, 6}.Clearly, A = C

2.4(a) S = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}

(b) S = {(x,y ) | 1 ≤ x,y ≤ 6}.

2.5 S = {1HH, 1HT, 1TH, 1TT, 2H, 2T, 3HH, 3HT, 3TH, 3TT, 4H, 4T, 5HH, 5HT, 5TH, 5TT, 6H, 6T }

2.6 S = {A1 A2 ,A1 A3 ,A1 A4 ,A2 A3 ,A2 A4 ,A3 A4 }

2.7 S1 = {MMMM,MMMF,MMFM,MFMM,FMMM,MMFF,MFMF,MFFM, FMFM,FFMM,FMMF,MFFF,FMFF,FFMF,FFFM,FFFF }. S2 = {0, 1, 2, 3, 4}

2.8(a) A = {(3, 6), (4, 5), (4, 6), (5, 4), (5, 5), (5, 6), (6, 3), (6, 4), (6, 5), (6, 6)} (b) B = {(1, 2), (2, 2), (3, 2), (4, 2), (5, 2), (6, 2), (2, 1), (2, 3), (2, 4), (2, 5), (2, 6)}

Chapter2 Probability

(d) A ∩ C = {(5, 4), (5, 5), (5, 6), (6, 3), (6, 4), (6, 5), (6, 6)}.

(e) A ∩ B = φ

(f) B ∩ C = {(5, 2), (6, 2)}.

(g)AVenndiagramisshownnext.

2.9(a) A = {1HH, 1HT, 1TH, 1TT, 2H, 2T }

(b) B = {1TT, 3TT, 5TT }.

(d) A ∩ B = {3TT, 5TT }.

(e) A ∪ B = {1HH, 1HT, 1TH, 1TT, 2H, 2T, 3TT, 5TT }

2.10(a) S = {FFF,FFN,FNF,NFF,FNN,NFN,NNF,NNN }

(b) E = {FFF,FFN,FNF,NFF }.

(c)Thesecondriverwassafeforﬁshing.

2.11(a) S = {M1 M2 ,M1

, F2 F1 }

(b) A = {M1 M2 ,M1 F1 ,M1 F2 ,M2 M1 ,M2 F1 ,M2 F2 }

(d) C = {F1 F2 ,F2 F1 }

(e) A ∩ B = {M1 F1 ,M1 F2 ,M2 F1 ,M2 F2 }.

(f) A ∪ C = {M1 M2 ,M1 F1 ,M1 F2 ,M2 M1 ,M2 F1 ,M2 F2 ,F1 F2 ,F2 F1 }

2.12(a) S = {ZYF,ZNF,WYF,WNF,SYF,SNF,ZYM }

(b) A ∪ B = {ZYF,ZNF,WYF,WNF,SYF,SNF } = A.

2.13AVenndiagramisshownnext.

2.14(a) A ∪ C = {0, 2, 3, 4, 5, 6, 8}

(b) A ∩ B = φ.

(d) C ∩ D = {1, 6, 7},so(C ∩ D ) ∪ B = {1, 3, 5, 6, 7, 9}.

(e)(S ∩ C ) = C = {0, 1, 6, 7, 8, 9}

(f) A ∩ C = {2, 4},so A ∩ C ∩ D = {2, 4}

2.15(a) A = {nitrogen, potassium, uranium, oxygen}

(b) A ∪ C = {copper, sodium, zinc, oxygen}.

(c) A ∩ B = {copper, zinc} and C = {copper, sodium, nitrogen, potassium, uranium, zinc}; so(A ∩ B ) ∪ C = {copper, sodium, nitrogen, potassium, uranium, zinc}.

(d) B ∩ C = {copper, uranium, zinc}.

(e) A ∩ B ∩ C = φ.

(f) A ∪ B = {copper, nitrogen, potassium, uranium, oxygen, zinc} and A ∩ C = {oxygen};so,(A ∪ B ) ∩ (A ∩ C )= {oxygen}

2.16(a) M ∪ N = {x | 0 <x< 9}

(b) M ∩ N = {x | 1 <x< 5}

2.17AVenndiagramisshownnext.

(a)FromtheaboveVenndiagram,(A ∩ B ) containstheregionsof1,2and4.

(b)(A ∪ B ) containsregion1.

(c)AVenndiagramisshownnext.

(A ∩ C ) ∪ B containstheregionsof3,4,5,7and8.

2.18(a)Notmutuallyexclusive.

(b)Mutuallyexclusive.

(c)Notmutuallyexclusive.

(d)Mutuallyexclusive.

2.19(a)Thefamilywillexperiencemechanicalproblemsbutwillreceivenoticketfortraﬃc violationandwillnotarriveatacampsitethathasnovacancies.

(b)Thefamilywillreceiveatraﬃcticketandarriveatacampsitethathasnovacanciesbut willnotexperiencemechanicalproblems.

(c)Thefamilywillexperiencemechanicalproblemsandwillarriveatacampsitethathas novacancies.

(d)Thefamilywillreceiveatraﬃcticketbutwillnotarriveatacampsitethathasno vacancies.

(e)Thefamilywillnotexperiencemechanicalproblems.

2.20(a)6; (b)2; (c)2,5,6; (d)4,5,7,8.

2.21With n1 =6sightseeingtourseachavailableon n2 =3diﬀerentdays,themultiplicationrule gives n1 n2 =(6)(3)=18waysforapersontoarrangeatour.

2.22With n1 =8bloodtypesand n2 =3classiﬁcationsofbloodpressure,themultiplicationrule gives n1 n2 =(8)(3)=24classiﬁcations.

2.23Sincethediecanlandin n1 =6waysandalettercanbeselectedin n2 =26ways,the multiplicationrulegives n1 n2 =(6)(26)=156pointsin S

2.24Sinceastudentmaybeclassifiedaccordingto n1 =4classstandingand n2 =2gender classifications,themultiplicationrulegives n1 n2 =(4)(2)=8possibleclassificationsforthe students.

2.25With n1 =5differentshoestylesin n2 =4differentcolors,themultiplicationrulegives n1 n2 =(5)(4)=20differentpairsofshoes.

2.26UsingTheorem2.8,weobtainthefollowings.

(a)Thereare 7 5 =21ways.

(b)Thereare 5 3 =10ways.

2.27Usingthegeneralizedmultiplicationrule,thereare n1 × n2

=(4)(3)(2)(2)=48 diﬀerenthouseplansavailable.

2.28With n1 =5differentmanufacturers, n2 =3differentpreparations,and n3 =2different strengths,thegeneralizedmultiplicationruleyields n1 n2 n3 =(5)(3)(2)=30differentways toprescribeadrugforasthma.

2.29With n1 =3racecars, n2 =5brandsofgasoline, n3 =7testsites,and n4 =2drivers,the generalizedmultiplicationruleyields(3)(5)(7)(2)=210testruns.

2.30With n1 =2choicesfortheﬁrstquestion, n2 =2choicesforthesecondquestion,andso forth,thegeneralizedmultiplicationruleyields n1 n2 n9 =29 =512waystoanswerthe test.

2.31Sincetheﬁrstdigitisa5,thereare n1 =9possibilitiesfortheseconddigitandthen n2 =8 possibilitiesforthethirddigit.Therefore,bythemultiplicationrulethereare n1 n2 =(9)(8)= 72registrationstobechecked.

2.32(a)ByTheorem2.3,thereare6!=720ways.

(b)Acertain3personscanfolloweachotherinalineof6peopleinaspeciﬁedorderis4 waysorin(4)(3!)=24wayswithregardtoorder.Theother3personscanthenbe placedinlinein3!=6ways.ByTheorem2.1,therearetotal(24)(6)=144waysto lineup6peoplewithacertain3followingeachother.

(c)Similarasin(b),thenumberofwaysthataspeciﬁed2personscanfolloweachotherin alineof6peopleis(5)(2!)(4!)=240ways.Therefore,thereare720 240=480ways ifacertain2personsrefusetofolloweachother.

2.33(a)With n1 =4possibleanswersfortheﬁrstquestion, n2 =4possibleanswersforthe secondquestion,andsoforth,thegeneralizedmultiplicationruleyields45 =1024ways toanswerthetest.

(b)With n1 =3wronganswersfortheﬁrstquestion, n2 =3wronganswersforthesecond question,andsoforth,thegeneralizedmultiplicationruleyields n1 n2 n3 n4 n5

=(3)(3)(3)(3)(3)=35 =243 waystoanswerthetestandgetallquestionswrong.

2.34(a)ByTheorem2.3,7!=5040.

(b)Sincetheﬁrstlettermustbe m,theremaining6letterscanbearrangedin6!=720 ways.

2.35Theﬁrsthousecanbeplacedonanyofthe n1 =9lots,thesecondhouseonanyofthe remaining n2 =8lots,andsoforth.Therefore,thereare9!=362, 880waystoplacethe9 homesonthe9lots.

2.36(a)Anyofthe6nonzerodigitscanbechosenforthehundredsposition,andoftheremaining 6digitsforthetensposition,leaving5digitsfortheunitsposition.So,thereare (6)(6)(5)=180threedigitnumbers.

(b)Theunitspositioncanbeﬁlledusinganyofthe3odddigits.Anyoftheremaining5 nonzerodigitscanbechosenforthehundredsposition,leavingachoiceof5digitsfor thetensposition.ByTheorem2.2,thereare(3)(5)(5)=75threedigitoddnumbers.

(c)Ifa4,5,or6isusedinthehundredspositionthereremain6and5choices,respectively, forthetensandunitspositions.Thisgives(3)(6)(5)=90threedigitnumbersbeginning witha4,5,or6.Ifa3isusedinthehundredsposition,thena4,5,or6mustbe usedinthetenspositionleaving5choicesfortheunitsposition.Inthiscase,thereare (1)(3)(5)=15threedigitnumberbeginwitha3.So,thetotalnumberofthreedigit numbersthataregreaterthan330is90+15=105.

2.37Theﬁrstseatmustbeﬁlledbyanyof5girlsandthesecondseatbyanyof4boys.Continuing inthismanner,thetotalnumberofwaystoseatthe5girlsand4boysis(5)(4)(4)(3)(3)(2)(2)(1)(1)= 2880.

2.38(a)8!=40320.

(b)Thereare4!waystoseat4couplesandtheneachmemberofacouplecanbeinterchanged resultingin24 (4!)=384ways.

(c)ByTheorem2.3,themembersofeachgendercanbeseatedin4!ways.Thenusing Theorem2.1,bothmenandwomencanbeseatedin(4!)(4!)=576ways.

2.39(a)Anyofthe n1 =8finalistsmaycomeinfirst,andofthe n2 =7remainingfinalistscan thencomeinsecond,andsoforth.ByTheorem2.3,there8!=40320possibleordersin which8finalistsmayfinishthespellingbee.

(b)Thepossibleordersfortheﬁrstthreepositionsare 8 P3 = 8! 5! =336.

2.40ByTheorem2.4, 8 P5 = 8! 3! =6720.

2.41ByTheorem2.4, 6 P4 = 6! 2! =360.

2.42ByTheorem2.4, 40 P3 = 40! 37! =59, 280.

2.43ByTheorem2.5,thereare4!=24ways.

2.44ByTheorem2.5,thereare7!=5040arrangements.

2.45ByTheorem2.6,thereare 8! 3!2! =3360.

2.46ByTheorem2.6,thereare 9! 3!4!2! =1260ways.

2.47ByTheorem2.8,thereare 8 3 =56ways.

2.48AssumeFebruary29thasMarch1stfortheleapyear.Therearetotal365daysinayear. Thenumberofwaysthatallthese60studentswillhavediﬀerentbirthdates(i.e,arranging 60from365)is 365 P60 .Thisisaverylargenumber.

2.49(a)Sumoftheprobabilitiesexceeds1.

(b)Sumoftheprobabilitiesislessthan1.

(c)Anegativeprobability.

(d)Probabilityofbothaheartandablackcardiszero.

2.50Assumingequalweights

(a) P (A)= 5 18 ; (b) P (C )= 1 3 ; (c) P (A ∩ C )= 7 36 .

2.51 S = {$10, $25, $100} withweights275/500=11/20,150/500=3/10,and75/500=3/20, respectively.Theprobabilitythattheﬁrstenvelopepurchasedcontainslessthan$100is equalto11/20+3/10=17/20.

2.52(a) P (S ∩ D )=88/500=22/125.

(b) P (E ∩ D ∩ S )=31/500.

2.53Considertheevents

S :industrywilllocateinShanghai, B :industrywilllocateinBeijing.

(a) P (S ∩ B )= P (S )+ P (B ) P (S ∪ B )=0 7+0 4 0 8=0 3. (b) P (S ∩ B )=1 P (S ∪ B )=1 0.8=0.2.

2.54Considertheevents

B :customerinvestsintax-freebonds, M :customerinvestsinmutualfunds.

(a) P (B ∪ M )= P (B )+ P (M ) P (B ∩ M )=0 6+0 3 0 15=0 75. (b) P (B ∩ M )=1 P (B ∪ M )=1 0.75=0.25.

2.55ByTheorem2.2,thereare N =(26)(25)(24)(9)(8)(7)(6)=47, 174, 400possiblewaystocode theitemsofwhich n =(5)(25)(24)(8)(7)(6)(4)=4, 032, 000beginwithavowelandendwith anevendigit.Therefore, n N = 10 117 .

2.56(a)Let A =Defectinbrakesystem; B =Defectinfuelsystem; P (A ∪ B )= P (A)+ P (B ) P (A ∩ B )=0 25+0 17 0 15=0 27.

(b) P (Nodefect)=1 P (A ∪ B )=1 0.27=0.73.

2.57(a)Since5ofthe26lettersarevowels,wegetaprobabilityof5/26.

(b)Since9ofthe26lettersprecede j ,wegetaprobabilityof9/26.

(c)Since19ofthe26lettersfollow g ,wegetaprobabilityof19/26.

2.58(a)Ofthe(6)(6)=36elementsinthesamplespace,only5elements(2,6),(3,5),(4,4),(5,3), and(6,2)addto8.Hencetheprobabilityofobtainingatotalof8isthen5/36.

(b)Tenofthe36elementstotalatmost5.Hencetheprobabilityofobtainingatotalofat mostis10/36=5/18.

2.59(a) (4 3)(48 2 ) (52 5 ) = 94 54145

(b) (13 4 )(13 1 ) (52 5 ) = 143 39984 .

2.60(a) (1 1)(8 2) (9 3) = 1 3

(b) (5 2)(3 1) (9 3) = 5 14

2.61(a) P (M ∪ H )=88/100=22/25; (b) P (M ∩ H )=12/100=3/25;

2.62(a)9; (b)1/9.

2.63(a)0.32; (b)0.68; (c)oﬃceorden.

2.64(a)1 0 42=0 58; (b)1 0.04=0.96.

2.65 P (A)=0.2and P (B )=0.35

(a) P (A )=1 0.2=0.8; (b) P (A ∩ B )=1 P (A ∪ B )=1 0.2 0.35=0.45; (c) P (A ∪ B )=0 2+0 35=0 55.

2.66(a)0 02+0 30=0 32=32%;

(b)0 32+0 25+0 30=0 87=87%; (c)0.05+0.06+0.02=0.13=13%; (d)1 0.05 0.32=0.63=63%.

2.67(a)0 12+0 19=0 31;

(b)1 0.07=0.93;

2.68(a)1 0 40=0 60.

(b)Theprobabilitythatallsixpurchasingtheelectricovenorallsixpurchasingthegasoven is0.007+0.104=0.111.Sotheprobabilitythatatleastoneofeachtypeispurchased is1 0.111=0.889.

2.69(a) P (C )=1 P (A) P (B )=1 0.990 0.001=0.009;

(b) P (B )=1 P (B )=1 0 001=0 999;

2.70(a)($4 50 $4 00) × 50, 000=$25, 000;

(b)Sincetheprobabilityofunderfillingis0.001,wewouldexpect50, 000 × 0.001=50boxes tobeunderfilled.So,insteadofhaving($4.50 $4.00) × 50=$25profitforthose50 boxes,therearealossof$4.00 × 50=$200duetothecost.So,thelossinprofitexpected duetounderfillingis$25+$200=$250.

2.71(a)1 0.95 0.002=0.048;

(b)($25 00 $20 00) × 10, 000=$50, 000;

(c)(0.05)(10, 000) × $5.00+(0.05)(10, 000) × $20=$12, 500.

2.72 P (A ∩ B )=1 P (A ∪ B )=1 (P (A)+ P (B ) P (A ∩ B )=1+ P (A ∩ B ) P (A) P (B ).

2.73(a)Theprobabilitythataconvictwhopusheddope,alsocommittedarmedrobbery.

(b)Theprobabilitythataconvictwhocommittedarmedrobbery,didnotpushdope.

(c)Theprobabilitythataconvictwhodidnotpushdopealsodidnotcommitarmedrobbery.

2.74 P (S | A)=10/18=5/9.

2.75Considertheevents:

M :apersonisamale; S :apersonhasasecondaryeducation; C :apersonhasacollegedegree.

(a) P (M | S )=28/78=14/39; (b) P (C | M )=95/112.

2.76Considertheevents:

A:apersonisexperiencinghypertension, B :apersonisaheavysmoker, C :apersonisanonsmoker.

(a) P (A | B )=30/49; (b) P (C | A )=48/93=16/31. Copyright c 2012PearsonEducation,Inc.PublishingasPrenticeHall.

2.77(a) P (M ∩ P ∩ H )= 10 68 = 5 34 ; (b) P (H ∩ M | P )= P (H ∩M ∩P ) P (P ) = 22 10 100 68 = 12 32 = 3 8

2.78(a)(0 90)(0 08)=0 072; (b)(0.90)(0.92)(0.12)=0.099.

2.79(a)0.018;

(b)0 22+0 002+0 160+0 102+0 046+0 084=0 614;

2.80Considertheevents:

C :anoilchangeisneeded, F :anoilﬁlterisneeded.

(a) P (F | C )= P (F ∩C ) P (C ) = 0 14 0 25 =0 56.

(b) P (C | F )= P (C ∩F ) P (F ) = 0 14 0 40 =0.35.

2.81Considertheevents:

H :husbandwatchesacertainshow, W :wifewatchesthesameshow.

(a) P (W ∩ H )= P (W )P (H | W )=(0 5)(0 7)=0 35.

(b) P (W | H )= P (W ∩H ) P (H ) = 0 35 0.4 =0.875.

2.82Considertheevents:

H :thehusbandwillvoteonthebondreferendum, W :thewifewillvoteonthebondreferendum.

Then P (H )=0.21, P (W )=0.28,and P (H ∩ W )=0.15.

(a) P (H ∪ W )= P (H )+ P (W ) P (H ∩ W )=0.21+0.28 0.15=0.34.

(b) P (W | H )= P (H ∩W ) P (H ) = 0 15 0 21 = 5 7

2.83Considertheevents:

A:thevehicleisacamper, B :thevehiclehasCanadianlicenseplates.

(a) P (B | A)= P (A∩B ) P (A) = 0 09 0 28 = 9 28

(b) P (A | B )= P (A∩B ) P (B ) = 0 09 0 12 = 3 4 .

2.84Deﬁne

H :headofhouseholdishome, C :achangeismadeinlongdistancecarriers.

P (H ∩ C )= P (H )P (C | H )=(0 4)(0 3)=0 12.

2.85Considertheevents:

A:thedoctormakesacorrectdiagnosis, B :thepatientsues.

P (A ∩ B )= P (A )P (B | A )=(0 3)(0 9)=0 27.

2.86(a)0.43; (b)(0 53)(0 22)=0 12; (c)1 (0 47)(0 22)=0 90.

2.87Considertheevents:

A:thehouseisopen, B :thecorrectkeyisselected.

P (A)=0.4, P (A )=0.6,and P (B )= (1 1)(7 2) (8 3) = 3 8 =0.375.

So, P [A ∪ (A ∩ B )]= P (A)+ P (A )P (B )=0 4+(0 6)(0 375)=0 625.

2.88Considertheevents:

F :failedthetest, P :passedthetest.

(a) P (failedatleastonetests)=1 P (P1 P2 P3 P4 )=1 (0 99)(0 97)(0 98)(0 99)=1 0 93=0 07,

(b) P (failed2or3)=1 P (P2 P3 )=1 (0 97)(0 98)=0 0494. (c)100 × 0.07=7. (d)0 25.

2.89Let A and B representtheavailabilityofeachﬁreengine.

(a) P (A ∩ B )= P (A )P (B )=(0.04)(0.04)=0.0016.

(b) P (A ∪ B )=1 P (A ∩ B )=1 0 0016=0 9984.

2.90(a) P (A ∩ B ∩ C )= P (C | A ∩ B )P (B | A)P (A)=(0.20)(0.75)(0.3)=0.045. (b) P (B ∩ C )= P (A ∩ B ∩ C )+ P (A ∩ B ∩ C )= P (C | A ∩ B )P (B | A)P (A)+ P (C | A ∩ B )P (B | A )P (A )=(0.80)(1 0.75)(0.3)+(0.90)(1 0.20)(1 0.3)=0.564. (c)Usesimilarargumentasin(a)and(b), P (C )= P (A ∩ B ∩ C )+ P (A ∩ B ∩ C )+ P (A ∩ B ∩ C )+ P (A ∩ B ∩ C )=0 045+0 060+0 021+0 504=0 630. (d) P (A | B ∩ C )= P (A ∩ B ∩ C )/P (B ∩ C )=(0 06)(0 564)=0 1064.

2.91(a) P (Q1 ∩ Q2 ∩ Q3 ∩ Q4 )= P (Q1 )P (Q2 | Q1 )P (Q3 | Q1 ∩ Q2 )P (Q4 | Q1 ∩ Q2 ∩ Q3 )= (15/20)(14/19)(13/18)(12/17)=91/323. Copyright c 2012PearsonEducation,Inc.PublishingasPrenticeHall.

(b)Let A betheeventthat4goodquartsofmilkareselected.Then P (A)= 15 4 20 4 = 91 323

2.92 P =(0 95)[1 (1 0 7)(1 0 8)](0 9)=0 8037.

2.93Thisisaparallelsystemoftwoseriessubsystems.

(a) P =1 [1 (0 7)(0 7)][1 (0 8)(0 8)(0 8)]=0 75112.

(b) P = P (A ∩C ∩D ∩E ) P systemworks = (0 3)(0 8)(0 8)(0 8) 0 75112 =0 2045.

2.94Deﬁne S :thesystemworks.

P (A | S )= P (A ∩S ) P (S ) = P (A )(1 P (C ∩D ∩E )) 1 P (S ) = (0 3)[1 (0 8)(0 8)(0 8)] 1 0 75112 =0 588.

2.95Considertheevents:

C :anadultselectedhascancer, D :theadultisdiagnosedashavingcancer.

P (C )=0.05, P (D | C )=0.78, P (C )=0.95and P (D | C )=0.06.So, P (D )= P (C ∩ D )+ P (C ∩ D )=(0.05)(0.78)+(0.95)(0.06)=0.096.

2.96Let S1 ,S2 ,S3 ,and S4 representtheeventsthatapersonisspeedingashepassesthrough therespectivelocationsandlet R representtheeventthattheradartrapsisoperating resultinginaspeedingticket.Thentheprobabilitythathereceivesaspeedingticket: P (R)= 4 i=1

2.97 P (C | D )= P (C ∩D ) P (D ) = 0 039 0 096 =0 40625.

2.98 P (S2 | R)= P (R∩ S2 ) P (R) = 0 03 0 27 =1/9.

2.99Considertheevents:

A:noexpirationdate,

B1 :Johnistheinspector, P (B1 )=0 20and P (A | B1 )=0 005, B2 :Tomistheinspector, P (B2 )=0.60and P (A | B2 )=0.010, B3 :Jeﬀistheinspector, P (B3 )=0.15and P (A | B3 )=0.011, B4 :Patistheinspector, P (B4 )=0.05and P (A | B4 )=0.005, P (B1 | A)= (0 005)(0 20) (0 005)(0 20)+(0 010)(0 60)+(0 011)(0 15)+(0 005)(0 05) =0.1124.

2.100Considertheevents

E :amalfunctionbyotherhumanerrors,

A:station A, B :station B ,and C :station C

P (C | E )=

/10)(10/43)

/43)+(7/15)(15/43)+(5/10)(10/43) = 0 1163 0 4419 =0.2632.

2.101Considertheevents:

A:acustomerpurchaseslatexpaint,

A :acustomerpurchasessemiglosspaint,

B :acustomerpurchasesrollers.

P (A | B )=

2.102Ifweusetheassumptionsthatthehostwouldnotopenthedooryoupickednorthedoor withtheprizebehindit,wecanuseBayesruletosolvetheproblem.Denotebyevents A, B , and C ,thattheprizeisbehinddoors A, B ,and C ,respectively.Ofcourse P (A)= P (B )= P (C )=1/3.Denoteby H theeventthatyoupickeddoor A andthehostopeneddoor B , whilethereisnoprizebehindthedoor B .Then

Henceyoushouldswitchdoor.

2.103Considertheevents:

G:guiltyofcommittingacrime, I :innocentofthecrime, i:judgedinnocentofthecrime, g :judgedguiltyofthecrime.

P (I | g )=

2.104Let Ai betheeventthatthe ithpatientisallergictosometypeofweek.

(a)

P (A1 )P (A2 )P (A3 )P (A4 )=(4)(1

(b)

2.105Nosolutionnecessary.

2.106(a)0.28+0.10+0.17=0.55.

(b)1 0.17=0.83.

2.107Thenumberofhands= 13 4 13 6 13 1 13 2 .

2.108(a) P (M1 ∩ M2 ∩ M3 ∩ M4 )=(0 1)4 =0 0001,where Mi representsthat ithpersonmakea mistake.

(b) P (J ∩ C ∩ R ∩ W )=(0 1)(0 1)(0 9)(0 9)=0 0081.

2.109Let R, S ,and L representtheeventsthataclientisassignedaroomattheRamadaInn, Sheraton,andLakeviewMotorLodge,respectively,andlet F representstheeventthatthe plumbingisfaulty.

(a) P (F )= P (F | R)P (R)+ P (F | S )P (S )+ P (F | L)P (L)=(0 05)(0 2)+(0 04)(0 4)+ (0.08)(0.3)=0.054.

(b) P (L | F )= (0 08)(0 3) 0 054 = 4 9

2.110Denoteby R theeventthatapatientsurvives.Then P (R)=0.8.

Chapter2 Probability

2.111Considerevents

M :aninmateisamale, N :aninmateisunder25yearsofage.

2.112Thereare 4 3 5 3 6 3 =800possibleselections.

2.113Considertheevents:

Bi :ablackballisdrawnonthe ithdrawl, Gi :agreenballisdrawnonthe ithdrawl.

(a) P (B1 ∩ B2 ∩ B3 )+ P (G1 ∩ G2 ∩ G3 )=(6/10)(6/10)(6/10)+(4/10)(4/10)(4/10)=7/25. (b)Theprobabilitythateachcolorisrepresentedis1 7/25=18/25.

2.114Thetotalnumberofwaystoreceive2or3defectivesetsamong5thatarepurchasedis 3 2 9 3 + 3 3 9 2 =288.

2.115Considertheevents:

O :overrun,

A:consultingﬁrm A,

B :consultingﬁrm B , C :consultingﬁrm C .

(a) P (C | O )=

0 0375 0 0680 =0.5515.

(b) P (A | O )= (0 05)(0 40) 0 0680 =0 2941.

2.116(a)36; (b)12; (c)orderisnotimportant.

2.117(a) 1 (36 2 ) =0 0016; (b) (12 1 )(24 1 ) (36 2 ) = 288 630 =0 4571.

2.118Considertheevents:

C :awomanover60hasthecancer, P :thetestgivesapositiveresult.

So, P (C )=0.07, P (P | C )=0.1and P (P | C )=0.05. P (C | P )= P (P |

2.119Considertheevents:

A:twonondefectivecomponentsareselected, N :alotdoesnotcontaindefectivecomponents, P (N )=0.6, P (A | N )=1, Copyright c

O :alotcontainsonedefectivecomponent, P (O )=0 3, P (A | O )= (19 2 ) (20 2 ) = 9 10 ,

T :alotcontainstwodefectivecomponents,P (T )=0.1, P (A | T )= (18 2 ) (20 2 ) = 153 190 .

(a) P (N | A)=

= (1)(0 6) (1)(0 6)+(9/10)(0 3)+(153/190)(0 1) = 0 6 0.9505 =0.6312;

(b) P (O | A)= (9/10)(0.3) 0 9505 =0 2841; (c) P (T | A)=1 0.6312 0.2841=0.0847.

2.120Considerevents:

D :apersonhastheraredisease, P (D )=1/500, P :thetestshowsapositiveresult, P (P | D )=0.95and P (P | D )=0.01.

P (D | P )= P (P | D )P (D ) P (P | D )P (D )+P (

)P (D ) = (0 95)(1/500) (0 95)(1/500)+(0 01)(1 1/500) =0.1599.

2.121Considertheevents: 1:engineer1, P (1)=0 7,and2:engineer2, P (2)=0 3, E :anerrorhasoccurredinestimatingcost, P (E | 1)=0 02and P (E | 2)=0 04. P (1 | E )= P (E | 1)P (1) P (E | 1)P (1)+P (E | 2)P (2) = (0 02)(0 7) (0 02)(0 7)+(0 04)(0 3) =0 5385,and P (2 | E )=1 0 5385=0 4615.So,morelikelyengineer1didthejob.

2.122Considertheevents: D :anitemisdefective

(a) P (D1 D2 D3 )= P (D1 )P (D2 )P (D3 )=(0.2)3 =0.008. (b) P (threeoutoffouraredefectives)= 4 3 (0.2)3 (1 0.2)=0.0256.

2.123Let A betheeventthataninjuredworkerisadmittedtothehospitaland N betheevent thataninjuredworkerisbacktoworkthenextday. P (A)=0.10, P (N )=0.15and P (A ∩ N )=0.02.So, P (A ∪ N )= P (A)+ P (N ) P (A ∩ N )=0.1+0.15 0.02=0.23.

2.124Considertheevents: T :anoperatoristrained, P (T )=0 5, M anoperatormeetsquota, P (M | T )=0.9and P (M | T )=0.65. P (T | M )= P (M | T )P (T ) P (M | T )P (T )+P (M | T )P (T ) = (0 9)(0 5) (0.9)(0.5)+(0.65)(0.5) =0.581.

2.125Considertheevents:

A:purchasedfromvendor A, D :acustomerisdissatisﬁed.

Then P (A)=0.2, P (A | D )=0.5,and P (D )=0.1. So, P (D | A)= P (A | D )P (D ) P (A) = (0 5)(0 1) 0 2 =0.25.

2.126(a) P (Unionmember | Newcompany(sameﬁeld))= 13 13+10 = 13 23 =0.5652. (b) P (Unemployed | Unionmember)= 2 40+13+4+2 = 2 59 =0.034.

2.127Considertheevents:

C :thequeenisacarrier, P (C )=0.5, D :aprincehasthedisease, P (D | C )=0.5.

P (C | D1 D2 D3 )= P (D1 D2 D3 | C )P (C ) P (D1 D2 D3 | C )P (C )+P (D1 D2 D3 | C )P (C ) = (0 5)3 (0 5) (0 5)3 (0 5)+1(0 5) = 1 9 .