hapter 4: Proalty and Proalty Dstrutons 4.1 a Ths experment nvolves tossng a sngle de and oservng the outcome. The sample space for ths experment conssts of the followng smple events: E 1 : Oserve a 1 E 4 : Oserve a 4 E 2 : Oserve a 2 E 5 : Oserve a 5 E 3 : Oserve a 3 E 6 : Oserve a 6 Events A through F are composed of the smple events n the followng manner: A: (E 2 ) B: (E 2, E 4, E 6) D: (E 2) E: (E 2, E 4, E 6 ) : (E 3, E 4, E 5, E 6) F: contans no smple events c Snce the smple events E, = 1, 2, 3,, 6 are equally lkely, PE ( ) 1 6. d To fnd the proalty of an event, we sum the proaltes assgned to the smple events n that event. For example, 1 P( A) P( E2) 6 3 1 4 2 Smlarly, P( D) 1 6; P( B) P( E) P( E2 ) P( E4 ) P( E6 ) ; and P ( ). Snce 6 2 6 3 event F contans no smple events, PF ( ) 0. 4.2 a It s gven that P( E1) P( E2) 0.15 and PE ( 3) 0.40. Snce PE ( ) 1, we know that P( E4) P( E5) 10.15 0.15 0.40 0.30 () Also, t s gven that P( E4) 2 P( E5) () We have two equatons n two unknowns whch can e solved smultaneously for P(E 4 ) and P(E 5 ). Susttutng equaton () nto equaton (), we have 2 P( E ) P( E ) 0.3 5 5 3 P( E ) 0.3 so that P( E ) 0.1 5 5 Then from (), P( E4) 2 P( E5) 0.2. To fnd the necessary proaltes, sum the proaltes of the smple events: P( A) P( E ) P( E ) P( E ) 0.15 0.4 0.2 0.75 1 3 4 P( B) P( E2) P( E3) 0.15 0.4 0.55 c d The followng events are n ether A or B or oth: { E 1, E 2, E 3, E 4 }. Only event E 3 s n oth A and B. 4.3 It s gven that PE ( 1) 0.45 and that 3 PE ( 2) 0.45, so that PE ( 2) 0.15. Snce PE ( ) 1, the remanng 8 smple events must have proaltes whose sum s P( E ) P( E )... P( E ) 10.45 0.15 0.4 3 4 10 Snce t s gven that they are equproale, 0.4 P( E ) 0.05 for 3, 4,..., 10 8 4.4 a It s requred that PE ( ) 1. Hence, PE ( 2) 10.49 0.21 0.09 0.21 S The player wll ht on at least one of the two free throws f he hts on the frst, the second, or oth. The assocated smple events are E 1, E 2, and E 3 and P(hts on at least one) P( E ) P( E ) P( E ) 0.91 1 2 3 S S NEL opyrght 2014 Nelson Educaton Lmted 4-1
Instructor s Solutons Manual to Accompany Introducton to Proalty and Statstcs, 3E 4.9 The four possle outcomes of the experment, or smple events, are represented as the cells of a 22tale, and have proaltes as gven n the tale. a P(adult judged to need glasses) = 0.44 + 0.14 = 0.58 P(adult needs glasses ut does not use them) = 0.14 c P(adult uses glasses) = 0.44 + 0.02 = 0.46 4.14 a Vsualze a tree dagram wth four stages selectng the runner who places frst, second, thrd, and fourth, respectvely. There are four choces at the frst stage, three choces (ranches) at the second stage, two choces (ranches) at the thrd stage, and only one choce (ranch) for the last runner. The total numer of smple events s 4(3)(2)(1) 24. The smple events are lsted elow: JBED JEBD JEDB JBDE JDEB JDBE BJED BEJD BEDJ BJDE BDEJ BDJE EBJD EJBD EJDB EBDJ EDJB EDBJ DBEJ DEBJ DEJB DBJE DJEB DJBE If all runners are equally qualfed, the smple events wll e equally lkely wth proalty PE ( ) 1 24. c e Sum the proaltes of the approprate smple events: 6 1 2 1 6 1 P(Dave wns), P(Dave s frst, John s second), and P(Ed s last) 24 4 24 12 24 4 4.15 Smlar to Exercse 4.9. The four possle outcomes of the experment, or smple events, are represented as the cells of a 22tale, and have proaltes (when dvded y 300) as gven n the tale. a P(normal eyes and normal wng sze) 140 300 0.467 P(vermllon eyes) (3151) 300 154 300 0.513 c P(ether vermllon eyes or mnature wngs or oth) (3151 6) 300 160 300 0.533 4.40 Each smple event s equally lkely, wth proalty 1 5. a A E2, E4, E5 P A 35 AB E 1 P AB 15 c B E 4 PB 15 d A B S E1, E2, E3, E4, E5 P AB 1 e B E 4 P( B ) 1 2 f A B E P( A B) 1 4 g A B S 1 P A B 1 h AB E, E, E, E P AB 45 4.41 a P A P A 2 3 4 5 2 3 1 1 5 5 1 4 1 1 5 5 P A B P A B 4.42 a P A B P( A B) 1 5 1 PB ( ) 4 5 4 4-2 opyrght 2014 Nelson Educaton Lmted NEL
Instructor s Solutons Manual to Accompany Introducton to Proalty and Statstcs, 3E PB P( B ) 1 5 1 P ( ) 2 5 2 4.43 a P A B P( A) P( B) P( A B) 2 5 4 51 5 5 5 1 P( A B) P( A B) P( B) 1 44 5 1 5 c P( B ) P( B ) P( ) 1 22 5 1 5 4.44 a From Exercse 4.40, P A B 1 4 whle PA ( ) 2 5. Therefore, A and B are not ndependent. From Exercse 4.40, P( AB) 1 5. Snce P( AB) 0, A and B are not mutually exclusve. 4.47 a Snce A and B are ndependent, P( A B) P( A) P( B) 0.4(0.2) 0.08. P( A B) P( A) P( B) P( A B) 0.4 0.2 (0.4)(0.2) 0.52 4.48 a Snce A and B are mutually exclusve, P( AB) 0. P( A B) P( A) P( B) P( A B) 0.3 0.5 0 0.8 4.49 a Use the defnton of condtonal proalty to fnd P( A B) 0.12 PB A 0.3 PA ( ) 0.4 Snce P( AB) 0, A and B are not mutually exclusve. c If PB ( ) 0.3, then P( B) P( B A), whch means that A and B are ndependent. 4.50 a The event A wll occur whether event B occurs or not (B ). Hence, P( A) P( A B) P( A B ) 0.34 0.15 0.49 Smlar to part a. P( B) P( A B) P( A B) 0.34 0.46 0.80 c The contents of the cell n the frst row and frst column s P( AB) 0.34. d P( A B) P( A) P( B) P( A B) 0.49 0.80 0.34 0.95 e Use the defnton of condtonal proalty: P( A B) 0.34 P A B 0.425 PB ( ) 0.80 f Smlar to part e. P( A B) 0.34 PB A 0.694 PA ( ) 0.49 4.51 a From Exercse 4.50, snce P( AB) 0.34, the two events are not mutually exclusve. From Exercse 4.50, P A B 0.425 and PA ( ) 0.49. The two events are not ndependent. 4.52 Defne the followng events: P: test s postve for drugs N: test s negatve for drugs D: employee s a drug user It s gven that, on a gven test, P( P D) 0.98, P( N D ) 0.98. a For a gven test, P( P D ) 10.98 0.02. Snce the tests are ndependent P(fal oth test D ) (0.02)(0.02) 0.0004 P(detecton D) P( N P D) P( P N D) P( P P D) (0.98)(0.02) (0.02)(0.98) (0.98)(0.98) 0.9996 c P(pass oth D) P( N N D) (0.02)(0.02) 0.0004 NEL opyrght 2014 Nelson Educaton Lmted 4-3
Instructor s Solutons Manual to Accompany Introducton to Proalty and Statstcs, 3E 4.53 Defne the followng events: A: project s approved for fundng D: project s dsapproved for fundng For the frst group, PA ( 1) 0.2 and PD ( 1) 0.8. For the second group, P(same decson as frst group) 0.7 and P(reversal) 0.3. That s, P( A2 A1 ) P( D2 D1 ) 0.7 and P( A D ) P( D A ) 0.3. 2 1 2 1 a P( A1 A2 ) P( A1 ) P( A2 A1 ) 0.2(0.7) 0.14 P( D1 D2 ) P( D1 ) P( D2 D1 ) 0.8(0.7) 0.56 c P D1 A2 P A1 D2 P D1 P A2 D1 P A1 P D2 A1 ( ) ( ) ( ) ( ) ( ) ( ) 0.8(0.3) 0.2(0.3) 0.30 4.58 Defne the followng events: S: student chooses Starucks T: student chooses Tm Hortons D: student orders a decaffenated coffee Then P( S) 0.3; P( T) 0.7; P( D S) P( D T) 0.60 a Usng the gven proaltes, P( T D) P( T) P( D T) 0.7(0.6) 0.42 Snce PD ( ) 0.6 regardless of whether the student vsts Starucks or Tm Hortons, the two events are ndependent. P( S D) P( S) P( D S) c P( S D) P( S) 0.3 P( D) P( D) d P( T D) P( T) P( D) P( T D) 0.7 0.6 (0.7)(0.6) 0.88 4.61 Defne the events: D: person des S: person smokes It s gven that PS ( ) 0.2, PD ( ) 0.006, and P( D S) 10 P( D S ). The proalty of nterest s P( D S ). The event D, whose proalty s gven, can e wrtten as the unon of two mutually exclusve ntersectons. That s, D ( D S) ( D S ). Then, usng the Addton and Multplcaton Rules, PD P( D S) P( D S ) P( D S) P( S) P( D S ) P( S ) P( D S)(0.2) 1 10 P( D S) (0.8) Snce PD ( ) 0.006, the aove equaton can e solved for P( D S ) P( D S)(0.2 0.08) 0.006 P( D S) 0.006 0.28 0.0214 4.63 Defne A: smoke s detected y devce A B: smoke s detected y devce B If t s gven that P( A) 0.95, P( B) 0.98, and P( A B) 0.94 a P( A B) P( A) P( B) P( A B) 0.95 0.98 0.94 0.99 ( P A B ) 1 P( A B) 10.99 0.01 4.64 Usng the proalty tale gven n the queston, we fnd a PA ( ) 0.42 P( AG) 0.22 0.22 c P( A G) 0.4231 0.52 d PG ( ) 0.52 e P( AG) P( A) P( G) P( AG) 0.42 0.52 0.22 0.72 f 0 P( D O) 0 0.06 4-4 opyrght 2014 Nelson Educaton Lmted NEL
Instructor s Solutons Manual to Accompany Introducton to Proalty and Statstcs, 3E g If events A and G are ndependent then must P( A G) P( A) e true. From parts c and a, we can see these proaltes are almost equal. Thus, A and G are ndependent. h D and O are mutually exclusve snce P( D O) 0.. D and O are not ndependent. If these two events are ndependent then P( D O) P( D) must e true. PD ( ) 0.09 whereas P( D O) 0. Also, mutually exclusve events are always dependent. j. If A and G are mutually exclusve, then P( A G) must e zero. Ths proalty s 0.22; thus, these two events are not mutually exclusve. 4.67 Smlar to Exercse 4.54. 54 a PA ( ) 1029 517 PF ( ) 1029 37 c P AF 1029 P( F A) 37 1029 37 d PF A PA ( ) 54 1029 54 e PF B f PF g P M h P( F B) 64 1029 64 PB ( ) 99 1029 99 P( F ) 138 1029 138 P ( ) 241 1029 241 P( M ) 103 1029 103 PM ( ) 512 1029 512 99 930 310 P( B ) 1 P( B) 1 1029 1029 343 4.74 Use the Law of Total Proalty: P( A) P( S1) P( A S1) P( S2) P( A S2) 0.6(0.3) 0.4(0.5) 0.38 4.75 Defne the followng events: V: crme s volent R: crme s reported It s gven that ( ) 0.2, ( P V P V ) 0.8, P( R V) 0.9, P( R V ) 0.7. a The overall reportng rate for crmes s P( R) P( V) P( R V) P( V ) P( R V ) 0.2(0.9) 0.8(0.7) 0.74 c Use Bayes Rule: P( V ) P( R V ) 0.2(0.9) P( V R) 0.24 PR ( ) 0.74 P( V ) P( R V ) 0.8(0.7) and P( V R) 0.76 PR ( ) 0.74 Notce that the proporton of non-volent crmes (0.8) s much larger than the proporton of volent crmes (0.2). Therefore, when a crme s reported, t s more lkely to e a non-volent crme. 4.76 Defne A: machne produces a defectve tem B: worker follows nstructons Then ( ) 0.01, ( ) 0.90, ( P A B P B P A B ) 0.03, P( B ) 0.10 NEL opyrght 2014 Nelson Educaton Lmted 4-5
Instructor s Solutons Manual to Accompany Introducton to Proalty and Statstcs, 3E The proalty of nterest s P( A) P( A B) P( A B ) P( A B) P( B) P( A B ) P( B ) 0.01(0.90) 0.03(0.10) 0.012 4.77 Defne the followng events: A: passenger uses arport A B: passenger uses arport B : passenger uses arport D: a weapon s detected Suppose that a passenger s carryng a weapon. It s gven that P( D A) 0.9 P( A) 0.5 P( D B) 0.5 P( B) 0.3 P( D ) 0.4 P( ) 0.2 The proalty of nterest s P( A) P( D A) 0.5(0.9) P( A D) 0.6618 P ( A ) P ( D A ) P ( B ) P ( D B ) P ( ) P ( D ) 0.5(0.9) 0.3(0.5) 0.2(0.4) 0.2(0.4) 0.08 Smlarly, P ( D ) 0.1176 0.5(0.9) 0.3(0.5) 0.2(0.4) 0.68 4.80 The proalty of nterest s P( A H ), whch can e calculated usng Bayes Rule and the proaltes gven n the exercse. P( A) P( H A) P( A H) P( A) P( H A) P( B) P( H B) P( ) P( H ) 0.01(0.90) 0.009 0.3130 0.01(0.90) 0.005(0.95) 0.02(0.75) 0.02875 4.82 a Usng the proalty tale, PD ( ) 0.08 0.02 0.10 P( D ) 1 P( D) 10.10 0.90 c P( N D ) 0.85 P( N D) 0.02 P( N D ) 0.94 P( N D) 0.20 PD ( ) 0.90 PD ( ) 0.10 P( D) P( N D) 0.10(0.20) Usng Bayes Rule, P( D N) 0.023 P( D) P( N D) P( D ) P( N D ) 0.10(0.20) 0.90(0.94) Usng the defnton of condtonal proalty, P( N D) 0.02 P( D N) 0.023 PN ( ) 0.87 d P e P f P( P D ) 0.05 false postve P( P D ) 0.056 PD ( ) 0.90 P( N D) 0.02 false negatve P( N D) 0.20 PD ( ) 0.10 The proalty of a false negatve s qute hgh, and would cause concern aout the relalty of the screenng method. 4-6 opyrght 2014 Nelson Educaton Lmted NEL