4: Probability and Probability Distributions

Transcription

1 : Probability and Probability Distributions. a This experiment involves tossing a single die and observing the outcome. The sample space for this experiment consists of the following simple events: E : Observe a E : Observe a E : Observe a E 5 : Observe a 5 E : Observe a E 6 : Observe a 6 b Events A through F are compound events and are composed in the following manner: A: (E ) B: (E, E, E 6) D: (E ) E: (E, E, E 6 ) : (E, E, E 5, E 6) F: contains no simple events c Since the simple events E i, i =,,,, 6 are equally likely, PE ( i ) = 6. d To find the probability of an event, we sum the probabilities assigned to the simple events in that event. For example, PA ( ) = PE ( ) = 6 Similarly, PD ( ) = 6; PB ( ) = PE ( ) = PE ( ) + PE ( ) + PE ( 6) = = ; and 6 contains no simple events, PF ( ) = 0. P ( ) = =. Since event F 6. a It is given that PE ( ) = PE ( ) =.5 and PE ( ) =.0. Since PE ( i ) =, we know that PE ( ) + PE ( ) = =.0 5 S (i) Also, it is given that PE ( ) = PE ( 5 ) (ii) We have two equations in two unknowns which can be solved simultaneously for P(E ) and P(E 5). Substituting equation (ii) into equation (i), we have PE ( ) + PE ( ) =. 5 5 PE ( ) =. so that PE ( ) =. 5 5 Then from (i), PE ( ) +. =. and PE ( ) =.. b To find the necessary probabilities, sum the probabilities of the simple events: PA ( ) = PE ( ) + PE ( ) + PE ( ) = =.75 PB ( ) = PE ( ) + PE ( ) =.5 +. =.55 c-d The following events are in either A or B or both: { E, E, E, E }. Only event E is in both A and B.. It is given that PE ( ) =.5and that PE ( ) =.5, so that PE ( ) =.5. Since PE ( i ) =, the remaining 8 simple events must have probabilities whose sum is PE ( ) + PE ( ) PE ( 0) =.5.5 =.. Since it is given that they are equiprobable,. PE ( i ) = =.05 for i=,,...,0 8 i S. a It is required that PE ( ) =. Hence, PE ( ) = =. 80 S

2 b The player will hit on at least one of the two freethrows if he hits on the first, the second, or both. The associated simple events are E, E, and E and P(hits on at least one) = P( E ) + P( E ) + P( E ) =.9.5 a The experiment consists of choosing three coins at random from four. The order in which the coins are drawn is unimportant. Hence, each simple event consists of a triplet, indicating the three coins drawn. Using the letters N, D, Q, and H to represent the nickel, dime, quarter, and half-dollar, respectively, the four possible simple events are listed below. E : (NDQ) E : (NDH) E : (NQH) E : (DQH) b The event that a half-dollar is chosen is associated with the simple events E, E, and E. Hence, P[choose a half-dollar] = P ( E) + P( E) + P( E) = + + = since each simple event is equally likely. c The simple events along with their monetary values follow: E NDQ $0.0 E NDH 0.65 E NQH 0.80 E DQH 0.85 Hence, P[total amount is $0.60 or more] = P( E) + P( E) + P( E) =..6 a The experiment consists of selecting one of 5 students and recording the student s gender as well as whether or not the student had gone to preschool. b The experiment is accomplished in two stages, as shown in the tree diagram below. Gender Preschool Simple Events Probability Yes E : Male, Preschool 8/5 Male No E : Male, No preschool 6/5 Female Yes E : Female, Preschool 9/5 No E : Female, No preschool /5 c Since each of the 5 students are equally likely to be chosen, the probabilities will be proportional to the number of students in each of the four gender-preschool categories. These probabilities are shown in the last column of the tree diagram above. d P(Male) = P( E 8 6 ) + P( E) = + = ; P(Female and no preschool) = P( E ) = Label the five balls as R, R, R, Y and Y. The selection of two balls is accomplished in two stages to produce the simple events in the tree diagram on the next page. 8

3 First Ball Second Ball Simple Events First Ball Second Ball Simple Events R R R R Y R R R R R Y R R Y R Y Y R Y R Y R Y Y Y Y R R R R Y R R R R R Y R Y R Y R Y R Y R Y R Y Y Y Y R R R R R R R Y R Y Y R Y.8 When the first ball is replaced before the second ball is drawn, five additional simple events become possible: R R, R R, R R, Y Y, and Y Y.9 The four possible outcomes of the experiment, or simple events, are represented as the cells of a table, and have probabilities as given in the table. a P[adult judged to need glasses] =. +. =.58 b P[adult needs glasses but does not use them] =. c P[adult uses glasses] =. +.0 =.6.0 a There are 8 simple events, each corresponding to a single outcome of the wheel s spin. The 8 simple events are indicated below. E : Observe a E : Observe a M E 6 : Observe a 6 E 7 : Observe a 0 E 8 : Observe a 00 b Since any pocket is just as likely as any other, PE ( i ) = 8. c The event A contains two simple events, E 7 and E 8. Then PA ( ) = PE ( ) + PE ( ) = 8= d Define event B as the event that you win on a single spin. Since you have bet on the numbers through 8, event B contains 8 simple events, E, E,, E 8. Then PB ( ) = PE ( ) + PE ( ) + L + PE ( ) = 8 8 = a Experiment: Select three people and record their gender (M or F). b Extend the tree diagram in Figure. of the text to include one more coin toss (a total of n = ). Then replace the H and T by M and F to obtain the 8 possible simple events shown below: FFF FMM MFM MMF MFF FMF FFM MMM c Since there are N = 8 equally likely simple events, each is assigned probability, PE ( i ) = N= 8. d-e Sum the probabilities of the appropriate simple events: P(only one man) = P( MFF) + P( FMF) + P( FFM ) = = 8 8 P(all three are women) = P( FFF) = 8. a Similar to Exercise.7. List the six jurors as M, M, M, M, F and F. The simple events can be found using a tree diagram and are shown below: 8

4 b M M M M M M M F M F M M M M M F M F M M M F M F M F M F F F Sum the probabilities of the appropriate simple events: P(both jurors are women) = P( FF ) = 5. a Experiment: A taster tastes and ranks three varieties of tea A, B, and, according to preference. b Simple events in S are in triplet form. E :(,,) E :(,,) E : (,, ) E : (,,) 5 E : (,, ) E6 : (,, ) Here is assigned to the most desirable, to the next most desirable, and to the least desirable. c Define the events D: variety A is ranked first F: variety A is ranked third Then PD ( ) = PE ( ) + PE ( ) = 6+ 6= The probability that A is least desirable is PF ( ) = PE ( ) + PE ( ) = 6+ 6= 5 6. a Visualize a tree diagram with four stages selecting the runner who places first, second, third and fourth, respectively. There are four choices at the first stage, three choices (branches) at the second stage, two choices (branches) at the third stage, and only one choice (branch) for the last runner. The total number of simple events is ()()() =. The simple events are listed below: JBED JEBD JEDB JBDE JDEB JDBE BJED BEJD BEDJ BJDE BDEJ BDJE EBJD EJBD EJDB EBDJ EDJB EDBJ DBEJ DEBJ DEJB DBJE DJEB DJBE b If all runners are equally qualified, the simple events will be equally likely with probability PE ( i ) =. c-e Sum the probabilities of the appropriate simple events: 6 6 P (Dave wins) = = P (Dave is first, John is second) = = and P (Ed is last) = =.5 Similar to Exercise.9. The four possible outcomes of the experiment, or simple events, are represented as the cells of a table, and have probabilities (when divided by 00) as given in the table. a P (normal eyes and normal wing size) = 0 00 =.67 b P (vermillion eyes) = ( + 5) 00 = 5 00 =.5 c P (either vermillion eyes or miniature wings or both) = ( ) 00 = =.5.6 a-b The five opinions available to each person are the simple events. They are not equally likely, but have the probabilities given in the table. c-d Sum the appropriate simple event probabilities: P(at least somewhat favorably) = P(somewhat favorably) + P(very favorably) =. +.0 =.5 P(very unfavorably) =.0.7 Use the mn Rule. There are 0(8) = 80 possible pairs..8 Use the extended mn Rule. There are (7)() = 8 possible triplets. 8

5 5 5! 0 0!.9 a P = = 5()() = 60 b P 9 = =, 68,800!! 6 6! 0 0! c P 6 = = 6! = 70 d P = = 0 0! 9! 5 5! 5().0 a = = = 0 b!! () c 6! 6!0! 6 6 = = d 0! 9!! 0 9 = = 0!!9! 0 0 = = 0 8 8!. Since order is important, you use permutations and P 5 = = 8(7)(6)(5)() = 670.! 0 0! 0(9). Since order is unimportant, you use combinations and = = = 90.!8! (). Use the extended mn Rule. The first die can fall in one of 6 ways, and the second and third die can each fall in one of 6 ways. The total number of simple events is 6(6)(6) 6 =.. Use the extended mn Rule. Each coin can fall in one of n i = ways, and the total number of simple events is ()()() = = ! 0(9)(8).5 Since order is unimportant, you use combinations and = = = 0.!7! ()().6 Use the extended mn Rule. There are m = pairs of jeans, n = T-shirts and t = pairs of sneakers. The total number of outfits is mnt = ()() = 9..7 This exercise involves the arrangement of 6 different cities in all possible orders. Each city will be visited once and only once. Hence, order is important and elements are being chosen from a single set. Permutations are used and the number of arrangements is 6 6! P 6 = = 6! = 6(5)()()()() = 70 0!.8 Use the extended mn Rule. The total number of options are (5)() = 60.9 a Each student has a choice of 5 cards, since the cards are replaced between selections. The mn Rule allows you to find the total number of configurations for three students as 5(5)(5) = 0,608. b Now each student must pick a different card. That is, the first student has 5 choices, but the second and third students have only 5 and 50 choices, respectively. The total number of configurations is found using the mn Rule or the rule for permutations: 5 5! mnt = 5(5)(50) =, 600 or P = =, ! c Let A be the event of interest. Since there are 5 different cards in the deck, there are 5 configurations in which all three students pick the same card (one for each card). That is, there are n A = 5 ways for the event A to occur, out of a total of N = 0, 608 possible configurations from part a. The probability of interest is n 5 PA ( ) = A = =.0007 N 0, 608 8

6 d Again, let A be the event of interest. There are n A =,600 ways (from part b) for the event A to occur, out of a total of N = 0,608 possible configurations from part a, and the probability of interest is n A, 600 PA ( ) = = =.9 N 0, Use the extended mn Rule. The total number of options are ()() = 8.. a Since the order of selection for the five-card hand is unimportant, use combinations to find the 5 5! 5(5)(50)(9)(8) number of possible hands as N = 5 = = =,598,960. 5!7! 5()()()() b Since there are only four different suits, there are n A = ways to get a royal flush. c From parts a and b, n P(royal flush) = A = = N,598,960. a From Exercise.a, N =,598,960. b The count is accomplished in two stages. There are different ways to get four cards of the same face value (four Aces, four twos, and so on). Once these four cards are chosen, there are 8 cards left from which to choose the fifth card. Using the mn Rule, the total number of ways is (8) = 6. c From parts a and b, n 6 P(four of a kind) = A = =.000 N,598,960. Notice that a sample of 0 nurses will be the same no matter in which order they were selected. Hence, order is unimportant and combinations are used. The number of samples of 0 selected from a total of 90 is 9 ( ) 6 ( ) 90! !80! = = = ( ) a Since the order of selection for the two-person subcommittee is unimportant, use combinations to find 5 5! 5() the number of possible subcommittees as N = = = = 0!! ()() b Since there is only n = subcommittee made up of Smith and Jones, the probability of this choice is 0. A.5 a Use the mn Rule. The Western conference team can be chosen in one of m = 7 ways, while there are 6 ways to choose the Eastern conference team, for a total of N = mn= 7(6) = possible pairings. b You must choose Los Angeles from the first group and New York from the second group, so that n A = ()() = and the probability is na N =. c Since there are two alifornia teams in the Western conference, there are two choices for the first team and eight choices for the second team. Hence n A = ()(6) = and the probability is n N = = / 7. A.6 a There are P = ()()() = possible orders of finish. b There is only one choice for the first position (the winner must be Dave) and, since the remaining runners may occupy any of the remaining three positions, there are P = ()() = 6 possible arrangements. The probability that Dave wins is na N = 6=. 85

7 c There is only one choice for the first position (Dave), one choice for the second position (John), and P = ways to arrange the other two runners. The probability that Dave wins and John finishes second is then n N = =. A ( ) d There is only choice for the last position (Ed) and, since the remaining runners may occupy any of the remaining three positions there are P = ()() = 6 possible arrangements. The probability that Ed is last is na N = 6=..7 The situation presented here is analogous to drawing 5 items from a jar (the five members voting in favor of the plaintiff). If the jar contains 5 red and white items (5 women and men), what is the probability that all five items are red? That is, if there is no sex bias, five of the eight members are randomly chosen to be those voting for the plaintiff. What is the probability that all five are women? There are 8 8! N = 5 = = 56 5!! simple events in the experiment, only one of which results in choosing 5 women. Hence, P (five women) = The instructor can select five out of the list of ten questions in 5 ways. If the student is to be able to answer all five questions, the instructor must choose five questions out of the six which the student can answer, and none of the four questions which the student cannot answer. The number of ways in which this 6 event can occur is 5 0. Hence the probability that the student can answer all five questions is = 5 5 =.9 The monkey can place the twelve blocks in any order. Each arrangement will yield a simple event and hence the total number of simple events (arrangements) is P =! It is necessary to determine the number of simple events in the event of interest (that he draws three of each kind, in order). First, he may draw the four different types of blocks in any order. Thus we need the number of ways of arranging these four items, which is P =! Once this order has been chosen, the three squares can be arranged in P =! ways, the three triangles can be arranged in P =! ways, and so on. Thus the total number of simple events in the event of interest is ( ) P P P P and the associated probability is ( P ) =!(!)!.0 Follow the instructions given in the MyPersonal Trainer section. The answers are given in the table. P(A) P(B) onditions for events A and B P(A B) P(A B) P(A B).. Independent.. +. (.)(.) = Mutually exclusive Mutually exclusive = Independent (.)(.5) = (.)(.5) =

8 . Follow the instructions given in the MyPersonal Trainer section. The answers are given in the table. P(A) P(B) onditions for events A and B P(A B) P(A B) P(A B).. Mutually exclusive = Independent.(.) =.. +. (.)(.) = Independent.(.5) = (.)(.5) = Mutually exclusive =.7 0. Each simple event is equally likely, with probability 5. a A = { E, E, E5} P( A ) = 5 b A B = { E } P( A B) = 5 c B = { E } P( B ) = 5 d A B = S = { E, E, E, E, E5} P( A B) = e B = { E } PB ( ) = f A B = { E } PA ( B ) = g A B S = P( A B ) = h ( A B) = { E, E, E, E 5 } P( A B) = 5. a P( A ) P( A) = = = 5 5 = = = 5 5 b P( A B) P( A B). a P( A B) b P( B ) PA ( B) 5 = = = PB ( ) 5 PB ( ) 5 = = = P ( ) 5.5 a P( A B) = P( A) + P( B) P( A B) = = 5 5= b PA ( B) = PA ( BPB ) ( ) = ( )( 5) = 5 c PB ( ) = PB ( P ) ( ) = ( )( 5) = 5.6 a From Exercise., ( ) P A B = while PA= ( ) 5. Therefore, A and B are not independent. b From Exercise., PA ( B) = 5. Since PA ( B) 0, A and B are not mutually exclusive..7 Refer to the solution to Exercise. where the six simple events in the experiment are given, with PE ( i ) = 6. a S = E, E, E, E, E, E and PS ( ) = 6 6= b P( A B) c { } 5 6 PA ( B) = = = PB ( ) { } B = E, E and PB ( ) = 6= 87

9 d A B contains no simple events, and P( A B ) = 0 e PA ( B) = PA ( BPB ) ( ) = ( ) = f A g B contains no simple events, and P( A ) = 0 contains no simple events, and P( B ) = 0 h A S i = and P( A ) = {,,,, } B = E E E E E and PB ( ) = a From Exercise.7, PA ( B) are not independent. b ( ) =, ( ) P A B =, ( ) PA=, P( A B) PA ( B) 0, so that A and B are not mutually exclusive. P A = P( A ) P( ) = 0, PA= ( ), PA ( ) 0 dependent. Since PA ( ) = 0, A and are mutually exclusive. =. Since P( A ).9 a Since A and B are independent, PA ( B) = PAPB ( ) ( ) =.(.) =.08 b PA ( B) = PA ( ) + PB ( ) PA ( B) =. +. (.)(.) =.5.50 a Since A and B are mutually exclusive, PA ( B) = 0. b PA ( B) = PA ( ) + PB ( ) PA ( B) = =.8.5 a Use the definition of conditional probability to find PA ( B). P( B A) = = =. PA ( ). b Since PA ( B) 0, A and B are not mutually exclusive. c If PB ( ) =., then PB ( ) = PB ( A) which means that A and B are independent..5 a The event A will occur whether event B occurs or not (B ). Hence, PA ( ) = PA ( B) + PA ( B ) =. +.5 =.9 b Similar to part a. PB ( ) = PA ( B) + PA ( B) =. +.6 =.80. c The contents of the cell in the first row and first column is PA ( B) =.. d PA ( B) = PA ( ) + PB ( ) PA ( B) = =.95. e Use the definition of conditional probability PA ( B). P( A B) = = =. 5 PB ( ).80 f Similar to part e. PA ( B). P( B A) = = =. 69 PA ( ).9. P( A), so that A and B.5 a From Exercise.5, since PA ( B) =., the two events are not mutually exclusive. ( ) P( A), A and are b From Exercise.5, P A B =.5 and PA= ( ).9. The two events are not independent..5 Define the following events: P: test is positive for drugs N: test is negative for drugs D: employee is a drug user It is given that, on a given test, PP ( D) =.98, PN ( D ) =.98. a For a given test, PP ( D ) =.98 =.0. Since the tests are independent 88

10 P(fail both test D ) = (.0)(.0) =.000 b P(detection D) = P( N P D) + P( P N D) + P( P P D) = (.98)(.0) + (.0)(.98) + (.98)(.98) =.9996 c P(pass both D) = P( N N D) = (.0)(.0) = Define the following events: A: project is approved for funding D: project is disapproved for funding For the first group, ( ).and ( ).8 PA = and P [ reversal ] =.. That is, PD =. For the second group, [ ] PA ( A) = PD ( D) =.7 and PA ( D) = PD ( A) =.. PA ( A) = PA ( ) PA ( A) =.(.7) =. a b PD ( D) = PD ( ) PD ( D) =.8(.7) =.56 c PD A PA D PD PA D PA PD A ( ) + ( ) = ( ) ( ) + ( ) ( ) =.8(.) +.(.) = P same decision as first group =.7.56 The two-way table in the text gives probabilities for events A, A, B, B in the column and row marked Totals. The interior of the table contains the four two-way intersections as shown below. A B A B A B A B The necessary probabilities can be found using various rules of probability if not directly from the table. a PA= ( ). b PB ( ) =.7 c P( A B) =.0 d P( A B) = =.67 e PA ( ) =. =.6 f P( A B) P( A B).67. = = = g P( A B) = P( A B) =.90 h P( A B) P A B P B i P( B A) P A B P A = ( ) ( ) =..7 =. 7 = ( ) ( ) =.. = Refer to Exercise.56. a From the table, P A B while ( ) =. ( ) ( B) =. ( ) ( B) b From the table, P A while PAPB ( ) A = (.)(.5) =. PBP ( ) A B = (.7)(..7) =. c From the table, P A = + + = while PA ( ) + PB ( ) PA ( B) = = a Define a birth date by using a three-digit number between 00 and 65. A simple event will be a pair of numbers, each between 00 and 65. A typical simple event might be (6, 0), for two people entering the room. b The probability that a person has a birthday on any specific date is /65. Hence, for two people the probability of a specific pair of dates is ( 65 ). c The simple events in A are {(00, 00), (00, 00),, (65, 65)}. There are 65 such events. PA= 65 ( ) (65) = 65, assuming all simple events are equally likely. d e 6 PA ( ) = PA ( ) = 65

11 .59 Fix the birth date of the first person entering the room. Then define the following events: A : second person s birthday differs from the first A : third person s birthday differs from the first and second A : fourth person s birthday differs from all preceding M A n : n th person s birthday differs from all preceding Then n + PA ( ) = PA ( ) PA ( ) LPA ( n ) = L since at each step, one less birth date is available for selection. Since event B is the complement of event A, PB ( ) = PA ( ) (6)(6) a For n =, PA ( ) = =.998 and PB ( ) =.998 =.008 (65) (6)(6)(6) b For n =, PA ( ) = =.986 and PB ( ) =.986 =.06 (65).60 Define the following events: S: student chooses Starbucks P: student chooses Peetes : student orders a café mocha Then PS ( ) =.7; PP ( ) =.; P ( S) = P ( P) =.60 a Using the given probabilities, PS ( ) = PSP ( ) ( S) =.7(.6) =. b Since P ( ) =.6regardless of whether the student visits Starbucks or Peetes, the two events are independent. PP ( ) PPP ( ) ( P) c PP ( ) = = = PP ( ) =. P ( ) P ( ) d PS ( ) = PS ( ) + P ( ) PS ( ) = (.7)(.6) =.88.6 Let events A and B be defined as follows: A: article gets by the first inspector B: article gets by the second inspector The event of interest is then the event A B, that the article gets by both inspectors. It is given that PA= ( )., and also that PB ( A ) =.5. Applying the Multiplication Rule, P( A B) = P( A) P( B A) = (.)(.5) =.05.6 Define the events: D: person dies S: person smokes It is given that PS ( ) =., PD ( ) =.006, and PD ( S) = 0 PD ( S ). The probability of interest is PD ( S). The event D, whose probability is given, can be written as the union of two mutually exclusive intersections. That is, D = ( D S) ( D S ) Then, using the Addition and Multiplication Rules, P( D) = PD ( S) + PD ( S ) = PD ( SPS ) ( ) + PD ( S ) PS ( ) = PD ( S)(.) + ( 0 ) PD ( S) (.8) Since PD ( ) =.006, the above equation can be solved for PD ( S) PD ( S)(. +.08) =.006 PD ( S) = =.0 90

12 .6 Define A: smoke is detected by device A B: smoke is detected by device B If it is given that PA ( ) =.95, PB ( ) =.98, and PA ( B) =.9. a PA ( B) = PA ( ) + PB ( ) PA ( B) = =.99 b ( PA B) = PA ( B) =.99=.0.6 a Each of the four cell events is equally likely with PE ( i ) =. Hence, the probability of at least one dominate allele is PrR ( ) + PRr ( ) + PRR ( ) = b Similar to part a. The probability of at least one recessive allele is PrR ( ) + PRr ( ) + Prr ( ) = c Define the events: A: plant has red flowers B: plant has one recessive allele Then PA ( B) PrR ( ) + PRr ( ) P( B A) = = = = PA ( ) PrR ( ) + PRr ( ) + PRR ( ).65 Similar to Exercise a PA= ( ) b c P( A G) 56 e P( G B) g P( P) = d ( ) PG ( B) 56 = = PB ( ) = f P( G ) P ( P) 56 = = = h PP ( ) PG ( ) = 56 PG ( A) P G A = = = 8 PA ( ) PG ( ) 56 = = = P ( ) PB ( ) = PB ( ) = = Similar to Exercise.56. a PF ( ) = =.7 b PG ( ) = =.9 c P( F M) e P( M F) PF ( M).5 = =. 6 PM ( ).55 PM ( F).5 = =. 9 PF ( ).7 = d P( F W) = e PW ( G) PF ( W).6 = = =. 80 PW ( ).5 PW ( G).09 = = =. PG ( ).9.67 Define the following events: A: player makes first free throw B: player makes second free throw The probabilities of events A and B will depend on which player is shooting. a PA ( B) = PAPB ( ) ( ) =.85(.85) =.75, since the free throws are independent. b The event that Shaq makes exactly one of the two free throws will occur if he makes the first and misses the second, or vice versa. Then P(makes exactly one) = P( A B ) + P( A B) c =.5(.5) +.5(.5) =.5 This probability is the intersection of the individual probabilities for both Shaq and Kobe. P(Shaq makes both and Kobe makes neither) = [.5(.5)][.5(.5)]=.00565

13 .68 Define the following events: A: player A wins the tournament B: player B enters the tournament It is given that PA ( B) = / 6; PA ( B ) = / ; and PB ( ) = /. PA ( ) = PA ( B) + PA ( B ) = PBPA ( ) ( B) + PB ( ) PA ( B ) 0 5 = 6 + = = a Use the Law of Total Probability, writing PA ( ) = PS ( ) PAS ( ) + PS ( ) PAS ( ) =.7(.) +.(.) =. b Use the results of part a in the form of Bayes Rule: PS ( i) PA ( Si) PS ( i A) = PS ( ) PA ( S) PS ( ) PA ( S) +.7(.). For i =, PS ( A ) = = = (.) +.(.)..(.).09 For i =, PS ( A ) = = =.9.7(.) +.(.). Use the Law of Total Probability..70 Use Bayes Rule: PS ( i) PA ( Si) PS ( i A) = PS ( ) PA ( S) + PS ( ) PA ( S) + PS ( ) PA ( S).(.).0 For i =, PS ( A ) = = =..(.) +.5(.) +.(.) For i =, PS ( A ) = =.778 and for i =, PS ( A ) = = Use the Law of Total Probability, writing PA ( ) = PS ( ) PAS ( ) + PS ( ) PAS ( ) =.6(.) +.(.5) =.8..7 Define the following events: V: crime is violent R: crime is reported It is given that PV ( ) =., PV ( ) =.8, PR ( V) =.9, P( R V ) =.7 a The overall reporting rate for crimes is PR ( ) = PV ( ) PRV ( ) + PV ( ) PRV ( ) =.(.9) +.8(.7) =.7 b Use Bayes Rule: PV ( ) P( R V).(.9) PV ( R) = = =. PR ( ).7 PV ( ) P( R V ).8(.7) and PV ( R) = = =.76 PR ( ).7 c Notice that the proportion of non-violent crimes (.8) is much larger than the proportion of violent crimes (.). Therefore, when a crime is reported, it is more likely to be a non-violent crime..7 Define A: machine produces a defective item B: worker follows instructions Then PA ( B) =.0, PB ( ) =.90, PA ( B) =.0, PB ( ) =.0. The probability of interest is 9

14 PA ( ) = PA ( B) + PA ( B) = PA ( BPB ) ( ) + PA ( B) PB ( ) =.0(.90) +.0(.0) =.0.7 Define the following events: A: passenger uses airport A B: passenger uses airport B : passenger uses airport D: a weapon is detected Suppose that a passenger is carrying a weapon. It is given that PD ( A) =.9 PA ( ) =.5 PD ( B) =.5 PB ( ) =. PD ( ) =. P ( ) =. The probability of interest is PAPD ( ) ( A).5(.9) PA ( D) = = =.668 PAPD ( ) ( A) + PBPD ( ) ( B) + PPD ( ) ( ).5(.9) +.(.5) +.(.) Similarly,.(.).08 P ( D ) = = =.76.5(.9) +.(.5) +.(.) Define L: play goes to the left R: play goes to the right S: right guard shifts his stance a It is given that PL ( ) =., PR ( ) =.7, PS ( R) =.8, ( PS L) =.9, PS ( L) =., and PS ( R) =.. Using Bayes Rule, PLPS ( ) ( L).(.9).7 PL ( S ).6585 PLPS ( ) ( L) PRPS ( ) ( R) = = +.(.9) +.7(.). b From part a, ( PR S ) = PL ( S ) =.6585 =.5. c Given that the guard takes a balanced stance, it is more likely (.6585 versus.5) that the play will go to the left..76 Define the events: A: athlete has been disqualified previously B: athlete is disqualified for the next 6 weeks It is given that PB ( A ) =.5, PB ( A ) =.50, PA= ( ).0. The event of interest is event B, which can be written as the union of two mutually exclusive events: B = ( A B) ( A B) Then, using the Law of Total Probability, P B = P A B + P A B = P B A P A + P B A P A = + = ( ) ( ) ( ) ( ) ( ) ( ) ( ).5(.).5(.7) The probability of interest is PA ( H) which can be calculated using Bayes Rule and the probabilities given in the exercise. PAPH ( ) ( A) PA ( H) = PAPH ( ) ( A) + PBPH ( ) ( B) + PPH ( ) ( ).0(.90).009 = = =.0.0(.90) +.005(.95) +.0(.75) Define the following events, under the assumptions that an incorrect return has been filed. G : individual guilty of cheating G : individual not guilty (filed incorrectly due to lack of knowledge) D: individual denies knowledge of error 9

15 It is given that PG ( ) =.05, PG ( ) =.0, PD ( G) =.80. Note that PD ( G ) = since if the individual has incorrectly filed due to lack of knowledge, he will, with probability deny knowledge of the error. Using Bayes Rule, PG ( ) PD ( G).05(.80) PG ( D) = = =.6667 PG ( ) PD ( G) + PG ( ) PD ( G).05(.80) +.0().79 a Using the probability table, PD ( ) = =.0 PD ( ) = PD ( ) =.0 =.90 b c PN ( D ).85 PN ( D ) = = =.9 PD ( ).90 PN ( D).0 PN ( D) = = =.0 PD ( ).0 PDPN ( ) ( D).0(.0) Using Bayes Rule, PD ( N) =.0 PDPN ( ) ( D) PD ( ) PN ( D ) = = +.0(.0) +.90(.9) Using the definition of conditional probability, PN ( D).0 PD ( N) = = =.0 PN ( ).87 PP ( D ).05 false positive = P( P D ) = = =.056 PD ( ).90 d P( ) PN ( D).0 e P( false negative ) = P( N D) = = =.0 PD ( ).0 f The probability of a false negative is quite high, and would cause concern about the reliability of the screening method..80 a The number of points scored is a discrete random variable taking the countably infinite number of values, 0,,, b Shelf life is a continuous random variable, since it can take on any positive real value. c Height is a continuous random variable, taking on any positive real value. d Length is a continuous random variable, taking on any positive real value. e Number of near collisions is a discrete random variable, taking the values 0,,,.8 a The increase in length of life achieved by a cancer patient as a result of surgery is a continuous random variable, since an increase in life (measured in units of time) can take on any of an infinite number of values in a particular interval. b The tensile strength, in pounds per square inch, of one-inch diameter steel wire cable is a continuous random variable. c The number of deer killed per year in a state wildlife preserve is a discrete random variable taking the values 0,,, d The number of overdue accounts in a department store at a particular point in time is a discrete random variable, taking the values 0,,,. e Blood pressure is a continuous random variable..8 a Since one of the requirements of a probability distribution is that px ( ) =, we need b p () = ( ) =.95 =.05 The probability histogram is shown in the figure on the next page. x 9

16 0. 0. p(x) x 5 c For the random variable x given here, µ = Ex ( ) = xpx ( ) = 0(.) + (.) + L + 5(.05) =.85 The variance of x is defined as σ E ( x µ ) = = ( x µ ) p( x) = (0.85) (.) + (.85) (.) + + (5.85) (.05) =.75 L and σ =.75 =.9. d The interval of interest is µ ± σ =.85±.8or.5 to.. This interval is shown on the probability histogram above. Then P[ x ] P[ x ].5. = 0 =.95. e Since the probability that x falls in the interval µ ± σ is.95 from part d, we would expect most of the observations to fall in this interval..8 a Since one of the requirements of a probability distribution is that px ( ) =, we need b p () = ( ) =.8 =. The probability histogram is shown below. x p(x) x c For the random variable x given here, µ = Ex ( ) = xpx ( ) = 0(.) + (.) + L + (.) =.9 The variance of x is defined as σ E ( x µ ) = = ( x µ ) p( x) = (0.9) (.) + (.9) (.) + + (.9) (.) =.9 L and σ =.9 =.6. 95

17 d Using the table form of the probability distribution given in the exercise, Px> ( ) =. +. =.. e Px ( ) = Px ( = ) =. =.9..8 a Since each of the six possible values, x =,,,, 5, 6 is equally likely with px ( ) = 6for all values of x, the graph of the probability distribution has a flat shape, called the discrete uniform probability distribution, shown in the figure below p(x) x 5 6 b c The average value of x is µ = Ex ( ) = xpx ( ) = = L 6 The variance of x is defined as σ E ( x µ ) = = ( x µ ) p( x) = (.5) + (.5) + + (6.5) =.967 L and σ =.967 =.7. d The interval of interest is µ ± σ =.5 ±. or.08 to 6.9. This interval is shown on the probability histogram above. Then P[ x ] P[ x ] = 6 =..85 For the probability distribution given in this exercise, µ = Ex ( ) = xpx ( ) = =.5. ( ) ( ) ( ) ( ).86 a Define D: person prefers David Letterman J: person prefers Jay Leno There are eight simple events in the experiment: DDD DDJ DJJ DJD JDJ JDD JJD JJJ and the probabilities for x = number who prefer Jay Leno = 0,,, are shown below. Px ( = 0) = PDDD ( ) = (.8) =.06 Px PDDJ PDJD PJDD ( = ) = ( ) + ( ) + ( ) = (.5)(.8) =.59 P x P DJJ P JJD P JDJ ( = ) = ( ) + ( ) + ( ) = (.5) (.8) =.89 Px PJJJ ( = ) = ( ) = (.5) =.06 96

18 b The probability histogram is shown below p(x) x c Px= ( ) =.59 d The average value of x is µ = Ex ( ) = xpx ( ) = =.56 The variance of x is σ = E ( x µ ) = ( x µ ) p( x) and ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) = (0.56).06 (.56).59 (.56).89 (.56).06 =.788 σ =.788 = a-b On the first try, the probability of selecting the proper key is /. If the key is not found on the first try, the probability changes on the second try. Let F denote a failure to find the key and S denote a success. The random variable is x, the number of keys tried before the correct key is found. The four associated simple events are shown below. E : S ( E x = x = ) : FFS ( ) x = ) : FFFS ( x = ) E : FS ( E c-d Then p() = P( x = ) = P( S) = ( )( ) ( )( )( ) ( )( )( )( ) p() = P( x = ) = P( FS) = P( F) P( S) = = p() = Px ( = ) = PFFS ( ) = PFPFPS ( ) ( ) ( ) = = p() = Px ( = ) = PFFFS ( ) = PFPFPFPS ( ) ( ) ( ) ( ) = = The probability distribution and probability histogram follow. x p(x) / / / / 97

19 p(x) x.88 If a $5 bet is placed on the number 8, the gambler will either win $75 ( 5 $5) with probability /8 or lose $5 with probability 7/8. Hence, the probability distribution for x, the gambler s gain is x p(x) 5 7/8 75 /8 µ = Ex ( ) = xpx ( ) = = $ 0.6. The expected gain is ( ) ( ) The expected gain is in fact negative, a loss of $ a-b Let W and W be the two women while M, M and M are the three men. There are 0 ways to choose the two people to fill the positions. Let x be the number of women chosen. The 0 equally likely simple events are: E : W W ( x = ) E6: W M ( x = ) E : W M ( x = ) E7: W M ( x = ) E : W M ( x = ) E8: M M ( x = 0) E : W M ( x = ) E9: M M ( x = 0) E 5 : W M ( x = ) E0: M M ( x = 0) The probability distribution for x is then ( ) is shown below. p(0) = 0, p = 6 0, p() = 0. The probability histogram p(x) x 98

20 .90 Similar to Exercise.89. The random variable x can take on the values 0,, or. The associated probabilities can be found by summing probabilities of the simple events for the respective numerical events or by using the laws of probability: P x = 0 = P nondefection on first selection P nondefective on second nondefective on first [ ] [ ] [ ] P[ nondefective on third nondefective on first and second] = P[ x = ] = P( DNN) + P( NDN) + P( NND) = + + = P[ x = ] = P( DDN) + P( DND) + P( NDD) = + + = The probability distribution for x and the probability histogram follow. x 0 p(x) /5 /5 / p(x) x.9 Let x be the number of drillings until the first success (oil is struck). It is given that the probability of striking oil is PO ( ) =., so that the probability of no oil is PN ( ) =.9 a [ ] p() P[ oil struck on second drilling] p() = P oil struck on first drilling = P( O) =. =. This is the probability that oil is not found on the first drilling, but is found on the second drilling. Using the Multiplication Law, p() = P( NO) = (.9)(.) =.09. Finally, p() = P( NNO) = (.9)(.9)(.) =.08. b-c For the first success to occur on trial x, (x ) failures must occur before the first success. Thus, x p( x) = P( NNNKNNO) = (.9) (.) since there are (x ) N s in the sequence. The probability histogram is shown below. p(x) x 99

21 .9 onsider the event x =. This will occur only if either A or B wins three sets in a row; that is, if the event AAA or BBB occurs. Then [ ] [ ] [ ] ( ) ( ) P x = = P( AAA) + P( BBB) = P( A) + P( B) =.6 +. =.8 onsider the event x =. This will occur if A-wins are spread over sets (with the last A-win in the fourth set) or if B-wins are similarly spread over sets. The associated simple events are ABAA BABB AABA BAAA ABBB BBAB and the probability that x = is p () = (.6) (.) + (.6)(.) =.7 The event x = 5 will occur if A-wins are spread over 5 sets (with the last A-win in the fifth set) or similarly for B. The associated simple events are ABBAA AABBA BBAAA BAABA ABABA BABAA ABBAB AABBB BBAAB BAABB ABABB BABAB and the probability that x = 5 is p (5) = 6(.6) (.) + 6(.6) (.) =.56 Notice that p() + p() + p(5) = = Refer to Exercise.9. For a general value PA ( ) distribution for x is () p = p + p ( ) ( ) ( ) ( ) ( ) () p = p p + p p p(5) = 6 p p + 6 p p = pand PB ( ) = p, we showed that the probability For the three values of p given in this exercise, the probability distributions and values of the table below. P( A ) =.6 P( A ) =.5 P( A ) =.9 x p(x) x p(x) x p(x) a Ex ( ) =.0656 b Ex ( ) =.5 c Ex ( ) =.86.9 a Ex ( ) = (.) + (.80) + (.06) + 5(.0) =.98 b Ex ( ) = (.) + (.80) + 5(.0) + 6(.0) =.9 c Ex ( ) = (.0) + 5(.80) + 6(.) + 7(.0) = 5.6 Ex ( ) are given in.95 The random variable G, total gain to the insurance company, will be D if there is no theft, but D 50,000 if there is a theft during a given year. These two events will occur with probability.99 and.0, respectively. Hence, the probability distribution for G is given below. G p(g) The expected gain is D.99 EG ( ) = GpG ( ) =.99 D+.0( D 50,000) = D 50, 000 D 50,000.0 In order that EG ( ) = 000, it is necessary to have 000 = D 500 or D = $ a Ex xpx ( ) ( ) L ( ) b µ = ( ) = ( ) = = 7.9 ( x ) ( ) ( 7.9) (.0) ( 7.9) σ = µ p x = + (.05) + L+ ( 7.9) (.0) =.7 σ =.7 = and

22 c alculate µ ± σ = 7.9 ±.50 or.55 to.5. Then, referring to the probability distribution of x, [ ] [ ] P.55 x.5 = P x = p() p() =.0 = a Similar to Exercise.9. For the first non-believer to be found on call x, (x ) people who do believe in heaven must be called before the first non-believer is found. Thus, x p( x) = P( NNN KNNY) = (.8) (.9) b As with other phone surveys, there is always a problem of non-response people who do not answer the telephone or decline to participate in the survey. Also, there is a problem of truthfulness of the response for a question such as this which may be a sensitive subject for some people..98 The company will either gain ($ ) if the package is delivered on time, or will lose $.80 if the package is not delivered on time. We assume that, if the package is not delivered within hours, the company does not collect the $5.50 delivery fee. Then the probability distribution for x, the company s gain is x p(x) and µ = Ex ( ) =.70(.98).80(.0) = The expected gain per package is $ We are asked to find the premium that the insurance company should charge in order to break even. Let c be the unknown value of the premium and x be the gain to the insurance company caused by marketing the new product. There are three possible values for x. If the product is a failure or moderately successful, x will be negative; if the product is a success, the insurance company will gain the amount of the premium and x will be positive. The probability distribution for x follows: x p(x) In order to break even, c.9 Ex ( ) = xpx ( ) = 0. 80,000 + c.0 5,000 + c.05 Therefore,.9( c) +.0( 80,000 + c) + (.05)( 5,000 + c) = ( ) c = 0 c = 050 Hence, the insurance company should charge a premium of $, Since there are four possible shapes for each of the three slots, there are a total of N = ()()() = 6 possible configurations for the three slots. Of these, only n A = (, LLL, SSS, or BBB) will result in a P win = n N = 6 = 6. win. Hence, the probability of winning is ( ).0 Define the following events: A: worker fails to report fraud B: worker suffers reprisal It is given that PB ( A ) =.and PA= ( ).69 A. The probability of interest is PA ( B) = PB ( A ) PA ( ) =.(.) =.07.0 a This experiment consists of two patients, each swallowing one of four tablets (two cold and two aspirin). There are four tablets to choose from, call them,, A and A. The resulting simple events are then all possible ordered pairs which can be formed from the four choices. ( ) ( ) (A ) (A ) ( A ) ( A ) (A ) (A ) ( A ) ( A ) (A A ) (A A ) Notice that it is important to consider the order in which the tablets are chosen, since it makes a difference, for example, which patient (A or B) swallows the cold tablet. 0

23 b A = {( ),( A )(, A )(, A ),( ),( A ) } c B = {( A),( A),( A),( A),( A ),( A ),( A ),( A ) } d = {( AA ),( AA) }.0 Refer to Exercise.0. There are simple events, each with equal probability /. By summing the probabilities of simple events in the events of interest we have PA= ( ) 6 = PA ( B) = 0 = 5 6 PB ( ) = 8 = P ( ) = = 6 PA ( B) = = PA ( ) = 0 PA ( ) = 8 =.0 Four customers enter a store which sells two styles of digital video recorders (DVRs). Each customer seeks to purchase one of the two styles. Let the number represent a customer seeking to purchase style and the number represent a customer seeking to purchase style. The sample space consists of the following 6 four-tuplets. The event A, all four customers seek to purchase the same style of DVR, consists of the two simple events and. Since the two styles are assumed to be in equal demand, we assume that each of the 6 simple events is equally likely, and we assign the probability /6 to each. Hence, PA= ( ) 6= Two systems are selected from seven, three of which are defective. Denote the seven systems as G, G, G, G, D, D, D according to whether they are good or defective. Each simple event will represent a particular pair of systems chosen for testing, and the sample space, consisting of pairs, is shown below. G G G D G D G D G G G G G D G D G G G D G D D D G G G D G D D D G G G D G D D D G G Note that the two systems are drawn simultaneously and that order is unimportant in identifying a simple event. Hence, the pairs G G and G G are not considered to represent two different simple events. The event A, no defectives are selected, consists of the simple events G G, G G, G G, G G, G G, G G. Since the systems are selected at random, any pair has an equal probability of being selected. Hence, the probability assigned to each simple event is / and PA= ( ) 6 = Define the random variable x to be daily sales; x can take the values $0, 50,000 or 00,000 depending on the number of customers the salesman contacts. The associated probabilities are shown below: P x = 0 = P contact one, fail to sell + P contact two, fail to sell [ ] [ ] [ ] = P[ contact one] P[ fail to sell] + P[ contact two] P[ fail with first] P[ fail with second] = ( )( 90) + ( )( 90)( 90) = = 5 00 Similarly, P x = 50,000 = P contact one, sell + P contact two, sell to one [ ] [ ] [ ] = P[ contact one, sell] + P[ contact two, sell to first only] + P[ contact two, sell to second only] = ( )( 0) + ( )( 0)( 9 0) + ( )( 9 0)( 0) = 6 00 Finally, P[ x = 00,000] = P[ contact two, sell to both] = ( )( 0)( 0) = 00 0

24 Then Ex ( ) = 0( 5 00) + 50, 000( 6 00) + 00, 000( 00) = 8.. Thus, the expected value of daily sales is $ The random variable x, defined as the number of householders insured against fire, can assume the values 0,,, or. The probability that, on any of the four draws, an insured person is found is.6; hence, the probability of finding an uninsured person is.. Note that each numerical event represents the intersection of the results of four independent draws. P x = 0 = (.)(.)(.)(.) =.056, since all four people must be uninsured. [ ] [ ] (.6)(.)(.)(.).56 P x = = = (Note: the appears in this expression because x = is the union of four mutually exclusive events. These represent the ways to choose the single insured person from the fours.) P x = = 6(.6)(.6)(.)(.) =.56, since the two insured people can be chosen in any of 6 ways. [ ] P[ x = ] = (.6) (.) =.56 and [ ] P x = = (.6) =.96. Then P at least three insured = p() + p() = =.75 [ ].08 In this exercise, x may take the values 0,,, or, and the probabilities associated with x are evaluated as in Exercise.07. a P[ x = ] = ( ) = P[ x ] ( )( ) P[ x ] ( ) ( ) P[ x = ] = ( ) = = =.8. =.096 = =.8. = The reader may verify that the probabilities sum to one and that 0 px ( ) for x = 0,,, and. The requirements for a probability distribution have been satisfied. b The alarm will function if x =,, or. Hence, P[ ] p p p c Ex xpx ( ) ( ) ( ) ( ) σ = ( x µ ) p( x) = (0.) (.008) + (.) (.096) ( ) ( ) alarm functions = () + () + () = =.99 µ = ( ) = ( ) = =..09 a Similar to Exercise.5. P( ) + (.).8 + (.).5 = cold = = = b Define: F: person has four or five relationships S: person has six or more relationships Then for the two people chosen from the total 76, ( one F and one S) = ( ) + ( ) P P F S P S F = + = P( three or fewer cold) Three or fewer cold = = = =.889 P cold c P( ) ( ) 0

25 .0 The completed table is shown below, and each of the possible pairings are equally likely with probability /6. ss yy Ss yy ssyy ss YY ss yy ss yy ss Yy ss YY Ss yy Ss yy Ss Yy Ss YY SS yy SS yy SS Yy SS YY a Smooth yellow peas result from all pairing having at least one S and at least one Y. Hence, P (smooth yellow) = 9 /6. b Smooth green peas result when the pairing has at least one S and the pair yy. Hence, P (smooth green) = /6. c Wrinkled yellow peas result when the pairing has at least one Y and the pair ss. Hence, P (wrinkled yellow) = /6 d Wrinkled green peas result only when the pairing is ss yy. Hence, P (wrinkled green) = /6. e Define: A: offspring has smooth yellow peas B: offspring has one s allele : offspring has one s allele and one y allele PA ( ) = 9 6; P A B = 6 6; P A = 6. Using the definition of conditional probability, Then ( ) ( ) PA ( B) 6 6 PB ( A) = = = and PA ( ) 9 6 PA ( ) 6 P ( A) = = = PA ( ) Similar to Exercise.7. An investor can invest in three of five recommended stocks. Unknown to him, only out of 5 will show substantial profit. Let P and P be the two profitable stocks. A typical simple event might be (P P N ), which represents the selection of two profitable and one nonprofitable stock. The ten simple events are listed below: E : (P P N ) E : (P P N ) E : (P P N ) E : (P N N ) E 5 : (P N N ) E 6 : (P N N ) E 7 : (N N N ) E 8 : (P N N ) E 9 : (P N N ) E 0 : (P N N ) P investor selects two profitable stocks = P E + P E + P E = 0 since the simple events are Then [ ] ( ) ( ) ( ) equally likely, with P( E i ) = 0. Similarly, [ ] ( ) ( ) ( ) ( ) P( E ) P( E ) P investor selects only one of the profitable stocks = P E + P E + P E + P E + + = a Experiment: Four union men, two from a minority group, are assigned to four one-man jobs, two of which are the most desirable and two of which are the least desirable. b Sample space: Let us assume that jobs and are the most desirable ones. Define M and M to be the minority workers and W and W to be the other two workers. A typical simple event is (M M W W ) which implies that minority workers and are assigned jobs and, while the other workers are assigned jobs and. There are simple events. E : (M M W W ) E 7 : (M M W W ) E : (W M M W ) E 9 : (W M M W ) E : (M W M W ) E 8 : (M M W W ) E : (W M W M ) E 0 : (W M W M ) E : (M W W M ) E 9 : (M W M W ) E 5 : (W M M W ) E : (W M M W ) E : (M W W M ) E 0 : (M W W M ) E 6 : (W M W M ) E : (W M W M ) E 5 : (M W M W ) E : (M W W M ) E 7 : (W W M M ) E : (W W M M ) E 6 : (M M W W ) E : (M W M W ) E 8 : (W W M M ) E : (W W M M ) c As jobs and are the least desirable (they correspond to positions and of one ordered -tuplet), the probability that the two men from the minority group are assigned to these jobs is P( E7 ) + P( E8 ) + P( E ) + P( E ) = = 6. a Define the following events: B : client buys on first contact B : client buys on second contact 0

26 Since the client may buy on either the first of the second contact, the desired probability is P client will buy = P client buys on first contact + P client doesn't buy on first, but buys on second [ ] [ ] [ ] = P( B) + ( P( B) ) P( B) =. + (.)(.55) =.7 b The probability that the client will not buy is one minus the probability that the client will buy, or.7 =.7.. Define the following events: B: man takes the bus S: man takes the subway L: the man is late It is given that PB ( ) =., PS ( ) =.7, PL ( B) =., PL ( S) =.. Using Bayes Rule, PL ( BPB ) ( ) (.)(.).09 PB ( L) =.9 PL ( BPB ) ( ) PL ( SPS ) ( ) = = = + (.)(.) + (.)(.7)..5 Define the following events: A: first system fails B: second system fails A and B are independent and PA ( ) = PB ( ) =.00. To determine the probability that the combined missile system does not fail, we use the complement of this event; that is, P system does not fail = P system fails = P A B [ ] [ ] ( ) = PAPB ( ) ( ) = (.00) = a If the fourth van tested is the last van with brake problems, then in the first three tests, we must find one van with brake problems and two without. That is, in choosing three from the six vans, we must find one faulty and two that are not faulty. Think of choosing three balls one white and two red from a total of six, and the probability can be calculated as (6) P( one faulty and two not) = = = Once this is accomplished, the van with brake problems must be chosen on the fourth test. Using the mn Rule, the probability that the fourth van tested is the last with faulty brakes is =. 5 5 b In order that no more than four vans must be tested, you must find one or both of the faulty vans in the first four tests. Proceed as in part a, this time choosing four from the six vans, and P( one or two faulty vans) = = = c If it is known that the first faulty van is found in the first two tests, there are four vans left from which to select those tested third and fourth. Of these four, only one is faulty. Hence, + P( one faulty and one not one faulty in first two tests) = = = 6.7 Each ball can be chosen from the set (, 6) and there are three such balls. Hence, there are a total of ()() = 8 potential winning numbers..8 a Since the onnecticut and Pennsylvania lotteries are independent, and the onnecticut lottery was fair (each ball can be chosen from a set of ten, 0,,, 9). P( 666 in onnecticut 666 in Pennsylvania) = P( 666 in onnecticut) = = ( ) 05

27 Since P (666 in Pennsylvania ) = 8, the probability of a 666 in both lotteries is = a Use the Law of Total Probability, with PB ( ) =.7, PF ( ) =.5, PML ( B) =.9, PAL ( B) =.6 PML ( F) =., and PAL ( F) =.67. Then P( ML) = P( ML B) + P( ML F) = P( B) P( ML B) + P( F) P( ML F) b c Use Bayes Rule, Use Bayes Rule, =.7(.9) +.5(.) =.58 PML ( F) PFPML ( ) ( F).5(.) PF ( ML) = = = =.88. P( ML) P( ML).58 PBPAL ( ) ( B).7(.6) PB ( AL) =.67 PBPAL ( ) ( B) PFPAL ( ) ( F) = = +.7(.6) +.5(.67).0 a Similar to Exercise.5. PT ( P) = 7 70 =.86 b PT ( P) = 0 c PT ( N) = 70 PT ( N) d P( P T) = = = PT ( ) 70 e P( false negative ) P( N T) ( N) 70 PT = = = = PT ( ) 70. a onsider a single trial which consists of tossing two coins. A match occurs when either HH or TT is observed. Hence, the probability of a match on a single trial is PHH ( ) + PTT ( ) = + =. Let MMM denote the event match on trials,, and. Then ( ) PM PM PM ( ) P MMM = ( ) ( ) ( ) = = 8. b On a single trial the event A, two trails are observed has probability PA ( ) = PTT ( ) =. Hence, in three trials ( ) PAAA ( ) = PAPAPA ( ) ( ) ( ) = = 6 c This low probability would not suggest collusion, since the probability of three matches is low only if we assume that each student is merely guessing at each answer. If the students have studied together or if they both know the correct answer, the probability of a match on a single trial is no longer /, but is substantially higher. Hence, the occurrence of three matches is not unusual.. Define A: union strike fund is adequate to support a strike : union-management team makes a contract settlement within weeks It is given that P ( ) =.5, PA ( ) =.6, PA ( ) =.. Then PA ( ). P ( A) = = =.5. PA ( ).6 Since P ( ) =.5and P ( A ) =.5, it appears that the settlement of the contract is independent of the ability of the union strike fund to support the strike.. Define R: the employee remains 0 years or more a The probability that the man will stay less than 0 years is PR ( ) = PR ( ) = 6= 5 6 b The probability that the man and the woman, acting independently, will both work less than 0 years is P( R R ) P( R ) P( R ) ( ) c = = 56 = 56 The probability that either or both people work 0 years or more is, 06

28 ( ) ( ) P R R = 5 6 = 5 6= 6. Let y represent the value of the premium which the insurance company charges and let x be the insurance company s gain. There are four possible values for x. If no accident occurs or if an accident results in no damage to the car, the insurance company gains y dollars. If an accident occurs and the car is damaged, the company will gain either y, 000 dollars, y.6(, 000) dollars, or y.(,000) dollars, depending upon whether the damage to the car is total, 60% of market value, or 0% of market value, respectively. The following probabilities are known. P accident occurs =.5 P total loss accident occurs =.08 [ ] [ ] [ ] P[ ] P 60% loss accident occurs =. 0% loss accident occurs =.80 Hence, P x = y,000 = P accident P total loss accident =.5(.08) =.0 Similarly, P x [ ] [ ] [ ] [ = y, 00 ] =.5(.) =.08 and P[ x y ] = 00 =.5(.80) =. The gain x and its associated probability distribution are shown below. Note that p(y) is found by subtraction. x p(x) y,000.0 y,00.08 y 00. y.85 Letting the expected gain equal zero, the value of the premium is obtained. Ex ( ) = xpx ( ) =.0( y,000) +.08( y, 00) +.( y 00) +.85y Ex ( ) = y ( ) = y 09.6 y = $ Define the events: A: the man waits five minutes or longer B: the woman waits five minutes or longer The two events are independent, and PA ( ) = PB ( ) =.. a P( A ) = P( A) =.8 b P( A B ) = P( A ) P( B ) = (.8)(.8) =.6 c P[ at least one waits five minutes or longer] [ ] P( A B ) = P neither waits five minutes or longer = =.6 =.6.6 This exercise provides an example of a lot acceptance sampling plan. Seven items are drawn from a large lot of bearings and we wish to calculate the probability of accepting the lot; that is, the probability of observing no defectives in the sample of seven. In order to obtain P[acceptance], it is necessary to assume that the lot is large enough so that the probability of getting a defective is not noticeably affected by repeated draws. For example, consider a lot which contains 0,000 items, 5000 of which are defective. The probability of getting a defective on the first draw is 5000/0,000 or /. Assume that a defective has been drawn. Then the probability of getting a defective on the second draw is 999/9999, which is not noticeably different from /. Define the following events: D: draw a defective G: draw a nondefective, where G = D and PG ( ) = PD ( ) A: the lot is accepted In each case, the desired probability is 07

29 [ ] 7 P( A) = P( GGGGGGG) = P( G) If all the items in the lot are nondefective, then PD ( ) = 0, PG ( ) = and the probability of acceptance is 7 PA= ( ). If /0 are defective, then PD ( ) =., PG ( ) =.9, and PA ( ) = (.9) =.78. If / are defective, then PD PG PA 7 ( ) =.5, ( ) =.5, and ( ) = (.5) = It is given that 0% of all people in a community favor the development of a mass transit system. Thus, given a person selected at random, the probability that the person will favor the system is.. Since the pollings are independent events, when four people are selected at random, P all favor the system = (.) =.056 [ ] Similarly, P [ none favor the system ] = (.) =.96.8 In this exercise, we assume that the presence or absence of disease in any pair of pairs of identical twins represent independent events. If there is no difference in the effect of the drugs, the probability that one drug causes a greater drop in blood pressure than the other is.5. Then P reading for drug A exceeds reading for drug B for all pairs = (.5) =.065 [ ] If we do observe the above event, we can reach one of two conclusions: The two drugs are equally effective and we have observed a rare event Drug B is more effective than drug A Since the probability of the above event is very small, we would draw the second conclusion..9 Since the first pooled test is positive, we are interested in the probability of requiring five single tests to detect the disease in the single affected person. There are (5)()()()() ways of ordering the five tests, and there are ()()() ways of ordering the tests so that the diseased person is given the final test. Hence, the desired probability is! =. 5! 5 If two people are diseased, six tests are needed if the last two tests are given to the diseased people. There are ()() ways of ordering the tests of the other three people and () ways of ordering the tests of the two diseased people. Hence, the probability that six tests will be needed is!! =. 5! 0.0 The objective is to determine how many times a coin must be tossed in order that the following inequality will be true: P observe at least one head.9 [ ] Using the complement of this event, we have: P observe no heads.9 or P observe no heads. [ ] [ ] Since the probability of observing a tail on a given toss is.5, P[ observe no heads in n tosses ] = (.5) n. Evaluating this probability for increasing values of n, we obtain the following table. n P[observe no heads] Note that the inequality will be satisfied if and only if n is greater than or equal to. Thus, the coin must be tossed four times.. The necessary probabilities can be found by summing the necessary cells in the probability table and dividing by 0, the total number of firms. 5 a PR ( ) = =.58 b PT ( F) = =

30 c 56 PF ( ) = =.709 d 0 PT ( F) PT ( F) = = = =.906 PF ( ) a Define the event R: subject chooses red and N: subject does not choose red. Then PR ( ) = and PN ( ) =. There are 8 simple events in the experiment: NNN ( 0 RRN x = Then x = ) ( ) x = ) ( x = ) x = ) ( x = ) x = ) ( x = ) RNN ( RNR NRN ( NRR NNR ( RRR 8 Px ( = 0) = PNNN ( ) = PNPNPN ( ) ( ) ( ) = = 7 Px ( = ) = PNPNPR ( ) ( ) ( ) = = 7 6 Px ( = ) = PNPRPR ( ) ( ) ( ) = = 7 Px ( = ) = PRRR ( ) = PRPRPR ( ) ( ) ( ) = = 7 The probability distribution for x is shown in the table. x 0 p(x) 8/7 /7 6/7 /7 b The probability histogram is shown below p(x) x. a Define P: shopper prefers Pepsi and : shopper prefers oke. Then if there is actually no difference in the taste, P(P) = P() = / and P(all four prefer Pepsi) = P( PPPP) = [ P( P)] = = = b P(exactly one prefers Pepsi) = P( P) + P( P) + P( P) + P( P) = PP ( )[ P ( )] =.5 = = 6 09

31 . Would it be unlikely, if three houses were chosen at random, to choose three adjacent houses in a row of The number of choices for the three houses is ()(0) = = 0 ()() There are 0 choices of three in which the houses are adjacent; that is {,, 56,, 0 } so that the probability of choosing three adjacent houses is 0 =.05 which is an unusual occurrence. 0.5 a There are six volunteers, from whom we must choose two people for the committee. The number of choices is then 6 6(5) = = 5 () If the number of women chosen from the two women is x, and the number of men chosen from the four men must be x. Then 0 6 Px ( = 0) = = Px ( = ) = = Px ( = ) = = 5 5 and the probability distribution for x is shown in the table. x 0 p(x) 6/5 8/5 /5 b µ = Ex ( ) = xpx ( ) = = = ( ) ( ) (0 ) ( ) ( ) σ = x µ p x = + + = = In an experiment that consists of tossing two dice, the sample space consists of 6 ordered pairs, the first stage being the number observed on the first die, and the second being the number observed on the second die. The simple events shown in the Tossing Dice applet can also be found using a tree diagram, in which the first branch represents the result of the first toss, and the second branching represents the results of the second toss. a lick the appropriate pairs of dice that have a sum of 7. There are 6 pairs, found in the center 6 diagonal of the grid, and P ( sum is 7) = =. There are pairs that have a sum of, found on a 6 6 diagonal in the lower right hand part of the grid, and P ( sum is ) = = 6 8 b lick the appropriate pairs of dice that have the same upper face. There are 6 pairs, and 6 P ( doubles) = = c There are 9 pairs, three in each of columns,, and 5, and P ( both odd) = =. 6.7 Refer to the Tossing Dice applet, in which the simple events for this experiment are displayed. Each simple event has a particular value of T associated with it, and by summing the probabilities of all simple events producing a particular value of T, the following probability distribution is obtained. The distribution is mound-shaped. 0

32 a-b T p(t) p(x) x a-b There are 8 equally likely simple events in the experiment: TTT ( HHT x = x = 0) ( ) x = ) ( x = ) x = ) ( x = ) x = ) ( x = ) HTT ( HTH THT ( THH TTH ( HHH and the probability distribution for x is shown in the table, and in the figure below. x 0 p(x) /8 /8 /8 / p(x) x c-d Answers will vary. A typical probability distribution is shown below.

33 .9 a-b When there is only one fair coin, Px ( = 0) = PT ( ) =.5and Px ( = ) = PT ( ) =. 5. The probability histogram is shown below, along with the simulation using the Flipping Fair oins applet..0 p(x) x.0 a Tails are more likely, since the probability of observing 0 heads or head are quite high. b P( no heads ) = P( TTT) = [ P( T) ] ( ) c From the simulation, P no heads.5. Then PT ( ).5 =.7, and PH ( ) =..