CHAPTER 6 PROBABILITY MULTIPLE CHOICE...
Transcript of CHAPTER 6 PROBABILITY MULTIPLE CHOICE...
CHAPTER 6 PROBABILITY
MULTIPLE CHOICE QUESTIONS
In the following multiple choice ques tions , please circle the correct answer .
1. An approach of assigning probabilities which assum e s that all outcom e s of the experimen t are equally likely is referred to as the:a. subjective approachb. objective approachc. classical approachd. relative frequency approachANSWER: c
2. If P(A) = 0.84, P(B) =0.76 and P(A or B) =0.90, then P(A and B) is:a. 0.06b. 0.14c. 0.70d. 0.83ANSWER: c
3. If P(A) = 0.35, P(B) = 0.45 and P(A and B) =0.25, then P(A/B) is:a. 1.4b. 1.8c. 0.714d. 0.556ANSWER: d
4. If P(A) = 0.20, P(B) = 0.30 and P(A and B) = 0.06, then A and B are:a. depend e n t eventsb. indepen d e n t eventsc. mutually exclusive eventsd. comple m e n t a ry eventsANSWER: b
5. If A and B are mutually exclusive events with P(A) = 0.70, then P(B):a. can be any value betwe e n 0 and 1b. can be any value betwe e n 0 and 0.70c. cannot be larger than 0.30d. cannot be determined with the information givenANSWER: c
6. If A and B are indepen d e n t events with P(A) = 0.60 and P(A/B) = 0.60, then P(B) is:a. 1.20
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b. 0.60c. 0.36d. Cannot be determined with the information givenANSWER: d
7. If P(A) = 0.65, P(B) = 0.58, and P(A and B) = 0.76, then P(A or B) is:a. 1.23b. 0.47c. 0.18d. 0.11ANSWER: b
8 . If you roll an unbiased die 50 times, you should expect an even number to appear :a. at least twice in the 50 rollsb. on every other rollc. 25 out of the 50 rollsd. on the averag e , 25 out of the 50 rollsANSWER: d
9. The collection of all possible outcom e s of an experimen t is called: a. a simple eventb. a sample spacec. a sampled. a populationANSWER: b
10. Which of the following is not an approach to assigning probabilities?a. Classical approachb. Trial and error approachc. Relative frequency approachd. Subjective approachANSWER: b
11. A useful graphical method of constructing the sample space for an experimen t is:a. a tree diagra mb. a pie chartc. a histogra md. an ogiveANSWER: a
12. A sample space of an experime n t consists of the following outcom e s: 1, 2, 3, 4, 5. Which of the following is a simple event? a. at least 3b. at most 2c. 3d. 15ANSWER: c
13. Suppose P(A) = 0.35. The probability of comple m e n t of A is:
Probability 73
a. 0.35b. 0.50c. 0.65d. 0.35ANSWER: c
14 . Assume that you invest ed $10,000 in each of three stocks. Each stock can increas e in value, decrea s e in value, or remain the same. Drawing a probability tree for this experime n t will show that the number of possible outcom e s is:a. 10,000b. 3c. 9d. 27ANSWER: d
15. An experimen t consists of tossing 3 unbiased coins simultaneo usly. Drawing a probability tree for this experimen t will show that the number of simple events in this experimen t is:a. 3b. 6c. 9d. None of the above answers is correct .ANSWER: d
16. If the events A and B are independ e n t with P(A) = 0.30 and P(B) = 0.40, then the probability that both events will occur simultaneo usly is:a. 0.10b. 0.12c. 0.70d. 0.75ANSWER: b
17. Two events A and B are said to be indepen d e n t if:a. P(A and B) = P(A) . P(B)b. P(A and B) = P(A) + P(B)c. P(A/B) = P(B)d. P(B /A) = P(A)ANSWER: a
18. Two events A and B are said to mutually exclusive if:a. P(A/B) = 1b. P(B /A) =1c. P(A and B) =1d. P(A and B) = 0ANSWER: d
19. Which of the following state m e n t s is always correct?a. P(A and B) = P(A) . P(B)b. P(A or B) = P(A) + P(B)
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c. P(A or B) = P(A) + P(B) + P(A and B)d. P(A) = 1 P ( )CAANSWER: d
20. Which of the following is a require m e n t of the probabilities assigned to the outcom e s iO ?
a. ( ) 0iP O ≤
b. ( ) 1iP O ≥
c. ( )0 1iP O≤ ≤ for each i
d. ( ) 1 ( )Ci iP O P O= +
ANSWER: c
21. An experime n t consists of three stages . There are two possible outcom e s in the first stage , three possible outcom e s in the second stage , and four possible outcom e s in the third stage. Drawing a tree diagra m for this experimen t will show that the total numb er of outcom e s is:a. 9b. 24c. 26d. 18ANSWER: b
22. Which of the following state m e n t s is correct given that the events A and B have nonzero probabilities?a. A and B cannot be both independ e n t and mutually exclusiveb. A and B can be both independ e n t and mutually exclusivec. A and B are always independ e n td. A and B are always mutually exclusiveANSWER: a
23. If A and B are mutually exclusive events , with P(A) = 0.20 and P(B) = 0.30, then P(A and B) is:a. 0.50b. 0.10c. 0.00d. 0.06ANSWER: c
24. If A and B are indepen d e n t events with P(A) = 0.60 and P(B) = 0.70, then the probability that A occurs or B occurs or both occur is:a. 1.30b. 0.88
Probability 75
c. 0.42d. 0.10ANSWER: b
25. If A and B are mutually exclusive events with P(A) = 0.30 and P(B) = 0.40, then P(A or B) is:a. 0.10b. 0.12c. 0.70d. None of the above answers is correct ANSWER: c
26. If A and B are indepen d e n t events with P(A) = 0.20 and P(B) =0.60, then P(A/B) is:a. 0.20b. 0.60c. 0.40d. 0.80ANSWER: a
27. If P(A) = 0.25 and P(B) = 0.65, then P(A and B) is:a. 0.25b. 0.40c. 0.90d. Cannot be determined from the information givenANSWER: d
28. If a coin is tossed three times and a statistician predicts that the probability of obtaining three heads in a row is 0.125, which of the following assum p tions is irrelevant to his prediction?a. The events are depend e n tb. The events are independ e n tc. The coin is unbiasedd. All of the above assu m ptions are relevant to his predictionANSWER: a
29. If an experimen t consists of five outcom e s with 1( )P O = 0.10, 2( )P O = 0.20,
3( )P O = 0.30, 4( )P O = 0.40, then 5( )P O isa. 0.50b. 0.25c. 1.00d. 0.00ANSWER: d
30. Of the last 500 custom er s entering a super m ark e t , 50 have purchas e d a wireless phone. If the classical approach for assigning probabilities is used, the probability that the next customer will purchase a wireless phone isa. 0.10b. 0.90
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c. 0.50d. None of the above answers is correctANSWER: c
TRUE/FALSE QUESTIONS
31. If the event of interes t is A, the probability that A will not occur is the comple m e n t of A.ANSWER: T
32. The probability of event A and event B occurring must be equal to 1.
ANSWER: F
33. The relative frequency approach to probability depends on the law of large numb ers .ANSWER: T
34. The annual estimat e of the number of deaths of infants is an example of the classical approach to probability. ANSWER: F
35. The outcom e of a game of roulet te based on historical data is not an example of the relative frequency approach to probability.ANSWER: F
36. You think you have a 90% chance of passing your next principles of accounting exam. This is an example of subjective approach to probability.ANSWER: T
37. Probability refers to a numb er betwe en 0 and 1, which express e s the chance that an event will occur.ANSWER: T
38. If event A does not occur, then its complem e n t CA will also not occur.ANSWER: F
39. Marginal probability is the probability that a given event will occur, with no other events taken into considera tion. ANSWER: T
40. Conditional probability is the probability that an event will occur, given that another event will also occur.ANSWER: F
41. When we wish to deter mine the probability that one or more of several events will occur in an experime n t , we would use addition rules.ANSWER: T
Probability 77
42. A physician has five choices for treating a patient ' s infection. After the first choice has been made , and becaus e of interaction betwe en the prescription drugs used, there are only three choices for the final stage of treat m e n t . Drawing a probability tree for this experimen t will show that the total numb er of possibilities for treating this patient is 10.ANSWER: F
43. Two or more events are said to be indepen d e n t when the occurrence of one event has no effect on the probability that another will occur.ANSWER: T
44. Five studen t s from a statistics class have formed a study group. Each may or may not attend a study session. Assuming that the memb ers will be making indepen d e n t decisions on whether or not to attend, there are 32 different possibilities for the composition of the study session.ANSWER: T
45. When events are mutually exclusive, two or more of them can happen at the same time.ANSWER: F
46. According to an old song lyric, "love and marriage go together like a horse and carriage." Let love be event A and marriage be event B. Events A and B are mutually exclusive.ANSWER: F
47. When it is not reason a ble to use the classical approach to assigning probabilities to the outcom e s of an experime n t , and there is no history of the outcom e s , we have no alterna tive but to employ the subjective approach.ANSWER: T
48. Bayes’ Law allows us to comput e conditional probabilities from other forms of probability.ANSWER: T
49. A useful graphical method of constructing the sample space for an experimen t is pie chart.ANSWER: F
50. An experimen t consists of tossing 3 unbiased coins simultaneo usly. Drawing a probability tree for this experime n t will show that the numb er of outcom e s is 9.ANSWER: F
51. Assume that A and B are independ e n t events with P(A) = 0.30 and P(B) = 0.50. The probability that both events will occur simultan eou sly is 0.80.ANSWER: F
52. Two events A and B are said to be indepen d e n t if P(A and B) = P(A) . P(B)ANSWER: T
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53. Two events A and B are said to mutually exclusive if P(A and B) = 0.ANSWER: T
54. If events A and B have nonzero probabilities, then they can be both indepen d e n t and mutually exclusive.ANSWER: F
55. If A and B are indepen d e n t events with P(A) = 0.35 and P(B) = 0.55, then P(A/B) is 0.20.ANSWER: F
56. An effective and simpler method of applying the probability rules is the probability tree, wherein the events in an experime n t are represe n t e d by lines.ANSWER: T
57. The probability of the union of two mutually exclusive events A and B is P(A or B) = 0.ANSWER: F
58. The relative frequency approach is useful to interpre t probability state m e n t s such as those heard from weather forecas t e r s or scientist s .ANSWER: T
59. Given that events A and B are independ e n t and that P( A) = 0.9 and P(B /A) = 0.5, then P(A and B) = 0.45.ANSWER: T
60. Jim and John go to a coffee shop during their lunch break and toss a coin to see who will pay. The probability that John will pay three days in a row is 0.125.ANSWER: T
Probability 79
TEST QUESTIONS
61. Abby, Brenda, and Cameron; three candida t e s for the presidency of a college’s studen t body, are to address a studen t forum. The forum’s organizer is to select the order in which the candida t e s will give their speech e s , and must do so in such a way that each possible order is equally likely to be selected.a. What is the random experimen t?b. List the outcom e s in the sample space.c. Assign probabilities to the outcom e s .d. What is the probability that Cameron will speak first?e. What is the probability that one of the women will speak first?f. What is the probability that Abby will speak before Cameron does?
ANSWERS:a. The random experime n t is to observe the order in which the three
candida t e s give their speech e s .b. S = {ABC, ACB, BAC, BCA, CAB, CBA}, where A=Abby, B=Brenda ,
C=Cam eron.c. The probability assigned to each outcom e is 1/6.d. 1/3e. 2/3f. 1/2
62. Suppose A and B are two indepen d e n t events for which P(A) = 0.20 and P(B) = 0.60.a. Find P(A/B ).b. Find P(B /A).c. Find P(A and B).d. Find P(A or B).
ANSWERS:a. 0.20b. 0.60c. 0.12d. 0.68
63. A Ph.D. gradua t e has applied for a job with two universities: A and B . The gradua t e feels that she has a 60% chance of receiving an offer from university A and a 50% chance of receiving an offer from university B . If she receives an offer from university B , she believes that she has an 80% chance of receiving an offer from university A.a. What is the probability that both universities will make her an offer?b. What is the probability that at least one university will make her an offer?c. If she receives an offer from university B , what is the probability that she
will not receive an offer from university A?
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ANSWERS:a. 0.4b. 0.7c. 0.2
64. There are three approach e s to determining the probability that an outcom e will occur: classical, relative frequency, and subjective. Which is most appropriate in deter mining the probability of the following outcom e s?a. The unem ploym e n t rate will rise next month.b. Five tosses of a coin will result in exactly two heads.c. An American will win the French Open Tennis Tournam e n t in the year
2000.d. A rando mly selected woman will suffer a breas t cancer during the coming
year.
ANSWERS:a. subjectiveb. classicalc. subjectived. relative frequency
65. Suppose P(A) = 0.50, P(B) = 0.40, and P(B/A ) = 0.30.a. Find P(A and B).b. Find P(A or B).c. Find P(A/B ).
ANSWERS:a. 0.15b. 0.75c. 0.375
66. At the beginning of each year, an investm e n t newslet t e r predicts whether or not the stock marke t will rise over the coming year. Historical evidence reveals that there is a 75% chance that the stock marke t will rise in any given year. The newslet t e r has predicted a rise for 80% of the years when the marke t actually rose, and has predicted a rise for 40% of the years when the marke t fell. Find the probability that the newslet t e r’s prediction for next year will be correct .
ANSWER:0.75
67. Suppose P( CA ) = 0.30, P( CB A/ ) = 0.40, and P( /C CB A ) = 0.50.a. Find P(A and B).b. Find P( CB )c. Find P(A or B).
ANSWERS:a. 0.42b. 0.43
Probability 81
c. 0.85
68. A stand ard admissions test was given at three locations. One thousand studen t s took the test at location A, 600 studen t s at location B , and 400 studen t s at location C. The percenta g e s of studen t s from locations A, B , and C, who passe d the test were 70%, 68%, and 77%, respec tively. One student is selected at rando m from among those who took the test . a. What is the probability that the selected student passed the test?b. If the selected student passed the test, what is the probability that the
studen t took the test at location B?c. What is the probability that the selected studen t took the test at location
C and failed?
ANWERS:a. 0.708b. 0.288c. 0.046
69. Sales records of an appliance store showed the following number of dishwash e r s sold weekly for each of the last 50 weeks.
Number of Dishwash er s
Sold
Number of Weeks
0 201 152 103 44 1
a. Define the rando m experimen t of interes t to the store.b. List the outcom e s in the sample spacec. Assign probabilities to the outcom e s .d. What approach have you used in deter mining the probabilities in part (c)?e. What is the probability of selling at least two dishwash er s in any given
week?
ANSWERS:a. The rando m experime n t consists of observing the numb er of dishwash e r s
sold in any given week.b. S = {0, 1, 2, 3, 4}c.
Outcome
Prob.
0 0.401 0.30
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2 0.203 0.084 0.02
d. The relative frequency approach was used.e. P{2, 3, 4} = 0.30
70. A woman is expec ting her second child. Her doctor has told her that she has a 50 50 chance of having another girl. If she has another girl, there is a 90% chance that she will be taller than the first. If she has a boy, however , there is only a 25% chance that he will be taller than the first child. Find the probability that the woman’s second child will be taller than the first.
ANSWER:0.675
71. A survey of a magazine’s subscribers indicates that 50% own a home, 80% own a car, and 90% of the homeown er s who subscribe also own a car. What proportion of subscribersa. own both a car and a house?b. own a car or a house, or both?c. own neither a car nor a house?
ANSWERS:a. 0.45b. 0.85c. 0.15
72. Suppose A and B are two mutually exclusive events for which P(A) = 0.30 and P(B) = 0.40.a. Find P(A and B).b. Find P(A or B).c. Are A and B indepen d e n t events? Explain using probabilities.
ANSWERS:a. 0.00b. 0.70c. No, since P(A and B) = 0.00 ≠ P(A ). P(B) = 0.12.
73. Suppose P(A) = 0.10, P(B /A ) = 0.20, and P(B / CA ) = 0.40.a. Find P(A and B).b. Find P(A and CB ).c. Find P( CB ).d. Find P(A or B).
ANSWERS:a. 0.02b. 0.08c. 0.62d. 0.46
Probability 83
74. An investor tells you that in his estimation there is 75% chance that a particular stock ‘s price over the next three weeks.a. Which approach was used to produce this figure?b. Interpre t the 75% probability.
ANSWERS:a. The relative frequency approach.b. We interpre t the 75% figure to mean that if we had an infinite numb er of
stocks with exactly the same econo mic and marke t charac t e ris tics as the one the investor will buy, 75% of them will increase in price over the next three weeks.
75. The sample space of the toss of a fair coin is S = {1, 2, 3, 4, 5, 6}. If the die is balanced, each simple event has the same probability. Find the probability of the following events .a. An odd numb erb. A numb er less than or equal to 3c. A numb er great e r than or equal to 5d. A numb er betwe en 2 and 5, inclusive.
ANSWERS:a. 3/6 b. 3/6 c. 2/6d. 4/6
76. Suppose P(A) = 0.4, P(B) = 0.5, and P(A and B) = 0.2.a. Find P(A or B).b. Are A and B mutually exclusive events? Explain.c. Are A and B indepen d e n t events? Explain.
ANSWERS:a. 0.70b. No, since P(A and B) = 0.20 > 0.c. Yes, since P(A and B) = 0.20 = P(A). P(B).
77. Suppose P(A) = 0.30, P(B) = 0.50, and P(B /A ) = 0.60.a. Find P(A and B).b. Find P(A or B).c. Find P(A /B ).
ANSWERS:a. 0.18b. 0.62c. 0.36
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78. Is it possible to have two events for which P(A) = 0.40, P(B) = 0.50, and P(A or B) = 0.30? Explain.
ANSWER:No, since P(A or B) must be at least as large as P(B).
79. A statistics professor classifies his student s according to their grade point averag e (GPA) and their gender . The accomp a nying table gives the proportion of student s falling into the various categories . One studen t is selected at rando m.
GPAGender
Under 2.0 2.0 – 3.0 Over 3.0
Male 0.05 0.25 0.10Female 0.10 0.30 0.20
a. If the studen t selected is female, what is the probability that her GPA is betwe en 2.0 and 3.0?
b. If the GPA of the student selected is over 3.0, what is the probability that the studen t is male?
c. What is the probability that the studen t selected is female or has a GPA under 2.0 or both?
d. Is GPA independ e n t of gender? Explain using probabilities.
ANSWERS:a. 0.50b. 0.333c. 0.65d. No; since P(male/GPA over 3.0) = 0.333 ≠ P(male) = 0.40
80. A pharm ac e u t ical firm has discovered a new diagnos tic test for a certain diseas e that has infected 1% of the population. The firm has announce d that 95% of those infected will show a positive test result, while 98% of those not infected will show a nega tive test result. What proportion of test results are correct?
ANSWER:0.9797
81. Suppose P(A) = 0.50, P(B) = 0.30, and P(A or B) = 0.80a. Find P(A and B).b. Find P(B/A ).c. Are A and B mutually exclusive events : Explain using probabilities.
Probability 85
ANSWERS:a. 0.0b. 0.0c. Yes; since P(A and B) = 0.0
82. Suppose P(A) = 0.40, P(B) = 0.50, and P(A or B) = 0.70.a. Find P(A and B).b. Find P(B/A ).c. Are A and B indepen d e n t events? Explain using probabilities.
ANSWERS:a. 0.20b. 0.50c. Yes; since P(B/A ) = 0.50 = P(B)
83. An accounting firm has recently recruited five gradua t e s : two men and three women. Two of the gradua t e s are to be selected at random to work in the firm’s suburban office.a. What is the probability that two women will be selected?b. What is the probability that at least one woman will be selected?
ANSWERS:a. 0.30b. 0.90
84. An insurance company has collected the following data on the gender and marital status of 300 custom ers .
Marital Status
Gender Single Married DivorcedMale 25 125 30Female 50 50 20
Suppose that a custom er is selected at rando m. Find the probability that the custom er selected is:a. a married femaleb. not singlec. married if the custom er is maled. female or divorcede. Are gender and marital status mutually exclusive? Explain using
probabilities.f. Is marital status independ e n t of gender? Explain using probabilities.
ANSWERS:a. 0.167b. 0.75c. 0.694
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d. 0.50e. No, since P(female and married) = 0.167 > 0.f. No, since P(married / male) = 0.694 ≠ P(married) = 0.583
QUESTIONS 85 THROUGH 94 ARE BASED ON THE FOLLOWING INFORMATION:
An ice cream vendor sells three flavors: chocolate , strawberry, and vanilla. Forty five percent of
the sales are chocolat e , while 30% are strawberry, with the rest vanilla flavored. Sales are by
the cone or the cup. The percent a g e s of cones sales for chocolate , strawberry, and vanilla,
are 75%, 60%, and 40%, respec tively. For a randomly selected sale, define the following events :
1A = chocolat e chosen
2A = strawberry chosen
3A = vanilla chosen B = ice crea m on a cone
CB = ice crea m in a cup
85. Find the probability that the ice crea m was sold on a cone and wasa. chocolate flavorb. strawberry flavorc. vanilla flavor
ANSWERS:a. P(B and 1A ) = P(B/ 1A ).P( 1A ) = (0.75)(0.45) = 0.3375
b. P(B and 2A ) = P(B/ 2A ).P( 2A ) = (0.60)(0.30) = 0.18
c. P(B and 3A ) = P(B/ 3A ).P( 3A ) = (0.40)(0.25) = 0.10
86. Find the probability that the ice crea m was sold in a cup and wasa. chocolate flavorb. strawberry flavorc. vanilla flavor
ANSWERS:P( CB and 1A ) = P( CB / 1A ).P( 1A ) = (0.25)(0.45) = 0.1125
P( CB and 2A ) = P( CB / 2A ).P( 2A ) = (0.40)(0.30) = 0.12
P( CB and 3A ) = P( CB / 3A ).P( 3A ) = (0.60)(0.25) = 0.15
87. Find the probability that the ice crea m was sold on a cone.
Probability 87
ANSWER:P(B) = P(B and 1A ) + P(B and 2A ) + P(B and 3A ) = 0.3375 + 0.18 + 0.10 = 0.6175
88. Find the probability that the ice crea m was sold in a cup.
ANSWER:P( CB ) = 1 – P(B) = 1 – 0.6175 = 0.3825
89. Find the probability that the ice crea m was chocolat e flavor, given that it was sold on a cone
ANSWER:P( 1A /B) = P( 1A and B) / P(B) = 0.3375 / 0.6175 = 0.5466
90. Find the probability that the ice crea m was strawberry flavor, given that it was sold on a cone
ANSWER:P( 2A /B) = P( 2A and B) / P(B) = 0.18 / 0.6175 = 0.2915
91. Find the probability that the ice crea m was vanilla flavor, given that it was sold on a cone
ANSWER:P( 3A /B) = P( 3A and B) / P(B) = 0.10 / 0.6175 = 0.1619
92. Find the probability that the ice crea m was chocolat e flavor, given that it was sold in a cup
ANSWER:P( 1A / CB ) = P( 1A and CB ) / P( CB ) = 0.1125 / 0.3825 = 0.2941
93. Find the probability that the ice crea m was strawberry flavor, given that it was sold in a cup
ANSWER:P( 2A / CB ) = P( 2A and CB ) / P( CB ) = 0.12 / 0.3825 = 0.3138
94. Find the probability that the ice crea m was vanilla flavor, given that it was sold in a cup
ANSWER:P( 3A / CB ) = P( 3A and CB ) / P( CB ) = 0.15 / 0.3825 = 0.3922
95. One card is rando mly selected from a deck of 52 playing cards. Let
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A = event card selected is a nineB = event card selected is a tenC = event card selected is a queen
Find ( or or )C C CP A B C using the addition rule.
ANSWER:1 – [(4/52) + (4/52) + (4/52)] = 40/52 = 0.7692
QUESTIONS 96 THROUGH 100 ARE BASED ON THE FOLLOWING INFORMATION:
A construc tion company has submitt ed bids on two separa t e state contract s , A and B . The company feels that it has a 60% chance of winning contract A, and a 50% chance of winning contrac t B . Furthermor e , the company believes that it has an 80% chance of winning contract A given that it wins contract B .
96. What is the probability that the company will win both contract s?
ANSWER:0.40
97. What is the probability that the company will win at least one of the two contract s?
ANSWER:0.70
98. If the company wins contract B , what is the probability that it will not win contract A?
ANSWER:0.20
99. What is the probability that the company will win at most one of the two contract s?
ANSWER:0.60
100. What is the probability that the company will win neither contract?
ANSWER:0.30
QUESTIONS 101 THROUGH 108 ARE BASED ON THE FOLLOWING INFORMATION:
An invest m e n t firm has classified its clients according to their gender and the composition of their investm e n t portfolio (primarily bonds, primarily stocks, or a
Probability 89
balanced mix of bonds and stocks). The proportions of clients falling into the various categories are shown in the following table: Portfolio Composition
Gender Bonds Stocks BalancedMale 0.18 0.20 0.25Female 0.12 0.10 0.15
One client is selected at random, and two events A and B are defined as follows:
A: The client select ed is male.B : The client selected has a balanced portfolio.
101. Find the following probabilities:a. P(A)b. P(B)c. P ( )CA
ANSWERS:a. 0.63b. 0.40c. 0.37
102. Express each of the following events in words:a. A or Bb. A and Bc. A and CBd. CA or CB
ANSWERS:a. The client selected either is male or has a balanced portfolio or both.b. The client selected is male and has a balanced portfolio.c. The client selected is male and has an unbalanc ed portfolio.d. The client selected either is female or has an unbalanc ed portfolio or both.
103. Find the following probabilities:a. P(A or B)b. P(A and B)c. P(A and CB )d. P( CA or CB )
ANSWERS:a. 0.78b. 0.25c. 0.38d. 0.75
90 Chapter Six
104. Express each of the following probabilities in words:a. P(A/B)b. P(B /A)c. P(A/ CB )d. P( CA /B)
ANSWERS:a. The probability that the employee selected is male, given that the
employee has a balanced portfolio.b. The probability that the employee selected has a balanced portfolio, given
that the employee is male.c. The probability that the employee selected is male, given that the
employee has an unbalanc ed portfolio.d. The probability that the employee selected is female, given that the
employee has a balanced portfolio.
105. Find the following probabilities:a. P(A/B)b. P(B /A)c. P(A/ CB )d. P( CA /B)
ANSWERS:a. 0.625b. 0.3968c. 0.633d. 0.375
106. Are A and B indepen d e n t events? Explain.
ANSWER:No, since P(A/B) = 0.625 ≠ P(A) = 0.63
107. Are A and CB indepen d e n t events? Explain.
ANSWER:No, since P(A/ CB ) = 0.633 ≠ P(A) = 0.63
108. Are A and CB mutually exclusive events? Explain.
Probability 91
ANSWER:No, since P(A and CB ) = 0.38 > 0
QUESTIONS 109 THROUGH 120 ARE BASED ON THE FOLLOWING INFORMATION:
A table of joint probabilities is shown below.
1A 2A 3A
1B 0.15 0.25 0.20
2B 0.10 0.15 0.15
109. Calculate the marginal probabilities of event A
ANSWER: P( 1A ) = 0.25, P( 2A ) = 0.40, P( 3A ) = 0.35
110. Calculate the marginal probabilities of event B
ANSWER:P( 1B ) = 0.60, P( 2B ) = 0.40
111. Calculate P( 1A / 1B )
ANSWER:P( 1A / 1B ) = P( 1A and 1B ) / P( 1B ) = 0.15/0.60 = 0.25
112. Calculate P( 2A / 1B )
ANSWER:P( 2A / 1B ) = P( 2A and 1B ) / P( 1B ) = 0.25 /0.60 = 0.4167
113. Calculate P( 3A / 1B )
ANSWER:P( 3A / 1B ) = P( 3A and 1B ) / P( 1B ) = 0.20 / 0.60 = 0.3333
114. Did your answers to Question 111, 112, and 113 sum to 1? Is this a coincidence? Explain.
92 Chapter Six
ANSWER:Yes, they sum to 1. This is not a coincidence . The reason is that P( 1A and 1B ) + P( 2A and 1B ) + P( 3A and 1B ) = P( 1B ); Therefore P( 1A / 1B ) + P( 2A / 1B ) + P( 3A / 1B ) = P( 1B ) / P( 1B ) = 1
115. Calculate P( 1A / 2B )
ANSWER:P( 1A / 2B ) = P( 1A and 2B ) / P( 2B ) = 0.10 / 0.40 = 0.25
116. Calculate P( 2B / 1A )
ANSWER:P( 2B / 1A ) = P( 2B and 1A ) / P( 1A ) = 0.10 / 0.25 = 0.40
117. Calculate P( 1A / 2A )
ANSWER:P( 1A / 2A ) = P( 1A and 2A ) / P( 2A ) = 0 / 0.40 = 0
118. Are the events A and B indepen d e n t? Explain
ANSWER:P( 2A / 1B ) = 0.4167, and P( 2A ) = 0.40. Since P( 2A / 1B ) ≠ P( 2A ), we conclude that the two events are depend e n t .
119. Calculate P( 1A or 1B )
ANSWER:P( 1A or 1B ) = P( 1A ) + P( 1B ) P( 1A and 1B ) = 0.25 + 0.60 – 0.15 = 0.70
120. Calculate P( 1A or 2B )
ANSWER:P( 1A or 2B ) = P( 1A ) + P( 2B ) P( 1A and 2B ) = 0.25 + 0.40 – 0.10 = 0.55