CHAPTER 5

A SURVEY OF PROBABILITY CONCEPTS

1. Outcome 1 21 A A2 A F3 F A4 F F

2. Outcome 1 21 A A2 A R3 A S4 R A5 R R6 R S7 S A8 S R9 S S

Chapter 552

3. a. 0.176 found by 6/34b. Empirical

4. a. 0.40, found by 2/5b. Classical

5. a. Empiricalb. Classicalc. Classicald. Empirical, based on seismological data.

6. a. Outcome 1st 2nd1 M M2 M F3 F F4 F M

b. Classical

7. a. The survey of 40 people about environmental issuesb. 26 or more, respond yes for examplec. 0.25 found by 10/40

Chapter 5 53

d. Empiricale. The events are not equally likely but they are mutually exclusive

8. a. Recording the number of violationsb. at least one violation, for examplec. 0.009, found by 18/2000d. Empirical

9. a. Answers will vary, here are some possibilities: 123, 124, 125, 999

b.

c. Classical

10. a. Answers will vary, value goes up 1/16, goes up 1/8, goes down 1b. Answer depends upon when the data are collectedc. Empirical

11.

12.

13. a. 0.51 found by 102/200

Chapter 554

b. 0.49 found by 61/200 + 37/200 = 0.305 + 0.185 Special rule of addition

14. a. 80%, found by 50% + 30%b. 80%, found by 100% – 20%

15. 0.75, found by 0.25 + 0.50

16. yes, no

17.

18.

19. When two events are mutually exclusive it means that if one occurs the other event cannot occur, Therefore, the probability of their joint occurrence is zero.

20.

21. a. 0.2b. 0.3c. No, because a store could have both. d. Joint probability

Chapter 5 55

e. 0.05, found by 1.0 – 0.95

22. a. 0.55, found by 0.50 + 0.40 – 0.35b. Joint probabilityc. No, a vacationer can visit both attractions

23.

24.

25. 0.90, found by (0.80 + 0.60) – 0.500.10, found by (1 – 0.90)

26. 5%, found by (1 – 0.95)

27. a. P(A1) = 3/10 = 0.30b. P(B1|A2) = 1/3 = 0.33c. P(B2 and A3) = 1/10 = 0.10

28. a. 6/380 or 0.01579, found by 3/20 x 2/19b. 272/380 or 0.7158, found by 17/20 x 16/19

Chapter 556

29. a. A contingency tableb. 0.27, found by 300/500 135 300c. A tree diagram would appear as:

30. Outcome A B C Outcome A B C Outcome A B C1 U U U 10 D U U 19 S U U2 U U D 11 D U D 20 S U D3 U U S 12 D U S 21 S U S4 U D U 13 D D U 22 S D U5 U D D 14 D D D 23 S D D6 U D S 15 D D S 24 S D S7 U S U 16 D S U 25 S S U8 U S D 17 D S D 26 S S D9 U S S 18 D S S 27 S S S

U = Stock is up D = Stock is down S = Stock is sameindicates the criteria that two stocks went up is met.2 of the stocks went up on seven of the 27 outcomes P(A) = 7/27 = 0.259.

31. Probability the 1st presentation wins 3/5 = 0.60

Chapter 5 57

Probability the 2nd presentation wins 2/5(3/4) = 0.30Probability the 3rd presentation wins (2/5)(1/4)(3/3) = 0.10

32. a.

b.

c.

33. 0.4286, found by:

34. 0.3636, found by:

35. 0.5645, found by:

36. 0.8571, found by:

37. 0.1053, found by:

38. 0.625 found by

39. a. 78,960,960b. 840, found by (7)(6)(5)(4). That is 7!/3!c. 10, found by 5!/3!2!

Chapter 558

40. a. 6,840b. 504c. 21

41. 210, found by (10)(9)(8)(7)/(4)(3)(2)

42. 10,000, found by (10)4

43. 120, found by 5!

44. 3003, found by

45. 10,897,286,400 found by

46. 210

47. a. Asking teenagers their reactions to a newly developed soft drink.b. Answers will vary, one possibility is more than half of the respondents like it.

48. Empirical

Chapter 5 59

49. Subjective

50. a. 0.10, found by 50/500b. Yes, mutually exclusive, because a given tip cannot fall in more than one category.c. 1.00d. 0.60, found by 300/500e. 0.90, found by 450/500 or 1 – (50/500)

51. a. The likelihood an event will occur, assuming that another event has already occurred.b. The collection of one or more outcomes of an experiment.c. A measure of the likelihood that two or more events will happen concurrently.

52. a. 4/52, or 0.077b. 3/51, or 0.059c. 0.0045, found by (4/52)(3/51)

53. a. 0.8145, found by (0.95)4

b. Special rule of multiplicationc. P(A and B and C and D) = P(A) x P(B) x P(C) x P(D)

54. a. Venn diagram

Chapter 560

b. Complement rulec. 1

55. a. 0.08, found by 0.80 x 0.10b. No, because if you know a person’s gender it changes the probability they attended

college.c.

d. Yes, because all the possible outcomes are shown on the tree diagram.

56. a. 0.6561 found by (0.9)(0.9)(0.9)(0.9)b. 0.0001 found by (0.1)(0.1)(0.1)(0.1)c. 0.3439 found by 1 0.6561

57. a. 0.57 found by 57/100b. 0.97 found by (57/100) + (40/100)c. Yes, because an employee cannot be both.d. 0.03 found by 1 0.97

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0.78

0.80 0.80 x 0.90 = 0.72

0.20

0.90

0.80 x 0.10 = 0.080

0.20 x 0.22 = 0.0440.22

0.10

0.20 x 0.78 = 0.156

58. a. empiricalb. 0.38, found by (0.335)3

c. 0.294, found by (1 – 0.335)3

d. 0.706, found by (1 – 0.294)

59. a. 0.4096 found by b. 0.0016 found by c. 0.9984 found by 1 – 0.0016

60. a. 0.3059 found by (50/90(49/89)b. 0.4994 found by (50/90)(40/89) + (40/90)(50/89)

61. a. 0.9039 found by b. 0.0961 found by 1 – 0.9039

62. a. 0.064 found by (0.4)3

b. 0.216 found by (0.6)3

c. 0.784 found by 1 – 0.216d. independent

63. a. 0.0333 found by (4/10)(3/9)(2/8)

Chapter 562

b. 0.1667 found by (6/10)(5/9)(4/8)c. 0.8333 found by 1 – 0.1667d. dependent

64. a. 0.50, found by 10/20b. 0.2368, found by (10/20)(9/19)c. 0.1053, found by (10/20)(9/18)(8/18)d. 0.7158, found by (17/20)(16/19)

65. a. 0.3818, found by (9/12)(8/11)(7/10)b. 0.6182, found by 1 – 0.3818

66. a. 0.30, found by 6/20b. 0.45, found by (6 + 7 - 4)/20c. 0.5714, found by 4/7d. 0.0789, found by (6/20)*(5/19)

67. a. 0.51, found by 0.60*0.85b. 0.09, found by 0.60*0.15

68. 0.2373, found by (¾)5

Chapter 5 63

69. a. 0.143, found by 1 – (857/1000)b. 0.2168, found by 31/143

70. a. 0.42b. 0.70(0.40) = 0.28c. 0.88

71. P(poor|profit) =

72. 0.4545, found by

73. a. P(P or D) =(1/50)(9/10) + (49/50)(1/10) = 0.116b. P(No) = (49/50)(9/10) = 0.882c. P(No on 3) = (0.882)3 = 0.686d. P(at least one prize) = 1 – 0.686 = 0.314

74. a. (10)(9)(8) = 720b. 0.00139, found by 1/720c. 0.99583, found by 1 – 3/720

Chapter 564

75. Yes, 256 is found by 28

76. 0.70, found by

77. 0.9744, found by 1 – (0.40)4

78. 15, found by 5 x 3

79. a. 0.185, found by (0.15)(0.95) + (0.05)(0.85)b. 0.0075, found by (0.15)(0.05)

80. 17,576,000 found by (26)(26)(26)(10)(10)(10)

81. a. P(F and > 60) = 0.25, found by solving with the general rule of multiplication:P(F) P(>60|F) = (0.50)(0.50)

b. 0c. 0.3333, found by 1/3

82. a. 0.333, found by (6/10)(5/9)b. 0.9286, found by 1 – [(6/10)(5/9)(4/8)(3/7)]c. dependent

Chapter 5 65

83. 456,976 found by 264

84. a. 2024, found by

b. 0.125, found by 1 – [(23/24)(22/23)(21/22)]

85. 3, 628,800 matches are possible so the probability is 1/3, 628,800

86. For the system to operate both components in the series must work. The probability they both work is 0.81, found by (0.90)(0.90)

87. a. The overall percent defective is 5.175,found by (0.20)(0.03) + (0.30)(0.04) + (0.25)(0.07) + (0.25)(0.065)

b. Using Bayes Rule:

Supplier Joint RevisedTyson 0.00600 0.1159Fuji 0.01200 0.2319Kirkpatricks 0.01750 0.3382Parts 0.01625 0.3140Total 0.05175 1.0000

88. a. P(G\A) =

Chapter 566

b. P(M|A) =

c. P(P|A) =

89. Answer depends on the game show host.

90. a. Using MINITABRows: Pool Column: Township

1 2 3 4 5 All0 9 8 7 11 3 381 6 12 18 18 13 67All 15 20 25 29 16 1051. P(1 or Pool) = 15/105 + 38/105 – 9/105 = 0.41902. P(Pool|3) = 7/25 = 0.283. P(3 and Pool) = 7/105 = 0.0667

b. Using MINITABRows: Garage Column: Township

1 2 3 4 5 All0 6 5 10 9 4 341 9 15 15 20 12 71All 15 20 25 29 16 1051. P(G) = 71/105 = 0.6762

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2. P(NG|5) = 4/16 = 0.253. P(G and 3) = 15/105 = 0.14294. P(NG or 2) = 34/105 + 20/105 – 5/105 = 49/105 = 0.4667

91. a. Winning Low Moderate HighSeason Attendance Attendance Attendance TotalNo 5 6 3 14Yes 1 12 3 16Total 6 18 6 30

1. 0.5333 found by 16/302. 0.6333 found by 16/30 + 6/30 3/30 = 19/303. 0.5000 found by 3/64. 0.1667 found by 5/30

b. Losing WinningSeason Season Total

Grass 12 15 27Artificial 2 1 3Total 14 16 301. 0.90 found by 27/302. Grass 0.5556 found by 15/27

Artificial 0.3333 found by 1/3 so grass appears better.3. 0.60 found by 16/30 + 3/30 1/30

92. a. Using MINITABRows: Pool Column: Township

1 2 3 4 5 All0 9 8 7 11 3 381 6 12 18 18 13 67All 15 20 25 29 16 1051. P(1 or Pool) = 15/105 + 67/105 – 6/105 = 0.72382. P(Pool|3) = 18/25 = 0.723. P(3 and Pool) = 18/105 = 0.1714

b. Using MINITABRows: Garage Column: Township

1 2 3 4 5 All0 6 5 10 9 4 341 9 15 15 20 12 71All 15 20 25 29 16 1051. P(G) = 71/105 = 0.67622. P(NG|5) = 4/16 = 0.253. P(G and 3) = 15/105 = 0.14294. P(NG or 2) = 34/105 + 20/105 – 5/105 = 49/105 = 0.4667

Chapter 568