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12.10: Chapter 10 Exercise Solutions

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Chapter 10: Probability

1. a. 613

b. 213

3. 150335=44.8%

5. 16

7. 2665

9. 36=12

11. 452=14

13. 1112=1112

15. 12565=4065

17. 1616=136

19. 1636=336=112

21. 17491648=174913=17147

23. a. 452452=162704=1169

b. 4524852=1922704=12169

c. 48524852=23042704=144169

d. 13521352=1692704=116

e. 48523952=18722704=117169

25. 452451=162652

27. a. 11251424=154600

b. 14251124=154600

c. 11251024=110600

d. 14251324=182600

e. no males = two females. Same as part d.

29. P(F and A)=1065

31. P(red or odd)=614+714314=1014. Or 6 red and 4 odd-numbered blue marbles is 10 out of 14.

33. P(F or B)=2665+2265465=4465. Or P(F or B)=18+4+10+1265=4465

35. P(King of Hearts or Queen)=152+452=552

37. a. P(even|red)=25

b. P(even|red)=26

39. P(Heads on second|Tails on first)=12. They are independent events.

41. P(speak French|female)=314

43. Out of 4,000 people, 10 would have the disease. Out of those 10, 9 would test positive, while 1 would falsely test negative. Out of the 3990 uninfected people, 399 would falsely test positive, while 3591 would test negative.

a. P(virus|positive)=99+399=9408=2.2%

b. P(no virus|negative)=35913591+1=35913592=99.97%

45. Out of 100,000 people, 300 would have the disease. Of those, 18 would falsely test negative, while 282 would test positive. Of the 99,700 without the disease, 3,988 would falsely test positive and the other 95,712 would test negative. P(disease|positive)=282282+3988=7207664=6.6%

47. Out of 100,000 women, 800 would have breast cancer. Out of those, 80 would falsely test negative, while 720 would test positive. Of the 99,200 without cancer, 6,944 would falsely test positive. P(cancer|positive)=720720+6944=7207664=9.4%

49. 2382=96 outfits

51. a. 444=64

b. 432=24

53. 262626101010=17,576,000

55. 4P4 or 4321=24 possible orders

57. Order matters. 7P4=840 possible teams

59. Order matters. 12P5=95,040 possible themes

61. Order does not matter. 12C4=495

63. 50C6=15,890,700

65. 27C1116=208,606,320

67. There is only 1 way to arrange 5 CD's in alphabetical order. The probability that the CD's are in alphabetical order is one divided by the total number of ways to arrange 5 CD's. Since alphabetical order is only one of all the possible orderings you can either use permutations, or simply use 5!. P(alphabetical)=15!=1(5P5)=1120.

69. There are 48C6 total tickets. To match 5 of the 6, a player would need to choose 5 of those 6, 6C5, and one of the 42 non-winning numbers, 42C1. 64212271512=25212271512

71. All possible hands is 52C5. Hands will all hearts is 13C5. 12872598960.

73. $3(337)+$2(637)+($1)(337)=$737+=$0.19

75. There are 23C6=100,947 possible tickets.

Expected value = $29,999(1100947)+($1)(100946100947)=$0.70

77. $48(0.993)+($302)(0.007)=$45.55


This page titled 12.10: Chapter 10 Exercise Solutions is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Darlene Diaz (ASCCC Open Educational Resources Initiative) via source content that was edited to the style and standards of the LibreTexts platform.

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