12.10: Chapter 10 Exercise Solutions

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Chapter 10: Probability

1. a. $\dfrac{6}{13}$

b. $\dfrac{2}{13}$

3. $\dfrac{150}{335} = 44.8 \%$

5. $\dfrac{1}{6}$

7. $\dfrac{26}{65}$

9. $\dfrac{3}{6} = \dfrac{1}{2} $

11. $\dfrac{4}{52} = \dfrac{1}{4}$

13. $1 - \dfrac{1}{12} = \dfrac{11}{12}$

15. $1 - \dfrac{25}{65} = \dfrac{40}{65}$

17. $\dfrac{1}{6} \cdot \dfrac{1}{6} = \dfrac{1}{36}$

19. $\dfrac{1}{6} \cdot \dfrac{3}{6} = \dfrac{3}{36} = \dfrac{1}{12} $

21. $\dfrac{17}{49} \cdot \dfrac{16}{48} = \dfrac{17}{49} \cdot \dfrac{1}{3} = \dfrac{17}{147} $

23. a. $\dfrac{4}{52} \cdot \dfrac{4}{52} = \dfrac{16}{2704} = \dfrac{1}{169} $

b. $\dfrac{4}{52} \cdot \dfrac{48}{52} = \dfrac{192}{2704} = \dfrac{12}{169} $

c. $\dfrac{48}{52} \cdot \dfrac{48}{52} = \dfrac{2304}{2704} = \dfrac{144}{169} $

d. $\dfrac{13}{52} \cdot \dfrac{13}{52} = \dfrac{169}{2704} = \dfrac{1}{16} $

e. $\dfrac{48}{52} \cdot \dfrac{39}{52} = \dfrac{1872}{2704} = \dfrac{117}{169} $

25. $\dfrac{4}{52} \cdot \dfrac{4}{51} = \dfrac{16}{2652}$

27. a. $\dfrac{11}{25} \cdot \dfrac{14}{24} = \dfrac{154}{600}$

b. $\dfrac{14}{25} \cdot \dfrac{11}{24} = \dfrac{154}{600}$

c. $\dfrac{11}{25} \cdot \dfrac{10}{24} = \dfrac{110}{600}$

d. $\dfrac{14}{25} \cdot \dfrac{13}{24} = \dfrac{182}{600}$

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

29. $P(\text{F} \text{ and } \text{A}) = \dfrac{10}{65} $

31. $P(\text{red} \text{ or } \text{odd}) = \dfrac{6}{14} + \dfrac{7}{14} - \dfrac{3}{14} = \dfrac{10}{14} $. Or $6$ red and $4$ odd-numbered blue marbles is $10$ out of $14$.

33. $P(\text{F} \text{ or } \text{B}) = \dfrac{26}{65} + \dfrac{22}{65} - \dfrac{4}{65} = \dfrac{44}{65}$. Or $P(\text{F} \text{ or } \text{B}) = \dfrac{18+4+10+12}{65} = \dfrac{44}{65} $

35. $P(\text{King of Hearts or Queen}) = \dfrac{1}{52} + \dfrac{4}{52} = \dfrac{5}{52} $

37. a. $P(\text{even} | \text{red}) = \dfrac{2}{5} $

b. $P(\text{even} | \text{red}) = \dfrac{2}{6} $

39. $P(\text{Heads on second} | \text{Tails on first}) = \dfrac{1}{2}$. They are independent events.

41. $P(\text{speak French} | \text{female}) = \dfrac{3}{14}$

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. $P(\text{virus} | \text{positive}) = \dfrac{9}{9 + 399} = \dfrac{9}{408} = 2.2 \%$

b. $P(\text{no virus} | \text{negative}) = \dfrac{3591}{3591 + 1} = \dfrac{3591}{3592} = 99.97 \%$

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. $P(\text{disease} | \text{positive}) = \dfrac{282}{282 + 3988} = \dfrac{720}{7664} = 6.6 \%$

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. $P(\text{cancer} | \text{positive}) = \dfrac{720}{720 + 6944} = \dfrac{720}{7664} = 9.4 \%$

49. $2 \cdot 3 \cdot 8 \cdot 2 = 96$ outfits

51. a. $4 \cdot 4 \cdot 4 = 64$

b. $4 \cdot 3 \cdot 2 = 24$

53. $26 \cdot 26 \cdot 26 \cdot 10 \cdot 10 \cdot 10 = 17,576,000$

55. $_4P_4 \text{ or } 4 \cdot 3 \cdot 2 \cdot 1 = 24$ possible orders

57. Order matters. $_7P_4 = 840$ possible teams

59. Order matters. $_{12}P_5 = 95,040$ possible themes

61. Order does not matter. $_{12}C_4 = 495$

63. $_{50}C_6 = 15,890,700$

65. $_{27}C_{11} \cdot 16 = 208,606,320$

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 $5!$. $P(\text{alphabetical}) = \dfrac{1}{5!} = \dfrac{1}{(_5P_5)} = \dfrac{1}{120}$.

69. There are $_{48}C_6$ total tickets. To match $5$ of the $6$, a player would need to choose $5$ of those $6$, $_6C_5$, and one of the $42$ non-winning numbers, $_{42}C_1$. $\dfrac{6 \cdot 42}{12271512} = \dfrac{252}{12271512}$

71. All possible hands is $_{52}C_5$. Hands will all hearts is $_{13}C_5$. $\dfrac{1287}{2598960}$.

73. $ $3 \left(\dfrac{3}{37} \right) + $2 \left(\dfrac{6}{37} \right) + (-$1) \left(\dfrac{3}{37} \right) = -$ \dfrac{7}{37} + = -$0.19$

75. There are $_{23}C_6 = 100,947$ possible tickets.

Expected value = $$29,999 \left(\dfrac{1}{100947} \right) + (-$1) \left(\dfrac{100946}{100947} \right) = -$0.70$

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

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