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# 12.10: Capítulo 10 Soluciones para ejercicios

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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 machos = dos hembras. Igual que la parte d.

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

31. $$P(\text{red} \text{ or } \text{odd}) = \dfrac{6}{14} + \dfrac{7}{14} - \dfrac{3}{14} = \dfrac{10}{14}$$. O los canicas azules$$6$$ rojos e$$4$$ impares están$$10$$ fuera de$$14$$.

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

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. $$P(\text{Heads on second} | \text{Tails on first}) = \dfrac{1}{2}$$. Son eventos independientes.

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

43. Fuera de la$$4,000$$ gente,$$10$$ tendría la enfermedad. De esos$$10$$,$$9$$ daría positivo, mientras que falsamente$$1$$ daría negativo. De las personas$$3990$$ no infectadas, falsamente$$399$$ daría positivo, mientras que$$3591$$ daría negativo.

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. Fuera de la$$100,000$$ gente,$$300$$ tendría la enfermedad. De esos, falsamente$$18$$ daría negativo, mientras que$$282$$ daría positivo. De los$$99,700$$ sin la enfermedad, falsamente$$3,988$$ daría positivo y el otro$$95,712$$ daría negativo. $$P(\text{disease} | \text{positive}) = \dfrac{282}{282 + 3988} = \dfrac{720}{7664} = 6.6 \%$$

47. Fuera de$$100,000$$ las mujeres,$$800$$ tendría cáncer de mama. De esos, falsamente$$80$$ daría negativo, mientras que$$720$$ daría positivo. De los$$99,200$$ sin cáncer, falsamente$$6,944$$ daría positivo. $$P(\text{cancer} | \text{positive}) = \dfrac{720}{720 + 6944} = \dfrac{720}{7664} = 9.4 \%$$

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

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. $$_4P_4 \text{ or } 4 \cdot 3 \cdot 2 \cdot 1 = 24$$pedidos posibles

57. Cuestiones de orden. $$_7P_4 = 840$$posibles equipos

59. Cuestiones de orden. $$_{12}P_5 = 95,040$$posibles temas

61. El orden no importa. $$_{12}C_4 = 495$$

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

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

67. Sólo hay$$1$$ forma de organizar los$$5$$ CD's en orden alfabético. La probabilidad de que los CD estén en orden alfabético es una dividida por el número total de formas de organizar los$$5$$ CD's. Dado que el orden alfabético es solo uno de todos los ordenamientos posibles puedes usar permutaciones, o simplemente usar$$5!$$. $$P(\text{alphabetical}) = \dfrac{1}{5!} = \dfrac{1}{(_5P_5)} = \dfrac{1}{120}$$.

69. Hay boletos$$_{48}C_6$$ totales. Para igualar$$5$$ del$$6$$, un jugador tendría que elegir entre$$5$$ esos$$6$$,$$_6C_5$$, y uno de los números$$42$$ no ganadores,$$_{42}C_1$$. $$\dfrac{6 \cdot 42}{12271512} = \dfrac{252}{12271512}$$

71. Todas las manos posibles es$$_{52}C_5$$. Las manos serán todos los corazones es$$_{13}C_5$$. $$\dfrac{1287}{2598960}$$.

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. Hay$$_{23}C_6 = 100,947$$ posibles boletos.

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

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

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