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351?e?2(1?1(12)2?1???142??12?2?2!?23!4!?5!)?0.99.
?1(131 ⑵ P(??31)?e22)?31!?0. 44.解:⑴ P(??1)?1?P(??0)?1?(0.3)3=0.97. ⑵ P(??2)?1?P(??3)?1?(0.7)3=0.66.
45. 解:P(??3)?C35(0.6)3(0.4)2?C445(0.6)0.4?C55(0.6)5=0.68.
2046. 解:??4P(??10)??4k?0.0081.
k?10k!e?4547.解:??6 ⑴ P(??5)?65!e?6?0.16.
(利用事件的独立性求概率) 48. 解:记Ai??第i家电视台在播放广告?,A为待求概率的事件.
⑴ A?A1A2A3,事件A1,A2,A3独立.
P(A)?P(A1)P(A2)P(A3)?0.1?0.15?0.2?0.003. ⑵ A?A1A2A3,事件A1,A2,A3独立,
P(A)?P(A1)P(A2)P(A3)?(1?0.1)(1?0.15)(1?0.2)?0.612. ⑶ A?A1?A2?A3?A1A2A3,P(A)?1?P(A1)P(A2)P(A3)?0.388. 49. 解:记Ai??第i个电器损坏?
(i?1,2,3),A为所求概率的事件.
⑴ A?A1A2A3,由题意,事件A1,A2,A3独立.
P(A)?P(A1)P(A2)P(A3)?0.3?0.4?0.5?0.06.
⑵ A?A1?A2?A3?A1A3A3, P(A)?1?0.06=0.94 50. 解:设A??目标被击中
?,Ai??第一组第i个射击手射中目标
?,
Bi??第二组第i个射击手中目标? (i=1,2,3,4,5),
则:A?A1?A2?A3?A4?A5,Ai(i?1,?,5)是独立的,
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?P(A)?1?P(A)?1?P(A1A2A3A4A5)?0.982.
同理:P(A)?1?P(B1B2B3B4B5)?0.9832. 所以第二组击中目标的概率大. 51. 解:设需n组系统,A??室内有灯照明
?,Ai??第i组系统正常?(i?1,?,n),
则:P(Ai)?0.6?0.5?0.3 A?A1???An,
P(A)?1?P(A)=1?P(A1)P(A2)?P(An)?1?(0.7)n?0.95 0.05?(0.7)n n? n?9. 52. 解:⑴ 记Ai?log0.05log0.7??1.3?0.1549?8.39
?第i架飞机投中目标?(i?1,?,5),
A?A1A2A3A4A5,Ai独立(i?1,?,5); (1)P(A)?(0.6)5?0.08.
(2)A?A1A2A3A4A5?...?A1A2A3A4A5,P(A)?0.6?(0.4)4?5?0.08. (3)设应有n架飞机去轰炸,
n P(A)?1?P(A)?1??P(Ai)?1?(0.4)n?0.9i?1(0.4)?0.1
n n?lg0.1lg0.4, n?3.
53.解:记Ai??第i名得满分?(i?1,?,4), 记A为所求事件.
⑴ P(A)?P(A1A2A3A4)?P(A1A2A3A4)?P(A1A2A3A4)?P(A1A2A3A4)
=0.04.
⑵ P(A)?0.1?0.2?0.3?0.4?0.0024. 54. 解:记Ai?A??第i道门被打开?(i?1,2,3),A1,A2,A3独立,
?此人进屋
?,A?A1A2A3,P(Ai)?13?13?13?12713,(i?1,2,3),
P(A)?P(A1)P(A2)P(A3)?.
55.解:记D为所求事件.
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A?B?C??乘公交车回家时间超过半小时乘地铁回家时间超过半小时
?,
???, ?,
乘出租车回家时间超过半小时
⑴P(D)?P(A?B?C)?1?P(A)P(B)P(C)=0.96. ⑵ D?ABC?ABC?ABC?ABC,
P(D)?P(ABC)?P(ABC)?P(ABC)?P(ABC)=0.7.
56. 解:记B={三个频道都在放广告}为所求事件,则
⑴ 记Ai??第i个频道在播广告?
16?112015??1?(i?1,2,3), 1 P(B)?P(A1A2A3)?⑵ P(B)?1?P(B)?1?
57. 解:记Ai?4120119120.
.
?第i个运动员能拿到锦旗?
(i?1,?,5),B??所求事件
?.
⑴ P(B)?1?P(B)?1?P(A1A2A3A4A5)?0.99. ⑵ B?A1A2A3A4A5?A1A2A3A4A5???A1A2A3A4A5,
P(B)?0.02.
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