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标题: prep-ps2-92 [打印本页]

作者: ferferzor    时间: 2009-8-7 20:33     标题: prep-ps2-92

On his drive to work, Leo listens to one of three radio stations, A, B, or C.  He first turns to A.  If A is playing a song he likes, he listens to it; if not, he turns to B.  If B is playing a song he likes, he listens to it; if not, he turns to C.  If C is playing a song he likes, he listens to it; if not, he turns off the radio.  For each station, the probability is 0.30 that at any given moment the station is playing a song Leo likes.  On his drive to work, what is the probability that Leo will hear a song he likes?

0.657

不知道怎么做,求解。


作者: federico0555    时间: 2009-8-8 04:03

每个台,Leo爱听的概率是0.3,不爱听的概率是0.7,那么三个台都不爱听的概率是0.7的三次方,即0.343,1-0.343=0.657,至少保证有台爱听。
作者: reginaer    时间: 2009-8-8 09:06

A 几率0.3

B 几率0.7*0.3(A没有好听歌的几率0.7)

C 几率0.7*0.7*0.3(同理)

三个相加,0.3+0.21+0.147=0.657


作者: ferferzor    时间: 2009-8-10 20:51

谢谢,

我自己想了一种,

解释eo可以在任一个station遇到他喜欢的,除非三个station都没有他喜欢的.


作者: vincentwh    时间: 2009-8-11 21:48

 Leo可以在任一个station遇到他喜欢的,除非三个station都没有他喜欢的.

这个解法的正确表达应该是1-0.7×0.7×0.7

就是所有的可能里面减去三个都没有喜欢的。所有的可能是1而不是你写的0.3*3


作者: ferferzor    时间: 2009-8-12 06:54

不好意思我的算术都很烂!!

明白了!






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