prep-ps2-92

ferferzor

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。

federico0555

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。

reginaer

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

ferferzor

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.

vincentwh

 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

ferferzor

不好意思我的算术都很烂！！

明白了！