PREP 中一题

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 stations, the probability is 0.30 that 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?

0.027, 0090, 0.417, 0.657,0,900

答案是0.657

1个不听的概率为.7,

3个电台都不听的概率为.7*.7*.7

所以听任意一个电台的可能性为1-.7*.7*.7=.657

问题的对立面是：Leo没有听到喜欢的Song，这种情况的概率是(1－0.3)(1－0.3)(1－0.3)=0.343

所以这道题的答案是1－0.343＝0.657

为什么不能用重复独立性实验的

3*0.3*0.7*0.7 呢？

因为这ABC并非独立发生的事件，而是相互影响的事件！！！

如果停留在A，则BC不发生，也就是根本不用考虑BC的概率，0.3

如果A不停留，停留在B，则C不发生则不用考虑C。0.7*0.3

如果AB不停留，则停留在C。0.7*0.7*0.3

再将三种情况相加得到.657

重点是要明白这不是独立互不影响的事件！！！而是相互影响的事件！！！

觉得很奇怪，GMAT中只考察independent的事件，而不会考察条件概率，也就是两件事是相互影响的！LZ哪里来的题目？？

我个人觉得这道题肯定是不会计分的题目。不用太在意！！！