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标题: Anyone would like to show me how to solve these problems? Thanks a lot. [打印本页]

作者: redpot    时间: 2002-8-10 08:35     标题: Anyone would like to show me how to solve these problems? Thanks a lot.

1. A的发生概率为0.6,B发生的概率为0.5,问A,B都不发生的最大概率?


2. There are 6 groups in a room. Each group consists of 3 men. How many handshakes will there be if each man only shakes hands with people who are outside his group?
作者: ssmcor    时间: 2002-8-10 12:40

当A包含B时,A,B都不发生的概率最大,为0.4
作者: carricact    时间: 2002-8-10 17:45

2.第一3人于其他15人握手,向下排,第二组人再向下于12人握手(因为第一组与第二组人握手已经算过了),如此下去,分别是9,6,3,0。加起来就可以了
作者: redpot    时间: 2002-8-11 05:10

As to 1, i am still a little bit confused.  Does it mean that 0.5 possibility for both A and B occur at the same time?  How about A with 0.6, B with 0.7, C with 0.8, then the possibility for all of them not occur?  Is it 0.2?  
Well, to 2, thank you very much.  I got it.  I can also get it by (C6,2)*(3*3).
作者: tongxun    时间: 2002-8-12 02:42

第一题由于不是独立事件,所以要考虑包含的问题。0.2。
若独立,则相成。
作者: redpot    时间: 2002-8-12 11:19

Multiple?  0.4*0.5=0.2?  or 0.2*0.3*0.4=0.024?
作者: tongxun    时间: 2002-8-12 18:19

Multiple? No.
独立事件的发生概率才要乘。
作者: redpot    时间: 2002-8-13 09:04

So how about A with 0.6, B with 0.7, C with 0.8, then the possibility for all of them not occur?  If they are dependent, the answer is 0.2.  What if they are independent?
作者: zhouxh    时间: 2002-8-13 21:53

答: 1.这道题本质上是一个集合的题。如果集合中只有A,B,且A包含B,都不发生的概率最大,0.4。 概率P=满足某个条件的所有可能情况数量/所有可能情况数量 性质 0<=P<=1 a1,a2为两两不相容的事件(即发生了a1,就不会发生a2) P(a1或a2)=P(a1)+P(a2) a1,a2不是两两不相容的事件,分别用集合A和集合B来表示 即集合A与集合B有交集,表示为A*B (a1发生且a2发生) 集合A与集合B的并集,表示为A U B (a1发生或a2发生) 则 P(A U B)= P(A)+P(B)-P(A*B)。。。。。。。。。。。。。。。。。公式2 还有就是条件概率: 考虑的是事件A已发生的条件下事件B发生的概率 定义:设A,B是两个事件,且P(A)>0,称 P(B|A)=P(A*B)/P(A)
作者: redpot    时间: 2002-8-18 11:11

Thanks so much.




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