一道让我烦的概率题，请帮助

steven.wei

4封放4个信封，只有一个放对的概率。
哪位G友帮我解释一下，特别是每一步的过程，谢谢。

tongxun

1、这个放对的信，可以是4封中的任何一个：4
2、剩下3封中取1封放：C3，1
3、放错信封的机会：2
4、剩下的2封只有1中放发：1
4*3*2*1=24

justag

to simplify the problem, let's say the four evelopes are 1, 2, 3, and 4 and the corresponding envelopes are 1', 2', 3' and 4' (let's say the envolpes are put on table in this order, so we only need to consider the letters' location)

total ways of allocation: P(4,4)
exactly one letter in the right envelop: 4 choices,1-1', or 2-2'...
(let's say, we put letter 4 right, ie. 4-4')
1, 2 and 3 are left and we need to put them all wrong. We only have two choices, 2,3,1 or 3,1,2. So taotal ways of allocation with only one letter int he right envelope: 4*2.

So the possibility should be: (4*2)/p(4,4)=8/24=1/3.

A professional way to solve this problem is:
[p(1,4)*p(1,2)*1]/p(4,4). The above example can help you understand why it's
p(1,2)*1 after p(1,4)

steven.wei

感谢二位仁兄的指点。一个在11点半，一个在凌晨。看来数学不拿5＊对不起帮助我的人，再次表示感谢。