prep 的一道数学题。麻烦各位高人解答一下!

Tiff323 · 发表于 2011-10-10 08:41:03

Tanya prepared 4 different letters to be sent to 4 different addresses. For each letter, she prepared an envelop with its correct address. If the 4 letters are to be put into the 4 envelops at random, what is the probability that only 1 letter will be put into the envelop with its correct address?

顺便问问如果题目是：what is the probability that only 2 letters will be put into the envelop with their correct addresses? 又该如何解答。
感谢各位朋友了

xiang5796 · 发表于 2011-10-10 08:46:13

1/3 .........？

tenby · 发表于 2011-10-10 09:06:25

C(4,1)*C(2,1)*C(1,1)/P(4,4)

Tiff323 · 发表于 2011-10-10 09:23:15

是的，请问怎么解？思路是？

Tiff323 · 发表于 2011-10-10 09:23:46

请问解题思路是？

tenby · 发表于 2011-10-10 09:39:20

随即选对一种的选发有C(4,1)，然后另外三个不能匹配共 C（2,1）C(1,1)
so total C(4,1)C（2,1）C(1,1)

4种随便投，等于排队 P（4,4）

so概率C(4,1)C（2,1）C(1,1)/P（4,4）=1/3
好好琢磨一下。

Tiff323 · 发表于 2011-10-10 09:57:11

不好意思，我还是没搞懂。分子弄懂了，分母为什么是p(4,4) 而不是16？

Tiff323 · 发表于 2011-10-13 05:39:53

maxawei · 发表于 2011-10-13 08:12:35

one letter correct: C(4,1)*C(2,1)*C(1,1)/P(4,4)
two letters correct: C(4,2)*C(1,1)/P(4,4)
three letters correct=four letters correct=1

prep 的一道数学题。麻烦各位高人解答一下!

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