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[组合] 概率一题

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guanmo1 Posted at 2013-9-12 20:44:19 |Read mode
4行4列的方阵中有16个数,从中任取3个数,则至少有2个数在同行或同列的概率是

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爪机专用 Posted at 2013-9-12 21:13:19
大概计算反面吧

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realnumber Posted at 2013-9-12 21:22:35
回复 2# 爪机专用


   \[ 那么就是 1-\frac{16}{16}\times\frac{9}{15}\times\frac{4}{14}?\]

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kuing Posted at 2013-9-12 21:41:39
反面就是三个数都不同行也不同列。
如果没有一个数在第一行,有 $4\times3\times2$ 种选择,同理,没有在其他行的也一样,所以总数是 $4\times4\times3\times2$。
基本事件总数 $C_{16}^3$,所以所求概率为 $1-4\times4\times3\times2/C_{16}^3=29/35$。

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 Author| guanmo1 Posted at 2013-9-12 21:50:43
应该是29/35,但答案是449/455,我想449/455应该是从中任取4个数,则至少有2个数在同行或同列的概率。

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kuing Posted at 2013-9-12 21:52:35
回复 5# guanmo1

是的

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