Question

一副四十张的牌中，数字 $1$ 、 $2$ 、 $\ldots$ 、 $10$ 各出现四次。先从牌中取走一对数字相同的牌，并且不将它们放回。设接着随机抽出的两张牌也组成一对的概率为 $m/n$ ，其中 $m$ 和 $n$ 是互质的正整数。求 $m + n$ 。

A deck of forty cards consists of four $1$ 's, four $2$ 's, $\ldots,$ and four $10$ 's. A matching pair (two cards with the same number) is removed from the deck. Given that these cards are not returned to the deck, let $m/n$ be the probability that two randomly selected cards also form a pair, where $m$ and $n$ are relatively prime positive integers. Find $m + n.$

答案

答案：758

知识点：基本概率组合

难度评级：2020

解答：

取走相同数字的一对后，还剩 $38$ 张牌：九个数字各有四张，一个数字只剩两张。能抽成一对的抽法数为 $9\binom{4}{2} + \binom{2}{2} = 54 + 1 = 55$ ，而所有等可能抽法数为 $\binom{38}{2} = 703$ 。

因为 $703 = 19 \cdot 37$ 与 $55 = 5 \cdot 11$ ，没有公因子，所以概率 $\frac{55}{703}$ 已是最简形式， $m + n = 55 + 703 = 758$ 。

After the matching pair is removed, $38$ cards remain: nine numbers with four cards each and one number with only two cards. The number of ways to draw a pair is $9\binom{4}{2} + \binom{2}{2} = 54 + 1 = 55,$ out of $\binom{38}{2} = 703$ equally likely draws.

Since $703 = 19 \cdot 37$ shares no factor with $55 = 5 \cdot 11,$ the probability $\frac{55}{703}$ is in lowest terms, and $m + n = 55 + 703 = 758.$

2000 AIME II 第 3 题

答案：758

难度评级：2020

解答：

其他年份的第 3 题