Question

Melinda 有三个空盒子和 $12$ 本课本，其中三本是数学课本。一个盒子能装任意三本课本，一个能装任意四本课本，一个能装任意五本课本。若 Melinda 按随机顺序把课本装进这些盒子，所有三本数学课本都在同一个盒子里的概率可写成 $\frac{m}{n}$ ，其中 $m$ 和 $n$ 是互质正整数。求 $m + n$ 。

Melinda has three empty boxes and $12$ textbooks, three of which are mathematics textbooks. One box will hold any three of her textbooks, one will hold any four of her textbooks, and one will hold any five of her textbooks. If Melinda packs her textbooks into these boxes in random order, the probability that all three mathematics textbooks end up in the same box can be written as $\frac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m + n.$

答案

答案：47

知识点：基本概率组合分类讨论

难度评级：2390

解答：

一次只看一个盒子。装 $k$ 本书的盒子得到 $12$ 本书中的一个等可能的 $k$ 元子集，因此它含有全部三本数学书的概率为 $\binom{9}{k-3}\big/\binom{12}{k}$ 。当 $k = 3, 4, 5$ 时，分别为 $\frac{1}{220}$ 、 $\frac{9}{495} = \frac{1}{55}$ 、 $\frac{36}{792} = \frac{1}{22}$ 。

这些事件互不相交，所以总概率为因此 $m + n = 3 + 44 = 47$ 。 $\begin{aligned} \frac{1}{220} + \frac{1}{55} + \frac{1}{22} &= \frac{1 + 4 + 10}{220} \\ &= \frac{15}{220} = \frac{3}{44}, \end{aligned}$

Focus on one box at a time. The box of $k$ books receives a uniformly random $k$ -subset of the $12$ books, so the probability that it contains all three math books is $\binom{9}{k-3}\big/\binom{12}{k}.$ For $k = 3, 4, 5$ this gives $\frac{1}{220},$ $\frac{9}{495} = \frac{1}{55},$ and $\frac{36}{792} = \frac{1}{22}.$

The events are disjoint, so the total probability is $\begin{aligned} \frac{1}{220} + \frac{1}{55} + \frac{1}{22} &= \frac{1 + 4 + 10}{220} \\ &= \frac{15}{220} = \frac{3}{44}, \end{aligned}$ and $m + n = 3 + 44 = 47.$

2013 AIME I 第 6 题

答案：47

难度评级：2390

解答：

其他年份的第 6 题