Question

一家旅馆为三位客人各打包了一份早餐。每份早餐本应包含三种小面包：坚果、奶酪和水果口味各一个。准备早餐的人把这九个小面包分别包好；包好后，这些小面包彼此无法区分。然后她随机给每位客人的袋子里放入三个小面包。已知每位客人都拿到每种口味各一个小面包的概率为 $\frac{m}{n}$ ，其中 $m$ 和 $n$ 是互质正整数，求 $m + n$ 。

A hotel packed a breakfast for each of three guests. Each breakfast should have consisted of three types of rolls, one each of nut, cheese, and fruit rolls. The preparer wrapped each of the nine rolls, and, once they were wrapped, the rolls were indistinguishable from one another. She then randomly put three rolls in a bag for each of the guests. Given that the probability that each guest got one roll of each type is $\frac{m}{n},$ where $m$ and $n$ are relatively prime positive integers, find $m + n.$

答案

答案：79

知识点：无放回抽样条件概率

难度评级：2170

解答：

逐个装第一位客人的袋子。第一个小面包可以是任意口味；第二个必须避开与第一个同口味的剩余 $2$ 个小面包，成功概率为 $\frac{6}{8}$ ；第三个必须是剩余 $7$ 个中缺少口味的 $3$ 个之一。因此第一个袋子含有每种口味各一个的概率为 $\frac{6}{8} \cdot \frac{3}{7} = \frac{9}{28}$ 。

在此条件下，剩下六个小面包，每种口味各两个，同样的论证给出第二个袋子的概率为 $\frac{4}{5} \cdot \frac{2}{4} = \frac{2}{5}$ 。此时第三个袋子自动是每种口味各一个。所求概率为 $\frac{9}{28} \cdot \frac{2}{5} = \frac{9}{70}$ ，所以 $m + n = 9 + 70 = 79$ 。

Fill the first guest's bag one roll at a time. The first roll can be anything; the second must avoid the $2$ remaining rolls of the first roll's type, succeeding with probability $\frac{6}{8};$ and the third must be one of the $3$ rolls of the missing type among the remaining $7.$ So the first bag has one roll of each type with probability $\frac{6}{8} \cdot \frac{3}{7} = \frac{9}{28}.$

Given that, six rolls remain, two of each type, and the same argument gives $\frac{4}{5} \cdot \frac{2}{4} = \frac{2}{5}$ for the second bag. The third bag is then automatically one of each type. The probability is $\frac{9}{28} \cdot \frac{2}{5} = \frac{9}{70},$ so $m + n = 9 + 70 = 79.$

2005 AIME II 第 2 题

答案：79

难度评级：2170

解答：

其他年份的第 2 题