Question

一组 $12$ 个筹码中有 $3$ 个红色、 $2$ 个白色、 $1$ 个蓝色和 $6$ 个黑色。这些筹码随机分给 $3$ 个游戏玩家，每人 $4$ 个筹码。某个玩家得到所有红色筹码，另一个玩家得到所有白色筹码，剩下的玩家得到蓝色筹码的概率可写成 $\dfrac{m}{n}$ ，其中 $m$ 和 $n$ 是互质的正整数。求 $m+n$ ？

A set of $12$ tokens — $3$ red, $2$ white, $1$ blue, and $6$ black — is to be distributed at random to $3$ game players, $4$ tokens per player. The probability that some player gets all the red tokens, another gets all the white tokens, and the remaining player gets the blue token can be written as $\dfrac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. What is $m+n?$

$387$

$388$

$389$

$390$

$391$

答案：C

知识点：基本概率多重集排列

难度评级：1820

解答：

把所有筹码视为可区分的；给每位玩家发 $4$ 个的总方式数是 $\binom{12}{4,4,4}=34650$ 。对有利事件，选择哪位玩家得到红色、白色和蓝色有 $3!=6$ 种方式。红色玩家还需 $1$ 个筹码，白色玩家还需 $2$ 个，蓝色玩家还需 $3$ 个，全部为黑色； $6$ 个黑色筹码按 $1,2,3$ 分配有 $\tfrac{6!}{1!\,2!\,3!}=60$ 种方式。概率为 $\dfrac{6\cdot60}{34650}=\dfrac{360}{34650}=\dfrac{4}{385}$ 。所以 $m+n=4+385=389$ 。因此正确答案是 C。

Treat all tokens as distinct; the total number of ways to deal $4$ to each player is $\binom{12}{4,4,4}=34650.$ For the favorable event, choose which player gets the reds, whites, and blue in $3!=6$ ways. The red player needs $1$ more token, the white player $2$ more, and the blue player $3$ more, all black; the $6$ black tokens split as $1,2,3$ in $\tfrac{6!}{1!\,2!\,3!}=60$ ways. So the probability is $\dfrac{6\cdot60}{34650}=\dfrac{360}{34650}=\dfrac{4}{385}.$ Then $m+n=4+385=389.$ Thus, the correct answer is C.

2024 AMC 12A 第 16 题

答案：C

难度评级：1820

解答：

其他年份的第 16 题