Question

将 $20$ 个球各自独立随机地投入 $5$ 个箱子之一。设 $p$ 为某个箱子最终有 $3$ 个球、另一个箱子有 $5$ 个球、其余三个箱子各有 $4$ 个球的概率。设 $q$ 为每个箱子最终都有 $4$ 个球的概率。 $\dfrac{p}{q}$ 是多少？

Each of $20$ balls is tossed independently and at random into one of $5$ bins. Let $p$ be the probability that some bin ends up with $3$ balls, another with $5$ balls, and the other three with $4$ balls each. Let $q$ be the probability that every bin ends up with $4$ balls. What is $\dfrac{p}{q}?$

$1$

$4$

$8$

$12$

$16$

答案：E

知识点：基本概率多重集排列

难度评级：1990

解答：

两个概率都除以 $5^{20}$ ，所以 $\dfrac{p}{q}$ 是排列计数之比。

对于 $q$ ，所有箱子都有 $4$ 个球，计数为 $\dfrac{20!}{(4!)^5}$ 。对于 $p$ ，选择哪个箱子有 $3$ 个、哪个箱子有 $5$ 个有 $5\cdot 4 = 20$ 种，再乘以 $\dfrac{20!}{3!\,5!\,(4!)^3}$ 。因此 $\begin{aligned} \frac{p}{q} &= 20 \cdot \frac{(4!)^5}{3!\,5!\,(4!)^3} \\ &= 20 \cdot \frac{(4!)^2}{3!\,5!} \\ &= 20 \cdot \frac{576}{720} \\ &= 16. \end{aligned}$

所以正确答案是 E。

Both probabilities divide by $5^{20},$ so $\dfrac{p}{q}$ is a ratio of arrangement counts.

For $q,$ all bins have $4:$ $\dfrac{20!}{(4!)^5}.$ For $p,$ choose which bin has $3$ and which has $5$ in $5\cdot 4 = 20$ ways, times $\dfrac{20!}{3!\,5!\,(4!)^3}.$ Therefore $\begin{aligned} \frac{p}{q} &= 20 \cdot \frac{(4!)^5}{3!\,5!\,(4!)^3} \\ &= 20 \cdot \frac{(4!)^2}{3!\,5!} \\ &= 20 \cdot \frac{576}{720} \\ &= 16. \end{aligned}$

Thus, the correct answer is E.

2021 AMC 12A Fall 第 18 题

答案：E

难度评级：1990

解答：

其他年份的第 18 题