Question

$20$ 个球各自独立且随机地投入 $5$ 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。求 $\dfrac{p}{q}$ ？

Each of $20$ balls is tossed independently and at random into one of the $5$ 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：组合基本概率

难度评级：1540

解答：

把球和箱子都看作可区分。 $5^{20}$ $q$ ， $\frac{20!}{(4!)^5}.$

两种情形都有三个箱子各含 $5\cdot4$ 个球，所以计算时可以先约去这 $3$ 个箱子的共同部分。对 $p$ 有 $5$ 种方法选择装个球的箱子，再有种方法选择装个球的箱子。最后有种方法分配相应的球。 $5\cdot4\cdot\frac{20!}{3!5!(4!)^3}.$

对约去个共同的装球数为的箱子后，有种方法确保剩余每个箱子装个球。 $\frac pq=20\cdot\frac{(4!)^2}{3!5!}=20\cdot\frac45=16.$

两种情况的总投放数相同，所以与之比就是有利投放数之比。因此满足所以正确答案是 E。

All $5^{20}$ assignments of the distinguishable balls to the labeled bins are equally likely. For $q,$ the number of assignments is $\frac{20!}{(4!)^5}.$

For $p,$ choose the bin with $3$ balls and the bin with $5$ balls in $5\cdot4$ ways. The number of assignments is then $5\cdot4\cdot\frac{20!}{3!5!(4!)^3}.$

The common probability denominator cancels, so $\frac pq=20\cdot\frac{(4!)^2}{3!5!}=20\cdot\frac45=16.$

Thus, E is the correct answer.

2021 AMC 10A Fall 第 21 题

答案：E

难度评级：1540

解答：

其他年份的第 21 题