2021 AMC 12A Fall 第 18 题

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

小提示：

两个概率都有相同的

5^{20}

分母，所以

\dfrac{p}{q}

是计数之比

Both probabilities share the same

5^{20}

denominator, so

\dfrac{p}{q}

is a ratio of counts

大提示：

q

使用

\dfrac{20!}{(4!)^5}

；

p

使用

5\cdot 4

种特殊箱子的选择，再乘以

\dfrac{20!}{3!\,5!\,(4!)^3}

q

uses

\dfrac{20!}{(4!)^5};

p

uses

5\cdot 4

choices of the special bins times

\dfrac{20!}{3!\,5!\,(4!)^3}

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}\text{。}

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

小提示：

大提示：

解答：

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