Question

四十张纸条放入一顶帽子中，每张纸条上写有 $1, 2, 3, 4, 5, 6, 7, 8, 9$ 或 $10$ ，且每个数字写在四张纸条上。不放回地随机抽出四张纸条。设 $p$ 为四张纸条都写有同一个数字的概率。设 $q$ 为其中两张写有数字 $a$ ，另外两张写有数字 $b \ne a$ 的概率。求 $\dfrac{q}{p}$ ？

Forty slips are placed into a hat, each bearing a number $1, 2, 3, 4, 5, 6, 7, 8, 9,$ or $10,$ with each number entered on four slips. Four slips are drawn from the hat at random and without replacement. Let $p$ be the probability that all four slips bear the same number. Let $q$ be the probability that two of the slips bear a number $a$ and the other two bear a number $b \ne a.$ What is the value of $\dfrac{q}{p}?$

$162$

$180$

$324$

$360$

$720$

答案：A

知识点：基本概率组合无放回抽样

难度评级：1660

解答：

两个事件都从 $\binom{40}{4}$ 个等可能选择中抽取，所以 $\dfrac{q}{p}$ 等于它们有利情况数量之比。

四张纸条都写同一个数字的抽法有 $10$ 种，每个数字一种。

若是两个 $a$ 和两个 $b$ ，先选两个数字，有 $\binom{10}{2}$ 种，再从四张 $a$ 纸条中选两张、从四张 $b$ 纸条中选两张： $\begin{aligned} \binom{10}{2}\binom{4}{2}\binom{4}{2} &= 45 \cdot 6 \cdot 6 \\ &= 1620. \end{aligned}$

因此 $\dfrac{q}{p} = \dfrac{1620}{10} = 162$ 。

所以正确答案是 A。

Both events draw from $\binom{40}{4}$ equally likely selections, so $\dfrac{q}{p}$ is the ratio of their favorable counts.

Exactly $10$ draws give four slips of the same number, one for each value.

For two $a$ 's and two $b$ 's, choose the two values in $\binom{10}{2}$ ways, then two of the four $a$ -slips and two of the four $b$ -slips: $\begin{aligned} \binom{10}{2}\binom{4}{2}\binom{4}{2} &= 45 \cdot 6 \cdot 6 \\ &= 1620. \end{aligned}$

Therefore $\dfrac{q}{p} = \dfrac{1620}{10} = 162.$

Thus, A is the correct answer.

2005 AMC 10B 第 21 题

答案：A

难度评级：1660

解答：

其他年份的第 21 题