Question

Al、Bill 和 Cal 将各自随机分到一个 $1$ 到 $10$ 之间的整数，且三人得到的数两两不同。Al 的数是 Bill 的数的正整数倍，并且 Bill 的数是 Cal 的数的正整数倍的概率是多少？

Al, Bill, and Cal will each randomly be assigned a whole number from $1$ to $10,$ inclusive, with no two of them getting the same number. What is the probability that Al's number will be a whole number multiple of Bill's and Bill's number will be a whole number multiple of Cal's?

$\dfrac{9}{1000}$

$\dfrac{1}{90}$

$\dfrac{1}{80}$

$\dfrac{1}{72}$

$\dfrac{2}{121}$

答案：C

知识点：基本概率整除性系统列举

难度评级：1600

解答：

设 Al、Bill、Cal 的数分别为 $(A,B,C)$ 。需要 $A$ 是 $B$ 的倍数，且 $B$ 是 $C$ 的倍数。

满足整除关系且三数不同的有序三元组共有 $9$ 个： $\begin{gathered} (4,2,1),(6,2,1),(8,2,1), \\ (10,2,1),(6,3,1),(9,3,1), \\ (8,4,1),(10,5,1),(8,4,2). \end{gathered}$

总分配数为 $10\cdot9\cdot8=720$ ，所以概率为 $\frac9{720}=\frac1{80}$ 。

所以正确答案是 C。

Let $(A,B,C)$ be the numbers assigned to Al, Bill, and Cal. We need $A$ to be a multiple of $B$ , and $B$ to be a multiple of $C$ , with all three numbers distinct.

The valid triples are $\begin{gathered} (4,2,1),(6,2,1),(8,2,1), \\ (10,2,1),(6,3,1),(9,3,1), \\ (8,4,1),(10,5,1),(8,4,2). \end{gathered}$ There are $9$ favorable assignments.

The total number of assignments is $10\cdot9\cdot8=720$ , so the probability is $\frac9{720}=\frac1{80}$ .

Thus, the correct answer is C.

2015 AMC 10B 第 16 题

答案：C

难度评级：1600

解答：

其他年份的第 16 题