Question

一个袋子里有四张纸片，分别标有数字 $1$ 、 $2$ 、 $3$ 、 $4$ ，没有重复。从中不放回地一次抽出三张，用来组成一个三位数。这个三位数是 $3$ 的倍数的概率是多少？

A bag contains four pieces of paper, each labeled with one of the digits $1,$ $2,$ $3$ or $4,$ with no repeats. Three of these pieces are drawn, one at a time without replacement, to construct a three-digit number. What is the probability that the three-digit number is a multiple of $3?$

$\dfrac{1}{4}$

$\dfrac{1}{3}$

$\dfrac{1}{2}$

$\dfrac{2}{3}$

$\dfrac{3}{4}$

答案：C

知识点：整除性基本概率数字

难度评级：1580

解答：

一个数能被 $3$ 整除，当且仅当它的数字和能被 $3$ 整除。

从 $1,2,3,4$ 中取三个不同数字，数字和能被 $3$ 整除的只有 $(1,2,3)$ 和 $(2,3,4)$ 。

因此所组成的数是 $3$ 的倍数，当且仅当袋中剩下的是 $1$ 或 $4$ 。

每种事件概率都是 $\dfrac{1}{4}$ ，总概率为 $2\cdot\dfrac{1}{4}=\dfrac{1}{2}$ 。

所以正确答案是 C。

Recall that a number is divisible by $3$ if the sum of its digits is divisible by $3$ .

The only triples of distinct digits from $1,2,3,4$ whose sum is divisible by $3$ are $(1,2,3)$ and $(2,3,4)$ .

This means the constructed number is a multiple of $3$ exactly when either $1$ or $4$ is left in the bag.

Each of these events has probability $\dfrac{1}{4}$ , for a total probability of $2\cdot\dfrac{1}{4}=\dfrac{1}{2}$ .

Thus, C is the correct answer.

2007 AMC 8 第 24 题

答案：C

难度评级：1580

解答：

其他年份的第 24 题