AMC 8 2007 Problem 24

AMC8 2007 年第 24 题

概率★★★★☆

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 ?

一个包中有 4 张分别标有数字 1，2，3，4 的纸片，每个数字都不重复。从中不放回的选择 3 张纸片，构成一个三位数。则这个三位数是 3 的倍数的概率是多少？

A. $\displaystyle \frac{1}{4}$
B. $\displaystyle \frac{1}{3}$
C. $\displaystyle \frac{1}{2}$
D. $\displaystyle \frac{2}{3}$
E. $\displaystyle \frac{3}{4}$

正确答案 · Correct Answer

解析 · Solution

解法1

常规解法 · 有序枚举组成三位数共

4\cdot3\cdot2=24

种。被 3 整除当且仅当数字和被 3 整除。从

\{1,2,3,4\}

取三个数，只有

\{1,2,3\}

（和 6）和

\{2,3,4\}

（和 9）合格，各能排出 3!=6 个数，共 12 个。概率

\dfrac{12}{24}=\dfrac12

。

Solution 1

Elementary · ordered enumeration There are

4\cdot3\cdot2=24

ordered three-digit numbers. A number is divisible by 3 iff its digit sum is. Among the size-3 subsets of

\{1,2,3,4\}

, only

\{1,2,3\}

(sum 6) and

\{2,3,4\}

(sum 9) qualify, each giving 3!=6 arrangements, so 12 work. The probability is

\dfrac{12}{24}=\dfrac12

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