2007 AMC 8 第 24 题

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

小提示：

一个数能被

3

整除，当且仅当它的数字和能被

3

整除

A number is divisible by

3

exactly when its digit sum is divisible by

3.

大提示：

不要先排列抽出的三个数字，先看留下了哪个数字

Instead of ordering the three drawn digits, look at which digit is left out.

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

小提示：

大提示：

解答：

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