2018 AMC 10B 第 6 题

Question

一个盒子里有

5

个筹码，编号为

1

、

2

、

3

、

4

和

5

。每次随机抽出一个筹码且不放回，直到抽出的编号之和超过

4

。需要抽

3

次的概率是多少？

A box contains

5

chips, numbered

1,

2,

3,

4,

and

5.

Chips are drawn randomly one at a time without replacement until the sum of the values drawn exceeds

4.

What is the probability that

3

draws are required?

\dfrac{1}{15}

\dfrac{1}{10}

\dfrac{1}{6}

\dfrac{1}{5}

\dfrac{1}{4}

答案：D

知识点：无放回抽样基本概率

难度评级：1290

小提示：

恰好需要第三次抽取，当且仅当前两个筹码的和仍不超过

4

。

A third draw is needed exactly when the first two chips still sum to at most

4

大提示：

哪些不同筹码对的和不超过

4

？在所有有序前两次抽取中计数。

Which pairs of distinct chips sum to

4

or less? Count them among all ordered first-two-draw outcomes

解答：

恰好需要第三次抽取，当且仅当前两个筹码的和仍不超过

4

。这样的无序对只有

\{1,2\}

和

\{1,3\}

。每一对都可以按两种顺序抽出，所以有利的有序前两次抽取共有

4

种。

设想事先就把五个筹码的完整随机顺序定好。这样前两个筹码的

5\cdot4=20

种有序结果都是等可能的，即使实际过程会在抽出第一个筹码后就停止也不影响。因此概率为

\frac{4}{20}=\frac{1}{5}

。正确答案是 D。

We need a third draw exactly when the first two chips still sum to

4

or less. The only such unordered pairs are

\{1,2\}

and

\{1,3\}.

Each can be drawn in either order, giving

4

favorable ordered prefixes.

Imagine that a complete random ordering of all five chips is chosen in advance. Then all

5\cdot4=20

ordered first-two-chip prefixes are equally likely, even when the actual process would stop after the first chip. Thus the probability is

\frac{4}{20}=\frac{1}{5}.

Therefore, the answer is D.

2018 AMC 10B 第 6 题

答案：D

难度评级：1290

小提示：

大提示：

解答：

其他年份的第 6 题