2008 AMC 12B 第 22 题

Question

一个停车场有一排

16

个车位。十二辆车先后到达，每辆车需要一个车位，司机从可用车位中随机选择停车。随后 Auntie Em 开着她的 SUV 到达，这辆车需要

2

个相邻车位。她能够停车的概率是多少？

A parking lot has

16

spaces in a row. Twelve cars arrive, each of which requires one parking space, and their drivers choose their spaces at random from among the available spaces. Auntie Em then arrives in her SUV, which requires

2

adjacent spaces. What is the probability that she is able to park?

\dfrac{11}{20}

\dfrac{4}{7}

\dfrac{81}{140}

\dfrac{3}{5}

\dfrac{17}{28}

答案：E

知识点：补集计数组合有限制的排列

难度评级：2110

小提示：

4

个空车位等可能地是

16

个车位中的任意

4

个；她失败恰好当没有两个空位相邻。

The

4

empty spaces are equally likely to be any

4

of the

16;

she fails exactly when no two empties are adjacent

大提示：

在

16

个车位中选择

4

个互不相邻的空位有

\binom{13}{4}

种；从

1

中减去这个概率。

The number of ways to choose

4

empty spaces among

16

with no two adjacent is

\binom{13}{4};

subtract that probability from

1

解答：

12

辆车停好后，有

4

个车位为空，等可能地是

16

个车位中的任意

4

个，共有

\binom{16}{4} = 1820

个等可能集合。

Auntie Em 不能停车恰好当没有两个空车位相邻。在

16

个车位中放置

4

个互不相邻的空位有

\binom{13}{4} = 715

种。

因此她能够停车的概率为

1 - \frac{715}{1820} = \frac{1105}{1820} = \frac{17}{28}\text{。}

因此，正确答案是 E。

After the

12

cars park,

4

spaces are empty, equally likely to be any

4

of the

16,

for

\binom{16}{4} = 1820

equally likely sets.

Auntie Em fails exactly when no two empty spaces are adjacent. The number of ways to place

4

non-adjacent empties among

16

is

\binom{13}{4} = 715.

So the probability she can park is

1 - \frac{715}{1820} = \frac{1105}{1820} = \frac{17}{28}.

Thus, the correct answer is E.

2008 AMC 12B 第 22 题

答案：E

难度评级：2110

小提示：

大提示：

解答：

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