Question

设 $S$ 为一个正五边形的边和对角线组成的集合。从 $S$ 中不放回随机选取两个元素。选出的两条线段长度相同的概率是多少？

Let $S$ be the set of sides and diagonals of a regular pentagon. A pair of elements of $S$ are selected at random without replacement. What is the probability that the two chosen segments have the same length?

$\dfrac{2}5$

$\dfrac{4}9$

$\dfrac{1}2$

$\dfrac{5}9$

$\dfrac{4}5$

答案：B

知识点：基本概率正多边形无放回抽样

难度评级：1220

解答：

正五边形有 $5$ 条边，长度相同；还有 $5$ 条对角线，长度也相同。

选出第一条后，剩下 $9$ 条线段，其中恰有 $4$ 条与第一条同长。

因此概率为 $\frac49$ 。

所以正确答案是 B。

A regular pentagon has $5$ sides of one length and $5$ diagonals of another length.

After the first segment is chosen, there are $9$ segments left, and exactly $4$ of them have the same length as the first chosen segment.

Therefore the probability is $\frac49$ .

Thus, the correct answer is B .

2013 AMC 10B 第 12 题

答案：B

难度评级：1220

解答：

其他年份的第 12 题