Question

一个正五边形的 $5$ 条边和 $5$ 条对角线各自独立随机染成红色或蓝色，且两种颜色概率相等。存在一个三角形，其顶点为该五边形的顶点，且三条边同色的概率是多少？

Each of the $5$ sides and the $5$ diagonals of a regular pentagon are randomly and independently colored red or blue with equal probability. What is the probability that there will be a triangle whose vertices are among the vertices of the pentagon such that all of its sides have the same color?

$\dfrac 23$

$\dfrac{105}{128}$

$\dfrac{125}{128}$

$\dfrac{253}{256}$

$1$

答案：D

知识点：图论对立事件概率

难度评级：2300

解答：

计算补事件：将 $10$ 条边组成的 $K_5$ 用两种颜色染色，且没有单色三角形。

在任意顶点，如果有 $3$ 条关联边同色，那么这三条边另一端之间的边都必须是另一种颜色；但这又会形成单色三角形。因此每个顶点恰有 $2$ 条红边和 $2$ 条蓝边。

所以红边构成一个 $2$ 正则图，顶点数为 $5$ ，只能是一个 $5$ 环。带标号的 $5$ 环共有 $\frac{(5-1)!}{2}=12$ 个。

总染色数为 $2^{10}=1024$ ，所以所求概率为 $1-\frac{12}{1024}=\frac{253}{256}.$

所以正确答案是 D。

Count the complement: colorings of the $10$ edges of $K_5$ with no monochromatic triangle.

At any vertex, if $3$ incident edges had the same color, then the edges among their other endpoints would all have to be the other color, making a monochromatic triangle. Thus each vertex has exactly $2$ red and $2$ blue incident edges.

So the red edges form a $2$ -regular graph on $5$ vertices, which must be a $5$ -cycle. The number of labeled $5$ -cycles is $\frac{(5-1)!}{2}=12.$

There are $2^{10}=1024$ total colorings, so the desired probability is $1-\frac{12}{1024}=\frac{253}{256}.$

Thus, the answer is D .

2021 AMC 10B Fall 第 23 题

答案：D

难度评级：2300

解答：

其他年份的第 23 题