AMC8 2026

AMC8 2026 · Q21

AMC8 2026 · Q21. It mainly tests Probability (basic).

Charlotte the spider is walking along a web shaped like a $5$-pointed star, shown in the figure below. The web has $5$ outer points and $5$ inner points. Each time Charlotte reaches a point, she randomly chooses a neighboring point and moves to that point. Charlotte starts at one of the outer points and makes $3$ moves (re-visiting points is allowed). What is the probability she is now at one of the outer points of the star?

蜘蛛Charlotte在一个形状如下图所示的五角星形网线上行走。该网有5个外部顶点和5个内部顶点。每次Charlotte到达一个顶点时，都会随机选择一个相邻的顶点移动过去。Charlotte从一个外部顶点开始，进行3次移动（允许重复访问顶点）。她现在在五角星的某个外部顶点的概率是多少？

(A) \frac{1}{5} \frac{1}{5}

(B) \frac{1}{4} \frac{1}{4}

(D) \frac{1}{2} \frac{1}{2}

(E) \frac{3}{5} \frac{3}{5}

Answer

Correct choice: (B)

正确答案：（B）

Solution

On the second turn, she cannot be on an outer point. Thus, for the first move she can do from an outer point to an inner, she has 2 ways to go. Then, from said point, she again has 2 ways to go. Finally, once more she has 2 ways to go that will end her up on an outer point. Then, there is just 8 configurations for her that are favorable. There are a total of $2 \times 4 \times 4 = 32$ possible movements. The probability is $\frac{8}{32} = \boxed{\frac{1}{4}}$.

在第二次移动时，她不可能处于外部顶点。因此，第一次从外部顶点移动到内部顶点时，她有2种选择。然后，从该内部顶点，她又有2种选择可以移动。最后，再次有2种选择使她最终回到外部顶点。于是，有8种有利的路径配置。移动的总可能数为 $2 \times 4 \times 4 = 32$。概率为 $\frac{8}{32} = \boxed{\frac{1}{4}}$。

Topics

CP.006 Probability (basic)