AMC 8 2026 Problem 21
AMC8 2026 年第 21 题
概率★★★☆☆
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?
Charlotte 是一只蜘蛛，她在一张形如五角星的网上爬行（如下图所示）。这张网有 5 个外顶点和 5 个内顶点。每次 Charlotte 到达一个顶点时，会随机选择一个相邻顶点并移动过去（允许重复访问）。她从一个外顶点出发，共移动 3 次。此时她位于外顶点的概率是多少？
- A.
- B.
- C.
- D.
- E.
正确答案 · Correct Answer
B
解析 · Solution
解法1
从外点出发，第一步必须到内点（外点的邻居全是内点），概率为 1。从内点出发，4 个邻居中 2 个内点、2 个外点，第二步走到内或外的概率各 。
第三步：若在內点，走到外点的概率为 ；若在外点，走到外点的概率为 0（只能去内点）。
总概率：1 × ( ) = 。
决策树法：第 1 步，外→内（概率 1）。第 2 步，从内点出发，4 个邻居 2 外 2 内，P(外)=P(内)= 。第 3 步，若在外点→必去内点（P(外)=0），若在内点→P(外)= 。唯一到达外点的路径为内→内→外，概率 = 1 × 。
Solution 1
From an outer vertex, the next move must go to an inner vertex (all neighbors of an outer vertex are inner). From an inner vertex, there are 2 inner and 2 outer neighbors, so P(inner) = and P(outer) = for the second step.
For the third step: if the spider is at an inner vertex, P(outer) = again. If at an outer vertex, P(outer) = 0 (all neighbors are inner).
Final probability: 1 × ( ) = .
Decision tree method: Step 1 — outer → inner (probability 1). Step 2 — from inner, 2 of 4 neighbors are outer, 2 are inner: P(outer)= , P(inner)= . Step 3 — if at outer: must go to inner (P(outer)=0). If at inner: P(outer)= . The only path ending at an outer point after 3 steps goes through inner→inner→outer, with probability 1 × .
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