2026 AMC 8 第 21 题

先试着解答 2026 AMC 8 第 21 题,然后核对你的答案与视频与精心整理的解答,解答来自 LIVE by Po-Shen Loh。你也可以参加完整限时模拟考试、查看全部 2026 AMC 8 解答,或核对答案。

所有题目均经美国数学协会(MAA)官方合法授权使用。

21.

蜘蛛 Charlotte 在图中形状像 55 角星的蛛网上随机行走。蛛网有 55 个外顶点和 55 个内顶点。每次到达一个顶点时,她都会随机选择一个相邻顶点并走过去。她从一个外顶点出发,走 33 步后在外顶点的概率是多少?
Charlotte the spider is walking along a web shaped like a 55-pointed star, shown in the figure below. The web has 55 outer points and 55 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 33 moves (re-visiting points is allowed). What is the probability she is now at one of the outer points of the star?
15\dfrac{1}{5}
14\dfrac{1}{4}
25\dfrac{2}{5}
12\dfrac{1}{2}
35\dfrac{3}{5}
答案:B
知识点:随机游走基本概率
小提示:
只需记录她在外顶点还是内顶点。
Track only outer versus inner points
大提示:
从内顶点出发,有一半的相邻顶点是外顶点。
From an inner point, half of the neighboring points are outer
视频讲解:
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文字解答:
从外顶点出发,Charlotte 第一步一定到内顶点。从内顶点出发,44 个相邻点中有 22 个是外顶点,所以走到外顶点的概率是 12\frac{1}{2}。一步后她在内顶点。两步后她有 12\frac{1}{2} 的概率在外顶点、12\frac{1}{2} 的概率在内顶点。第三步后在外顶点,只能来自两步后在内顶点的情况,这一步概率为 12\frac{1}{2},所以最终概率为 12⋅12=14\frac12\cdot\frac12=\frac14。
From an outer point, Charlotte must move to an inner point. From an inner point, 22 of the 44 neighboring points are outer, so the probability of moving to an outer point is 12\frac{1}{2}. After one move she is inner. After two moves she is outer with probability 12\frac{1}{2} and inner with probability 12\frac{1}{2}. After the third move, only the inner case can move to outer, with probability 12\frac{1}{2}, so the final probability is 12⋅12=14\frac12\cdot\frac12=\frac14.
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