2005 AMC 12B 第 25 题
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25.
六只蚂蚁同时站在一个正八面体的六个顶点上,每个顶点一只。它们同时且独立地从所在顶点移动到四个相邻顶点之一,每个选择概率相同。没有两只蚂蚁到达同一顶点的概率是多少?
Six ants simultaneously stand on the six vertices of a regular octahedron, with each ant at a different vertex. Simultaneously and independently, each ant moves from its vertex to one of the four adjacent vertices, each with equal probability. What is the probability that no two ants arrive at the same vertex?
答案:A
解答:
共有 种等可能移动。把顶点标为 ,其中带撇号的是对应对顶点。有效结果是一个排列 ,且 ,其他顶点同理。
有序对 有 个选择。其中 、 互为对顶点的有 种,相邻的有 种。
若 互为对顶点,例如 ,则 ,且 ,给出 种。
若 相邻,例如 ,则 中必须有一个是 ,有 个有序选择 ,每个给 留下 种,共 种。
概率为
所以正确答案是 A。
There are equally likely combinations of moves. Label the vertices where primed vertices are opposite the corresponding unprimed ones. An ant cannot move to its own vertex or the opposite one, so a valid outcome is a permutation with and similarly for each pair.
There are ordered choices for Of these, and are opposite in cases and adjacent in
If are opposite, say then and giving valid combinations.
If are adjacent, say then one of must be and there are ordered choices for each leaving for that is valid combinations.
Hence the probability is
Thus, the correct answer is A.
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