2005 AMC 12B 第 25 题

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所有题目均经美国数学协会(MAA)官方合法授权使用。

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?

5256\dfrac{5}{256}

211024\dfrac{21}{1024}

11512\dfrac{11}{512}

231024\dfrac{23}{1024}

3128\dfrac{3}{128}

答案:A
知识点:基本概率有限制的排列分类讨论
难度评级:2520
解答:

共有 464^6 种等可能移动。把顶点标为 A,B,C,A,B,CA, B, C, A', B', C',其中带撇号的是对应对顶点。有效结果是一个排列 ff,且 f(A){A,A}f(A) \notin \{A, A'\},其他顶点同理。

有序对 (f(A),f(A))(f(A), f(A'))43=124 \cdot 3 = 12 个选择。其中 f(A)f(A)f(A)f(A') 互为对顶点的有 44 种,相邻的有 88 种。

f(A),f(A)f(A), f(A') 互为对顶点,例如 B,BB, B',则 {f(C),f(C)}={A,A}\{f(C), f(C')\} = \{A, A'\},且 {f(B),f(B)}={C,C}\{f(B), f(B')\} = \{C, C'\},给出 422=164 \cdot 2 \cdot 2 = 16 种。

f(A),f(A)f(A), f(A') 相邻,例如 B,CB, C,则 f(B),f(B)f(B), f(B') 中必须有一个是 CC',有 44 个有序选择 (f(B),f(B))(f(B), f(B')),每个给 (f(C),f(C))(f(C), f(C')) 留下 22 种,共 842=648 \cdot 4 \cdot 2 = 64 种。

概率为 16+6446=804096=5256. \dfrac{16 + 64}{4^6} = \dfrac{80}{4096} = \dfrac{5}{256}.

所以正确答案是 A

There are 464^6 equally likely combinations of moves. Label the vertices A,B,C,A,B,C,A, B, C, A', B', C', 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 ff with f(A){A,A},f(A) \notin \{A, A'\}, and similarly for each pair.

There are 43=124 \cdot 3 = 12 ordered choices for (f(A),f(A)).(f(A), f(A')). Of these, f(A)f(A) and f(A)f(A') are opposite in 44 cases and adjacent in 8.8.

If f(A),f(A)f(A), f(A') are opposite, say B,B,B, B', then {f(C),f(C)}={A,A}\{f(C), f(C')\} = \{A, A'\} and {f(B),f(B)}={C,C},\{f(B), f(B')\} = \{C, C'\}, giving 422=164 \cdot 2 \cdot 2 = 16 valid combinations.

If f(A),f(A)f(A), f(A') are adjacent, say B,C,B, C, then one of f(B),f(B)f(B), f(B') must be CC' and there are 44 ordered choices for (f(B),f(B)),(f(B), f(B')), each leaving 22 for (f(C),f(C)):(f(C), f(C')): that is 842=648 \cdot 4 \cdot 2 = 64 valid combinations.

Hence the probability is 16+6446=804096=5256. \dfrac{16 + 64}{4^6} = \dfrac{80}{4096} = \dfrac{5}{256}.

Thus, the correct answer is A.

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