Question

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?

$\dfrac{5}{256}$

$\dfrac{21}{1024}$

$\dfrac{11}{512}$

$\dfrac{23}{1024}$

$\dfrac{3}{128}$

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

小提示：

共有 $4^6$ 种移动组合；每个顶点只与它的对顶点不相邻。

There are $4^6$ move combinations; each vertex is non-adjacent only to its opposite

大提示：

有效的最终分配是一个排列，且没有顶点被送到自身或对顶点；按对顶点的像是相对还是相邻分类。

A valid final assignment is a permutation with no vertex sent to itself or its opposite; split by whether opposite vertices’ images are opposite or adjacent

解答：

共有 $4^6$ 种等可能移动。把顶点标为 $A, B, C, A', B', C'$ ，其中带撇号的是对应对顶点。有效结果是一个排列 $f$ ，且 $f(A) \notin \{A, A'\}$ ，其他顶点同理。

有序对 $(f(A), f(A'))$ 有 $4 \cdot 3 = 12$ 个选择。其中 $f(A)$ 、 $f(A')$ 互为对顶点的有 $4$ 种，相邻的有 $8$ 种。

若 $f(A), f(A')$ 互为对顶点，例如 $B, B'$ ，则 $\{f(C), f(C')\} = \{A, A'\}$ ，且 $\{f(B), f(B')\} = \{C, C'\}$ ，给出 $4 \cdot 2 \cdot 2 = 16$ 种。

若 $f(A), f(A')$ 相邻，例如 $B, C$ ，则 $f(B), f(B')$ 中必须有一个是 $C'$ ，有 $4$ 个有序选择 $(f(B), f(B'))$ ，每个给 $(f(C), f(C'))$ 留下 $2$ 种，共 $8 \cdot 4 \cdot 2 = 64$ 种。

概率为 $\dfrac{16 + 64}{4^6} = \dfrac{80}{4096} = \dfrac{5}{256}\text{。}$

所以正确答案是 A。

There are $4^6$ equally likely combinations of moves. Label the vertices $A, B, C, A', B', C',$ 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 $f$ with $f(A) \notin \{A, A'\},$ and similarly for each pair.

There are $4 \cdot 3 = 12$ ordered choices for $(f(A), f(A')).$ Of these, $f(A)$ and $f(A')$ are opposite in $4$ cases and adjacent in $8.$

If $f(A), f(A')$ are opposite, say $B, B',$ then $\{f(C), f(C')\} = \{A, A'\}$ and $\{f(B), f(B')\} = \{C, C'\},$ giving $4 \cdot 2 \cdot 2 = 16$ valid combinations.

If $f(A), f(A')$ are adjacent, say $B, C,$ then one of $f(B), f(B')$ must be $C'$ and there are $4$ ordered choices for $(f(B), f(B')),$ each leaving $2$ for $(f(C), f(C')):$ that is $8 \cdot 4 \cdot 2 = 64$ valid combinations.

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

Thus, the correct answer is A.

2005 AMC 12B 第 25 题

答案：A

难度评级：2520

小提示：

大提示：

解答：

其他年份的第 25 题