2006 AMC 12A 第 20 题

先试着解答 2006 AMC 12A 第 20 题,然后核对你的答案与精心整理的解答,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 2006 AMC 12A 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。

20.

一只虫子从立方体的一个顶点出发,并按如下规则沿立方体的棱移动。在每个顶点,这只虫子会从该顶点发出的三条棱中选择一条前进。每条棱被选中的概率相等,且所有选择相互独立。七次移动后,这只虫子恰好访问每个顶点一次的概率是多少?
A bug starts at one vertex of a cube and moves along the edges of the cube according to the following rule. At each vertex the bug will choose to travel along one of the three edges emanating from that vertex. Each edge has equal probability of being chosen, and all choices are independent. What is the probability that after seven moves the bug will have visited every vertex exactly once?
12187\dfrac{1}{2187}
1729\dfrac{1}{729}
2243\dfrac{2}{243}
181\dfrac{1}{81}
5243\dfrac{5}{243}
答案:C
知识点:基本概率图论分类讨论
小提示:
共有 373^7 条等可能的 77-步路径;数出访问全部 88 个顶点的路径。
There are 373^7 equally likely 77-move walks; count those visiting all 88 vertices
大提示:
第一步有 33 种选择,第二步有 22 种选择,之后路径几乎被确定。
There are 33 choices for the first move and 22 for the second, after which the path is nearly forced
解答:
从起点出发共有 373^7 条等可能的 77-步路径。考虑一条访问全部 88 个顶点的路径:第一步有 33 种选择,第二步有 22 种选择。
用三位二进制串给立方体的顶点编号。由对称性,固定前两步之后,可设前三个顶点依次为 000,001,011000,001,011。逐个分支检验可知,恰好只有以下三种走完全程的方式:010,110,111,101,100,010,110,100,101,111,111,101,100,110,010。 \begin{aligned} &010,110,111,101,100,\\ &010,110,100,101,111,\\ &111,101,100,110,010 \end{aligned}\text{。} 因此这样的路径共有 3⋅2⋅3=183 \cdot 2 \cdot 3 = 18 条。
概率为 1837=182187=2243\dfrac{18}{3^7} = \dfrac{18}{2187} = \dfrac{2}{243}。
因此,正确答案是 C。
From the start there are 373^7 equally likely 77-move walks. For a walk visiting all 88 vertices, there are 33 choices for the first move and 22 for the second, since it cannot return to the starting vertex.
Label cube vertices by three-bit strings. By symmetry, after fixing those first two moves we may take the first three vertices to be 000,001,011.000,001,011. A branch check gives exactly these three completions: 010,110,111,101,100,010,110,100,101,111,111,101,100,110,010. \begin{aligned} &010,110,111,101,100,\\ &010,110,100,101,111,\\ &111,101,100,110,010. \end{aligned} Thus there are 3⋅2⋅3=183 \cdot 2 \cdot 3 = 18 such walks.
The probability is 1837=182187=2243.\dfrac{18}{3^7} = \dfrac{18}{2187} = \dfrac{2}{243}.
Thus, the correct answer is C.
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