AMC12 2005 B

AMC12 2005 B · Q25

AMC12 2005 B · Q25. It mainly tests Basic counting (rules of product/sum), Probability (basic).

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?

六只蚂蚁同时站在正八面体的六个顶点上，每只蚂蚁在不同的顶点。同时且独立地，每只蚂蚁从其顶点移动到四个相邻顶点之一，每种选择等概率。求没有两只蚂蚁到达同一顶点的概率。

(A) \dfrac{5}{256} \dfrac{5}{256}

(B) \dfrac{21}{1024} \dfrac{21}{1024}

(D) \dfrac{23}{1024} \dfrac{23}{1024}

(E) \dfrac{3}{128} \dfrac{3}{128}

Answer

Correct choice: (A)

正确答案：（A）

Solution

We approach this problem by counting the number of ways ants can do their desired migration, and then multiply this number by the probability that each case occurs. Let the octahedron be $ABCDEF$, with points $B,C,D,E$ coplanar. Then the ant from $A$ and the ant from $F$ must move to plane $BCDE$. Suppose, without loss of generality, that the ant from $A$ moved to point $B$. Then, we must consider three cases. - Case 1: Ant from point $F$ moved to point $C$ On the plane, points $B$ and $C$ are taken. The ant that moves to $D$ can come from either $E$ or $C$. The ant that moves to $E$ can come from either $B$ or $D$. Once these two ants are fixed, the other two ants must migrate to the "poles" of the octahedron, points $A$ and $F$. Thus, there are two degrees of freedom in deciding which ant moves to $D$, two degrees of freedom in deciding which ant moves to $E$, and two degrees of freedom in deciding which ant moves to $A$. Hence, there are $2 \times 2 \times 2=8$ ways the ants can move to different points. - Case 2: Ant from point $F$ moved to point $D$ On the plane, points $B$ and $D$ are taken. The ant that moves to $C$ must be from $B$ or $D$, but the ant that moves to $E$ must also be from $B$ or $D$. The other two ants, originating from points $C$ and $E$, must move to the poles. Therefore, there are two degrees of freedom in deciding which ant moves to $C$ and two degrees of freedom in choosing which ant moves to $A$. Hence, there are $2 \times 2=4$ ways the ants can move to different points. - Case 3: Ant from point $F$ moved to point $E$ By symmetry to Case 1, there are $8$ ways the ants can move to different points. Given a point $B$, there is a total of $8+4+8=20$ ways the ants can move to different points. We oriented the square so that point $B$ was defined as the point to which the ant from point $A$ moved. Since the ant from point $A$ can actually move to four different points, there is a total of $4 \times 20=80$ ways the ants can move to different points. Each ant acts independently, having four different points to choose from. Hence, each ant has probability $1/4$ of moving to the desired location. Since there are six ants, the probability of each case occuring is $\frac{1}{4^6} = \frac{1}{4096}$. Thus, the desired answer is $\frac{80}{4096}= \boxed{\frac{5}{256}}$

我们通过计数蚂蚁满足要求的迁移方式数目来解决此题，然后将该数目乘以每种情况发生的概率。设八面体为 $ABCDEF$，其中点 $B,C,D,E$ 共面。则从 $A$ 出发的蚂蚁和从 $F$ 出发的蚂蚁必须移动到平面 $BCDE$ 上。不失一般性，设从 $A$ 出发的蚂蚁移动到点 $B$。接下来分三种情况讨论。 - 情况 1：从点 $F$ 出发的蚂蚁移动到点 $C$ 在该平面上，点 $B$ 和 $C$ 已被占据。移动到 $D$ 的蚂蚁可以来自 $E$ 或 $C$。移动到 $E$ 的蚂蚁可以来自 $B$ 或 $D$。一旦这两只蚂蚁确定，另外两只蚂蚁必须迁移到八面体的“极点” $A$ 和 $F$。因此，决定哪只蚂蚁移动到 $D$ 有 $2$ 种自由度，决定哪只蚂蚁移动到 $E$ 有 $2$ 种自由度，决定哪只蚂蚁移动到 $A$ 有 $2$ 种自由度。故共有 $2 \times 2 \times 2=8$ 种方式使蚂蚁移动到不同的点。 - 情况 2：从点 $F$ 出发的蚂蚁移动到点 $D$ 在该平面上，点 $B$ 和 $D$ 已被占据。移动到 $C$ 的蚂蚁必须来自 $B$ 或 $D$，但移动到 $E$ 的蚂蚁也必须来自 $B$ 或 $D$。另外两只分别从 $C$ 和 $E$ 出发的蚂蚁必须移动到极点。因此，决定哪只蚂蚁移动到 $C$ 有 $2$ 种自由度，选择哪只蚂蚁移动到 $A$ 有 $2$ 种自由度。故共有 $2 \times 2=4$ 种方式使蚂蚁移动到不同的点。 - 情况 3：从点 $F$ 出发的蚂蚁移动到点 $E$ 与情况 1 对称，共有 $8$ 种方式使蚂蚁移动到不同的点。给定点 $B$，共有 $8+4+8=20$ 种方式使蚂蚁移动到不同的点。我们将正方形定向，使得点 $B$ 被定义为从点 $A$ 出发的蚂蚁所到达的点。由于从点 $A$ 出发的蚂蚁实际上可以移动到 $4$ 个不同的点，因此总共有 $4 \times 20=80$ 种方式使蚂蚁移动到不同的点。每只蚂蚁独立行动，有 $4$ 个不同的点可选。因此每只蚂蚁以概率 $1/4$ 移动到指定位置。由于共有 $6$ 只蚂蚁，每种情况发生的概率为 $\frac{1}{4^6}=\frac{1}{4096}$。因此所求概率为 $\frac{80}{4096}=\boxed{\frac{5}{256}} \Rightarrow \mathrm{(A)}$。

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