AMC10 2024 A

AMC10 2024 A · Q24

AMC10 2024 A · Q24. It mainly tests Probability (basic), Counting in geometry (lattice points).

A bee is moving in three-dimensional space. A fair six-sided die with faces labeled $A^+, A^-, B^+, B^-, C^+,$ and $C^-$ is rolled. Suppose the bee occupies the point $(a,b,c).$ If the die shows $A^+$, then the bee moves to the point $(a+1,b,c)$ and if the die shows $A^-,$ then the bee moves to the point $(a-1,b,c).$ Analogous moves are made with the other four outcomes. Suppose the bee starts at the point $(0,0,0)$ and the die is rolled four times. What is the probability that the bee traverses four distinct edges of some unit cube? Diagrams have been moved to the bottom of the solutions.

一只蜜蜂在三维空间移动。掷一个公平六面骰子，面标 $A^+, A^-, B^+, B^-, C^+,$ 和 $C^-$。蜜蜂在点 $(a,b,c)$。$A^+$ 时移到 $(a+1,b,c)$，$A^-$ 时移到 $(a-1,b,c)$，其他类似。从 $(0,0,0)$ 开始掷四次。蜜蜂遍历某个单位立方体四条不同边的概率是多少？ Diagrams have been moved to the bottom of the solutions.

(A) \frac{1}{54} \frac{1}{54}

(B) \frac{7}{54} \frac{7}{54}

(D) \frac{5}{18} \frac{5}{18}

(E) \frac{2}{5} \frac{2}{5}

Answer

Correct choice: (B)

正确答案：（B）

Solution

We start by imagining the three dimensional plane. Notice how the three dimensional plane is split into 8 different regions. There exists 8 cubes with one vertex on the point $(0,0,0)$. We need to consider each of these cases. We arbitrarily take a cube from one region of the three dimensional plane. Below is a sample drawing. Roll 1: From $(0,0,0)$, there are 3 favorable outcomes that can occur. They are listed below. Roll 2: WLOG, say the bee moved up one, there are then 2 ways the bee can go. Roll 3: WLOG, say the bee moved forward one. From this point, there is again 2 ways the bee can go. Roll 4: Finally, WLOG, say the bee moves to the left one. From that point, there are still 2 ways the bee can go. The total probability for one cube is $\frac{3 \cdot 2 \cdot 2 \cdot 2}{6^4} \Rightarrow \frac{1}{54}$. We have 8 of these cubes, which gives us a total probability of $\frac{8}{54}$. However, we have overcounted. Notice that if the bee moved down one instead of left one at roll 3, it would only have one way. This is because the bee has a choice between either the origin or a unique point. This will always occur on the third move, and because we have 8 cubes(The 3d graph has one for each quadrant), the probability of this happening is $\frac{3 \cdot 8}{6^4} = \frac{1}{54}$. Our probability is $\frac{8}{54} - \frac{1}{54} =$ $\boxed{\textbf{(B) }\frac{7}{54}}$.

三维空间分 8 个区域，每个区域有一个以 $(0,0,0)$ 为顶点的立方体。取一立方体：第 1 次：3 种有利结果。第 2 次：2 种。第 3 次：2 种。第 4 次：2 种。一立方体概率 $\frac{3 \cdot 2 \cdot 2 \cdot 2}{6^4} = \frac{1}{54}$。8 个立方体：$\frac{8}{54}$。但过计。第三次若选另一方向，仅 1 种（回到原点或唯一点）。8 个立方体，概率 $\frac{3 \cdot 8}{6^4} = \frac{1}{54}$。总概率 $\frac{8}{54} - \frac{1}{54} = \boxed{\textbf{(B) }\frac{7}{54}}$。

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