AMC10 2021 A

AMC10 2021 A · Q23

AMC10 2021 A · Q23. It mainly tests Probability (basic), Casework.

Frieda the frog begins a sequence of hops on a $3 \times 3$ grid of squares, moving one square on each hop and choosing at random the direction of each hop-up, down, left, or right. She does not hop diagonally. When the direction of a hop would take Frieda off the grid, she "wraps around" and jumps to the opposite edge. For example if Frieda begins in the center square and makes two hops "up", the first hop would place her in the top row middle square, and the second hop would cause Frieda to jump to the opposite edge, landing in the bottom row middle square. Suppose Frieda starts from the center square, makes at most four hops at random, and stops hopping if she lands on a corner square. What is the probability that she reaches a corner square on one of the four hops?

青蛙 Frieda 在一个 $3 \times 3$ 的方格网格上开始一系列跳跃，每次跳跃移动一个方格，并随机选择每个跳跃的方向——上、下、左或右。她不斜向跳跃。当跳跃方向会使 Frieda 离开网格时，她“环绕”并跳到对边。例如，如果 Frieda 从中心方格开始并向上跳两次，第一次跳会让她到顶行中间方格，第二次跳会让她跳到对边，落在底行中间方格。假设 Frieda 从中心方格开始，最多随机跳四次，并在落地到角落方格时停止跳跃。她在四次跳跃之一中到达角落方格的概率是多少？

(A) \frac{9}{16} \frac{9}{16}

(B) \frac{5}{8} \frac{5}{8}

(D) \frac{25}{32} \frac{25}{32}

(E) \frac{13}{16} \frac{13}{16}

Answer

Correct choice: (D)

正确答案：（D）

Solution

We will use complementary counting. First, the frog can go left with probability $\frac14$. We observe symmetry, so our final answer will be multiplied by 4 for the 4 directions, and since $4\cdot \frac14=1$, we will ignore the leading probability. From the left, she either goes left to another edge ($\frac14$) or back to the center ($\frac14$). Time for some casework. $\textbf{Case 1:}$ She goes back to the center. Now, she can go in any 4 directions, and then has 2 options from that edge. This gives $\frac12$. --End case 1 $\textbf{Case 2:}$ She goes to another edge (rightmost). Subcase 1: She goes back to the left edge. She now has 2 places to go, giving $\frac12$ Subcase 2: She goes to the center. Now any move works. $\frac14 \cdot \frac12 + \frac14 \cdot 1=\frac18 + \frac 14=\frac38$ for this case. --End case 2 She goes back to the center in Case 1 with probability $\frac14$, and to the right edge with probability $\frac14$ So, our answer is $\frac14 \cdot \frac12 + \frac14 \cdot \frac38=\frac14 (\frac12+\frac38)=\frac14 \cdot \frac78 = \frac7{32}$ But, don't forget complementary counting. So, we get $1-\frac7{32}=\frac{25}{32} \implies \boxed{D}$.

我们使用补集计数。首先，青蛙向左的概率为 $\frac14$。我们观察到对称性，因此最终答案将乘以 4 个方向，且由于 $4\cdot \frac14=1$，我们忽略前导概率。从左边，她要么向左到另一个边缘 ($\frac14$)，要么返回中心 ($\frac14$)。是时候进行分类讨论了。 $\textbf{情况 1:}$ 她返回中心。现在，她可以向任意 4 个方向，然后从那个边缘有 2 个选项。这给出 $\frac12$。--情况 1 结束 $\textbf{情况 2:}$ 她去到另一个边缘（最右）。子情况 1: 她返回左边缘。现在她有 2 个地方可去，给出 $\frac12$ 子情况 2: 她去到中心。现在任意移动都行。 $\frac14 \cdot \frac12 + \frac14 \cdot 1=\frac18 + \frac 14=\frac38$ 对于此情况。--情况 2 结束她在情况 1 中以概率 $\frac14$ 返回中心，在情况 2 中以概率 $\frac14$ 去到右边缘。因此，我们的答案是 $\frac14 \cdot \frac12 + \frac14 \cdot \frac38=\frac14 (\frac12+\frac38)=\frac14 \cdot \frac78 = \frac7{32}$ 但，别忘了补集计数。因此，我们得到 $1-\frac7{32}=\frac{25}{32} \implies \boxed{D}$。

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