Question

青蛙 Frieda 在一个 $3\times 3$ 方格网中开始一串跳跃，每次跳一格，并随机选择跳跃方向：上、下、左、右。她不斜着跳。当某次跳跃方向会让 Frieda 跳出方格网时，她会“绕回”并跳到相对的边。例如，如果 Frieda 从中心格开始并连续向上跳两次，第一次会到达上排中间格，第二次会让她跳到相对边，落在下排中间格。假设 Frieda 从中心格开始，最多随机跳四次，并且一旦落在角格就停止。她在四次跳跃中的某一次到达角格的概率是多少？

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 places her in the top row middle square, and the second hop causes her 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?

$\dfrac{9}{16}$

$\dfrac{5}{8}$

$\dfrac{3}{4}$

$\dfrac{25}{32}$

$\dfrac{13}{16}$

答案：D

知识点：递推概率随机游走

难度评级：2520

解答：

将方格分为中心格 $C$ 、边中格 $E$ 和角格（吸收状态）。从 $C$ 出发，每一步都会到达一个 $E$ 格。从一个 $E$ 格出发，四个相邻格中有两个是角格，一个是中心格，一个是另一个 $E$ 格，所以 $P(\text{corner}) = \tfrac12$ 、 $P(\text{center}) = \tfrac14$ 、 $P(E) = \tfrac14$ 。

设 $a_n$ 为从边格出发、在 $n$ 步内到达角格的概率，且 $c_n$ 为从中心出发的概率（第一步必到边格）。于是 $a_0=c_0=0$ ，并且 $c_n=a_{n-1}$ 。计算得 $a_n=\tfrac12+\tfrac14c_{n-1}+\tfrac14a_{n-1}$ ，且 $a_1=\tfrac12$ ， $a_2=\tfrac12+\tfrac14\cdot\tfrac12=\tfrac58$ ， $a_3 = \tfrac12 + \tfrac14\cdot\tfrac12 + \tfrac14\cdot\tfrac58 = \tfrac{25}{32}.$

从中心开始且有四步可用时，概率等于 $a_3 = \tfrac{25}{32}$ （第一步到达边格，剩下三步）。

因此，正确答案是 D。

Classify squares as center $C,$ edge-middle $E,$ or corner (absorbing). From $C,$ every hop lands on an $E$ square. From an $E$ square, two of the four neighbors are corners, one is the center, and one is another $E$ square, so $P(\text{corner}) = \tfrac12,$ $P(\text{center}) = \tfrac14,$ $P(E) = \tfrac14.$

Let $a_n$ be the probability of reaching a corner within $n$ hops starting from an edge square, and let $c_n$ be the corresponding probability from the center. Set $a_0=c_0=0.$ Then $c_n=a_{n-1}$ and $a_n=\tfrac12+\tfrac14c_{n-1}+\tfrac14a_{n-1}.$ Thus $a_1=\tfrac12,$ $a_2=\tfrac12+\tfrac14\cdot\tfrac12=\tfrac58,$ and $a_3 = \tfrac12 + \tfrac14\cdot\tfrac12 + \tfrac14\cdot\tfrac58 = \tfrac{25}{32}.$

Starting from the center with four hops available, the probability equals $a_3 = \tfrac{25}{32}$ (the first hop reaches an edge, leaving three hops).

Thus, the correct answer is D.

2021 AMC 12A Spring 第 23 题

答案：D

难度评级：2520

解答：

其他年份的第 23 题