Question

一只青蛙坐在点 $(1, 2)$ 处，开始一系列跳跃。每次跳跃都平行于一条坐标轴，长度为 $1$ ，且每次跳跃的方向（上、下、右、左）都独立随机选择。当青蛙到达顶点为 $(0, 0)$ 、 $(0, 4)$ 、 $(4, 4)$ 和 $(4, 0)$ 的正方形的一条边时，跳跃序列结束。跳跃序列在该正方形的竖直边上结束的概率是多少？

A frog sitting at the point $(1, 2)$ begins a sequence of jumps, where each jump is parallel to one of the coordinate axes and has length $1,$ and the direction of each jump (up, down, right, or left) is chosen independently at random. The sequence ends when the frog reaches a side of the square with vertices $(0, 0),$ $(0, 4),$ $(4, 4),$ and $(4, 0).$ What is the probability that the sequence of jumps ends on a vertical side of the square?

$\dfrac{1}{2}$

$\dfrac{5}{8}$

$\dfrac{2}{3}$

$\dfrac{3}{4}$

$\dfrac{7}{8}$

答案：B

知识点：随机游走方程组对称性

难度评级：1630

解答：

令 $P(x, y)$ 为在竖直边上结束的概率。在竖直边上 $P = 1$ ，在水平边上 $P = 0$ ，在内部点处 $P$ 是四个相邻点概率的平均。

由左右对称性， $P(2, 2) = \tfrac12$ 。令 $a = P(1, 2)$ ， $b = P(1, 1) = P(1, 3)$ ，且 $c = P(2, 1) = P(2, 3)$ 。则

$a = \tfrac14\left(1 + \tfrac12 + 2b\right)$ ， $\;b = \tfrac14(1 + c + a)$ ，且 $c = \tfrac14\left(2b + \tfrac12\right)$ 。

代入可得 $b = \tfrac12$ ，因此 $a = \tfrac38 + \tfrac12 b = \tfrac58$ 。

因此，正确答案是 B。

Let $P(x, y)$ be the probability of ending on a vertical side. On a vertical side $P = 1,$ on a horizontal side $P = 0,$ and at an interior point $P$ is the average of its four neighbors.

By left-right symmetry $P(2, 2) = \tfrac12.$ Let $a = P(1, 2),$ $b = P(1, 1) = P(1, 3),$ and $c = P(2, 1) = P(2, 3).$ Then

$a = \tfrac14\left(1 + \tfrac12 + 2b\right),$ $\;b = \tfrac14(1 + c + a),$ and $c = \tfrac14\left(2b + \tfrac12\right).$

Substituting gives $b = \tfrac12,$ hence $a = \tfrac38 + \tfrac12 b = \tfrac58.$

Thus, B is the correct answer.

2020 AMC 12A 第 11 题

答案：B

难度评级：1630

解答：

其他年份的第 11 题