Question

一只青蛙按如下规则在数轴上跳跃。它从 $0$ 开始。如果它在 $0$ ，那么它以概率 $\tfrac12$ 移动到 $1$ ，并以概率 $\tfrac12$ 消失。对于 $n = 1, 2$ 或 $3$ ，如果它在 $n$ ，那么它以概率 $\tfrac14$ 移动到 $n + 1$ ，以概率 $\tfrac14$ 移动到 $n - 1$ ，并以概率 $\tfrac12$ 消失。

青蛙到达 $4$ 的概率是多少？

A frog hops along the number line according to the following rules. It starts at $0.$ If it is at $0,$ then it moves to $1$ with probability $\tfrac12$ and it disappears with probability $\tfrac12.$ For $n = 1, 2,$ or $3,$ if it is at $n,$ then it moves to $n + 1$ with probability $\tfrac14,$ it moves to $n - 1$ with probability $\tfrac14,$ and it disappears with probability $\tfrac12.$

What is the probability that the frog reaches $4?$

$\dfrac{1}{101}$

$\dfrac{1}{100}$

$\dfrac{1}{99}$

$\dfrac{1}{98}$

$\dfrac{1}{97}$

答案：E

知识点：随机游走递推概率方程组

难度评级：2170

解答：

令 $p_n$ 为从位置 $n$ 出发到达 $4$ 的概率，且 $p_4 = 1$ 。规则给出 $p_0 = \tfrac12 p_1$ 、 $p_1 = \tfrac14 p_0 + \tfrac14 p_2$ 、 $p_2 = \tfrac14 p_1 + \tfrac14 p_3$ 、 $p_3 = \tfrac14 p_2 + \tfrac14$ 。向上求解可得 $p_1 = \tfrac27 p_2$ 和 $p_2 = \tfrac{7}{26} p_3$ 。继续代回得到 $p_3 = \tfrac{26}{97}$ 、 $p_2 = \tfrac{7}{97}$ 、 $p_1 = \tfrac{2}{97}$ ，最后 $p_0 = \tfrac{1}{97}$ 。因此正确答案是 E。

Let $p_n$ be the probability of reaching $4$ from position $n,$ with $p_4 = 1.$ The rules give $p_0 = \tfrac12 p_1,$ $p_1 = \tfrac14 p_0 + \tfrac14 p_2,$ $p_2 = \tfrac14 p_1 + \tfrac14 p_3,$ and $p_3 = \tfrac14 p_2 + \tfrac14.$ Work upward: $p_1 = \tfrac27 p_2$ and $p_2 = \tfrac{7}{26} p_3.$ These unwind to $p_3 = \tfrac{26}{97},$ $p_2 = \tfrac{7}{97},$ $p_1 = \tfrac{2}{97},$ and finally $p_0 = \tfrac{1}{97}.$ Therefore, the answer is E.

2025 AMC 10B 第 24 题

答案：E

难度评级：2170

解答：

其他年份的第 24 题