Question

一个小池塘里有十一片睡莲叶排成一行，标号为 $0$ 到 $10$ 。一只青蛙坐在标号 $1$ 的睡莲叶上。当青蛙在标号 $N$ 的叶子上，且 $0 \lt N \lt 10$ 时，它会以概率 $\dfrac{N}{10}$ 跳到 $N - 1$ ，并以概率 $1 - \dfrac{N}{10}$ 跳到 $N + 1$ 。每次跳跃都与之前的跳跃独立。如果青蛙到达标号 $0$ 的叶子，它会被一条耐心等待的蛇吃掉；如果到达标号 $10$ ，它就会离开池塘，不再回来。青蛙逃过被蛇吃掉的概率是多少？

In a small pond there are eleven lily pads in a row labeled $0$ through $10.$ A frog is sitting on pad $1.$ When the frog is on pad $N,$ $0 \lt N \lt 10,$ it will jump to pad $N - 1$ with probability $\dfrac{N}{10}$ and to pad $N + 1$ with probability $1 - \dfrac{N}{10}.$ Each jump is independent of the previous jumps. If the frog reaches pad $0$ it will be eaten by a patiently waiting snake. If the frog reaches pad $10$ it will exit the pond, never to return. What is the probability that the frog will escape being eaten by the snake?

$\dfrac{32}{79}$

$\dfrac{161}{384}$

$\dfrac{63}{146}$

$\dfrac{7}{16}$

$\dfrac{1}{2}$

答案：C

知识点：随机游走递推概率对称性

难度评级：2450

小提示：

由对称性，从标号 $5$ 的叶子出发的逃脱概率为 $\tfrac12$ 。

By symmetry the escape probability from pad $5$ is $\tfrac12$

大提示：

设 $p_j$ 为从标号 $j$ 的叶子出发的逃脱概率，并用相邻位置表示每个 $p_j$ 。

Let $p_j$ be the escape probability from pad $j$ and write each $p_j$ in terms of its neighbors

解答：

设 $p_j$ 为从标号 $j$ 的叶子出发最终到达标号 $10$ 的概率。由中心处跳跃规则的对称性， $p_5 = \tfrac12$ 。

每个内部位置满足 $p_j = \tfrac{10-j}{10}\,p_{j+1} + \tfrac{j}{10}\,p_{j-1}$ ，因此 $\begin{gathered} p_4 = \tfrac25 p_3 + \tfrac35 p_5, \\ \quad p_3 = \tfrac{3}{10} p_2 + \tfrac{7}{10} p_4 \end{gathered}\text{，}$ $p_2 = \tfrac15 p_1 + \tfrac45 p_3,\quad p_1 = \tfrac{9}{10} p_2\text{。}$

令 $d_j=p_j-p_{j-1}$ 。递推式等价于 $(10-j)d_{j+1}=jd_j$ ，所以 $\begin{gathered} d_2=\dfrac{d_1}{9},\quad d_3=\dfrac{d_1}{36}, \\ d_4=\dfrac{d_1}{84},\quad d_5=\dfrac{d_1}{126} \end{gathered}\text{。}$ 又因为 $p_0=0$ 且 $p_5=\frac{1}{2}$ ，所以 $\begin{aligned} \dfrac12&=d_1\biggl(1+\dfrac19+\dfrac1{36} \\ &\qquad+\dfrac1{84}+\dfrac1{126}\biggr) \\ &=d_1\cdot\dfrac{73}{63} \end{aligned}\text{。}$ 因此 $p_1=d_1=\frac{63}{146}$ 。

所以正确答案是 C。

Let $p_j$ be the probability of eventually reaching pad $10$ starting from pad $j.$ By the symmetry of the jump rule at the center, $p_5 = \tfrac12.$

Each interior pad satisfies $p_j = \tfrac{10-j}{10}\,p_{j+1} + \tfrac{j}{10}\,p_{j-1},$ which gives $\begin{gathered} p_4 = \tfrac25 p_3 + \tfrac35 p_5, \\ \quad p_3 = \tfrac{3}{10} p_2 + \tfrac{7}{10} p_4, \end{gathered}$ $p_2 = \tfrac15 p_1 + \tfrac45 p_3,\quad p_1 = \tfrac{9}{10} p_2.$

Put $d_j=p_j-p_{j-1}.$ The recurrence is equivalent to $(10-j)d_{j+1}=jd_j,$ so $\begin{gathered} d_2=\dfrac{d_1}{9},\quad d_3=\dfrac{d_1}{36}, \\ d_4=\dfrac{d_1}{84},\quad d_5=\dfrac{d_1}{126}. \end{gathered}$ Since $p_0=0$ and $p_5=\frac{1}{2},$ $\begin{aligned} \dfrac12&=d_1\biggl(1+\dfrac19+\dfrac1{36} \\ &\qquad+\dfrac1{84}+\dfrac1{126}\biggr) \\ &=d_1\cdot\dfrac{73}{63}. \end{aligned}$ Therefore $p_1=d_1=\frac{63}{146}.$

Thus, the correct answer is C.

2014 AMC 12B 第 22 题

答案：C

难度评级：2450

小提示：

大提示：

解答：

其他年份的第 22 题