Question

一只蟋蟀在 $4$ 片叶子之间随机跳跃，每次跳到另外 $3$ 片叶子中的一片，概率相等。跳 $4$ 次后，蟋蟀回到起始叶子的概率是多少？

A cricket randomly hops between $4$ leaves, on each turn hopping to one of the other $3$ leaves with equal probability. After $4$ hops, what is the probability that the cricket has returned to the leaf where it started?

$\displaystyle \dfrac{2}{9}$

$\displaystyle \dfrac{19}{80}$

$\displaystyle \dfrac{20}{81}$

$\displaystyle \dfrac{1}{4}$

$\displaystyle \dfrac{7}{27}$

答案：E

知识点：递推概率对称性

难度评级：1670

解答：

设 $p_n$ 为跳 $n$ 次后在起始叶子上的概率， $p_0=1$ 。

若蟋蟀在起点，下一跳一定离开；若不在起点，下一跳有 $3$ 个等可能选择，其中一个会回到起点。因此 $p_{n+1}=\frac{1-p_n}{3}.$

逐次计算得 $\begin{gathered} p_1=0,\quad p_2=\frac13,\quad p_3=\frac29,\quad \\ p_4=\frac{1-\frac29}{3}=\frac{7}{27}. \end{gathered}$

正确答案是 E。

Let $p_n$ be the probability that the cricket is on its starting leaf after $n$ hops. We have $p_0=1$ .

If the cricket is on the starting leaf, the next hop must leave it. If the cricket is not on the starting leaf, exactly one of the $3$ possible hops returns to the start. Therefore $p_{n+1}=\frac{1-p_n}{3}.$

Thus $\begin{gathered} p_1=0,\quad p_2=\frac13,\quad p_3=\frac29,\quad \\ p_4=\frac{1-\frac29}{3}=\frac{7}{27}. \end{gathered}$

Thus, the correct answer is E.

2022 AMC 8 第 25 题

答案：E

难度评级：1670

解答：

其他年份的第 25 题