2022 AMC 8 第 25 题

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}

只跟踪蟋蟀是否在起始叶子上

Track only whether the cricket is on the starting leaf

如果蟋蟀不在起点，下一跳以

\frac13

的概率回到起点

If the cricket is not at the start, the next hop returns with probability

\frac13

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}\text{。}

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}\text{。}

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.

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