AMC8 2022

AMC8 2022 · Q25

AMC8 2022 · Q25. It mainly tests Probability (basic), Casework.

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?

一只蟋蟀在4片叶子上随机跳跃，每次跳到其他3片叶子之一，概率相等。经过4次跳跃后，它回到起始叶子的概率是多少？

(A) \frac{2}{9} \frac{2}{9}

(B) \frac{19}{80} \frac{19}{80}

(D) \frac{1}{4} \frac{1}{4}

(E) \frac{7}{27} \frac{7}{27}

Answer

Correct choice: (E)

正确答案：（E）

Solution

Let $A$ denote the leaf where the cricket starts and $B$ denote one of the other $3$ leaves. Note that: - If the cricket is at $A,$ then the probability that it hops to $B$ next is $1.$ - If the cricket is at $B,$ then the probability that it hops to $A$ next is $\frac13.$ - If the cricket is at $B,$ then the probability that it hops to $B$ next is $\frac23.$ We apply casework to the possible paths of the cricket: 1. $A \rightarrow B \rightarrow A \rightarrow B \rightarrow A$ The probability for this case is $1\cdot\frac13\cdot1\cdot\frac13=\frac19.$ 2. $A \rightarrow B \rightarrow B \rightarrow B \rightarrow A$ The probability for this case is $1\cdot\frac23\cdot\frac23\cdot\frac13=\frac{4}{27}.$ Together, the probability that the cricket returns to $A$ after $4$ hops is $\frac19+\frac{4}{27}=\boxed{\textbf{(E) }\frac{7}{27}}.$

设$A$为起始叶子，$B$为其他3片之一之一。注意： - 在$A$时，下跳到$B$概率为$1$。 - 在$B$时，下跳到$A$概率为$\frac13$。 - 在$B$时，下跳到$B$概率为$\frac23$。按路径分类讨论： 1. $A \rightarrow B \rightarrow A \rightarrow B \rightarrow A$ 概率$1\cdot\frac13\cdot1\cdot\frac13=\frac19$。 2. $A \rightarrow B \rightarrow B \rightarrow B \rightarrow A$ 概率$1\cdot\frac23\cdot\frac23\cdot\frac13=\frac{4}{27}$。总概率$\frac19+\frac{4}{27}=\boxed{\textbf{(E) }\frac{7}{27}}$。

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