2022 AMC 8 Problems/Problem 25

Problem

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?

$\textbf{(A) }\frac{2}{9}\qquad\textbf{(B) }\frac{19}{80}\qquad\textbf{(C) }\frac{20}{81}\qquad\textbf{(D) }\frac{1}{4}\qquad\textbf{(E) }\frac{7}{27}$

Solution 1 (Casework)

Solution 2 (Recursion)

Denote $P_n$ to be the probability that the cricket would return back to the first point after $n$ Hops. Then, we get the recursive formula $P_n = \frac13(1-P_{n-1})$ because if the leaf is not on the target leaf, then there is a $\frac13$ probability that he’ll make it back. With this formula and the fact that $P_0=0,$ we have: $P_1 = \frac13, P_2 = \frac29, P_3 = \frac7{27},$ so our answer is $\boxed{\textbf{(E) }\frac7{27}}$

~wamofan

2022 AMC 8 (Problems • Answer Key • Resources)
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2022 AMC 8 Problems/Problem 25

Contents

Problem

Solution 1 (Casework)

Solution 2 (Recursion)

See Also