Question

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?

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

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

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

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

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

Solution:

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

Thus, the correct answer is E.

2022 AMC 8 Problem 25

Solution:

Problem 25 in Other Years