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Difference between revisions of "2022 AMC 8 Problems/Problem 25"

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A cricket randomly hops between <math>4</math> leaves, on each turn hopping to one of the other <math>3</math> leaves with equal probability. After <math>4</math> hops what is the probability that the cricket has returned to the leaf where it started?
 
A cricket randomly hops between <math>4</math> leaves, on each turn hopping to one of the other <math>3</math> leaves with equal probability. After <math>4</math> hops what is the probability that the cricket has returned to the leaf where it started?
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[[File:2022 AMC 8 Problem 25 Picture.jpg|center|600px]]
  
 
<math>\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}</math>
 
<math>\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}</math>

Revision as of 20:28, 28 January 2022

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?

2022 AMC 8 Problem 25 Picture.jpg

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

Solution

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

See Also

2022 AMC 8 (Problems • Answer Key • Resources)
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Problem 24 		Followed by
Last Problem
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All AJHSME/AMC 8 Problems and Solutions

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