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

(Let's prioritize the 8th grade solution. Appreciate it. :))
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==Solution 1 (Casework)==
 
==Solution 1 (Casework)==
  
 +
Let <math>A</math> denote the leaf where the cricket starts and <math>B</math> denote one of the other <math>3</math> leaves. Note that:
  
 +
* If the cricket is at <math>A,</math> then the probability that it hops to <math>B</math> next is <math>1.</math>
 +
 +
* If the cricket is at <math>B,</math> then the probability that it hops to <math>A</math> next is <math>\frac13.</math>
 +
 +
* If the cricket is at <math>B,</math> then the probability that it hops to <math>B</math> next is <math>\frac23.</math>
 +
 +
We apply casework to the possible paths of the cricket:
 +
<ol style="margin-left: 1.5em;">
 +
  <li><math>A \longrightarrow B \longrightarrow A \longrightarrow B \longrightarrow A</math> <p>
 +
The probability for this case is <math>1\cdot\frac13\cdot1\cdot\frac13=\frac19.</math></li><p>
 +
  <li><math>A \longrightarrow B \longrightarrow B \longrightarrow B \longrightarrow A</math> <p>
 +
The probability for this case is <math>1\cdot\frac23\cdot\frac23\cdot\frac13=\frac{4}{27}.</math></li><p>
 +
</ol>
 +
Together, the probability that the cricket returns to <math>A</math> is <math>\frac19+\frac{4}{27}=\boxed{\textbf{(E) }\frac{7}{27}}.</math>
 +
 +
~MRENTHUSIASM
  
 
==Solution 2 (Recursion)==
 
==Solution 2 (Recursion)==
  
−
Denote <math>P_n</math> to be the probability that the cricket would return back to the first point after <math>n</math> Hops. Then, we get the recursive formula <math>P_n = \frac13(1-P_{n-1})</math> because if the leaf is not on the target leaf, then there is a <math>\frac13</math> probability that he’ll make it back. With this formula and the fact that <math>P_0=0,</math> we have: <cmath>P_1 = \frac13, P_2 = \frac29, P_3 = \frac7{27},</cmath> so our answer is <math>\boxed{\textbf{(E) }\frac7{27}}</math>
+
Denote <math>P_n</math> to be the probability that the cricket would return back to the first point after <math>n</math> Hops. Then, we get the recursive formula <math>P_n = \frac13(1-P_{n-1})</math> because if the leaf is not on the target leaf, then there is a <math>\frac13</math> probability that he’ll make it back. With this formula and the fact that <math>P_0=0,</math> we have: <cmath>P_1 = \frac13, P_2 = \frac29, P_3 = \frac7{27},</cmath> so our answer is <math>\boxed{\textbf{(E) }\frac{7}{27}}</math>.
  
 
~wamofan
 
~wamofan

Revision as of 20:09, 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 1 (Casework)

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. $A \longrightarrow B \longrightarrow A \longrightarrow B \longrightarrow A$

    The probability for this case is $1\cdot\frac13\cdot1\cdot\frac13=\frac19.$

  2. $A \longrightarrow B \longrightarrow B \longrightarrow B \longrightarrow A$

    The probability for this case is $1\cdot\frac23\cdot\frac23\cdot\frac13=\frac{4}{27}.$

Together, the probability that the cricket returns to $A$ is $\frac19+\frac{4}{27}=\boxed{\textbf{(E) }\frac{7}{27}}.$

~MRENTHUSIASM

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 $\boxed{\textbf{(E) }\frac{7}{27}}$.

~wamofan

See Also

2022 AMC 8 (Problems • Answer Key • Resources)
Preceded by
Problem 24 		Followed by
Last Problem
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AJHSME/AMC 8 Problems and Solutions

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