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)==
  
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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:
  
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* If the cricket is at <math>A,</math> then the probability that it hops to <math>B</math> next is <math>1.</math>
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* If the cricket is at <math>B,</math> then the probability that it hops to <math>A</math> next is <math>\frac13.</math>
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* If the cricket is at <math>B,</math> then the probability that it hops to <math>B</math> next is <math>\frac23.</math>
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We apply casework to the possible paths of the cricket:
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<ol style="margin-left: 1.5em;">
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  <li><math>A \longrightarrow B \longrightarrow A \longrightarrow B \longrightarrow A</math> <p>
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The probability for this case is <math>1\cdot\frac13\cdot1\cdot\frac13=\frac19.</math></li><p>
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  <li><math>A \longrightarrow B \longrightarrow B \longrightarrow B \longrightarrow A</math> <p>
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The probability for this case is <math>1\cdot\frac23\cdot\frac23\cdot\frac13=\frac{4}{27}.</math></li><p>
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</ol>
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Together, the probability that the cricket returns to <math>A</math> is <math>\frac19+\frac{4}{27}=\boxed{\textbf{(E) }\frac{7}{27}}.</math>
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~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 21:09, 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 (ProblemsAnswer KeyResources)
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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