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

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We apply casework to the possible paths of the cricket:
 
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   <li><math>A \longrightarrow B \longrightarrow A \longrightarrow B \longrightarrow A</math> <p>
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   <li><math>A \rightarrow B \rightarrow A \rightarrow B \rightarrow A</math> <p>
 
The probability for this case is <math>1\cdot\frac13\cdot1\cdot\frac13=\frac19.</math></li><p>
 
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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   <li><math>A \rightarrow B \rightarrow B \rightarrow B \rightarrow A</math> <p>
 
The probability for this case is <math>1\cdot\frac23\cdot\frac23\cdot\frac13=\frac{4}{27}.</math></li><p>
 
The probability for this case is <math>1\cdot\frac23\cdot\frac23\cdot\frac13=\frac{4}{27}.</math></li><p>
 
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Revision as of 15:30, 29 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. $A \rightarrow B \rightarrow A \rightarrow B \rightarrow A$

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

  2. $A \rightarrow B \rightarrow B \rightarrow B \rightarrow A$

    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 it will 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

Solution 3 (Casework)

We can label the leaves as shown:

2022 AMC 8 Problem 25 Picture 2.png

Carefully counting cases, we see that there are $7$ ways for the cricket to return to leaf $A$ after four hops if its first hop was to leaf $B$:

\[A \Rightarrow B \Rightarrow C \Rightarrow D \Rightarrow A\] \[A \Rightarrow B \Rightarrow C \Rightarrow B \Rightarrow A\] \[A \Rightarrow B \Rightarrow D \Rightarrow C \Rightarrow A\] \[A \Rightarrow B \Rightarrow D \Rightarrow B \Rightarrow A\] \[A \Rightarrow B \Rightarrow A \Rightarrow B \Rightarrow A\] \[A \Rightarrow B \Rightarrow A \Rightarrow C \Rightarrow A\] \[A \Rightarrow B \Rightarrow A \Rightarrow D \Rightarrow A\]

Taking advantage of symmetry, this also means there are $7$ ways if the cricket's first hop was to leaf $C$.

Finally, if the cricket's first hop was to leaf $D$, we see that there are also $7$ ways:

\[A \Rightarrow D \Rightarrow B \Rightarrow C \Rightarrow A\] \[A \Rightarrow D \Rightarrow B \Rightarrow D \Rightarrow A\] \[A \Rightarrow D \Rightarrow C \Rightarrow B \Rightarrow A\] \[A \Rightarrow D \Rightarrow C \Rightarrow D \Rightarrow A\] \[A \Rightarrow D \Rightarrow A \Rightarrow B \Rightarrow A\] \[A \Rightarrow D \Rightarrow A \Rightarrow C \Rightarrow A\] \[A \Rightarrow D \Rightarrow A \Rightarrow D \Rightarrow A\]

So, in total, there are $21$ ways for the cricket to return to leaf $A$ after four hops.

Since there are $3^4 = 81$ possible ways altogether for the cricket to hop to any other leaf four times, the answer is $\frac{21}{81} = \boxed{\textbf{(E) } \frac{7}{27}}$.

~mahaler

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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