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

Revision as of 21:01, 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?

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

Video Solution

https://www.youtube.com/watch?v=85A6av3oqRo

~Mathematical Dexterity

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:

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

$A \rightarrow B \rightarrow B \rightarrow B \rightarrow A$
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 (Casework)

We can label the leaves as shown below:

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 $B$ :

$A \rightarrow B \rightarrow A \rightarrow B \rightarrow A$

$A \rightarrow B \rightarrow A \rightarrow C \rightarrow A$

$A \rightarrow B \rightarrow A \rightarrow D \rightarrow A$

$A \rightarrow B \rightarrow C \rightarrow B \rightarrow A$

$A \rightarrow B \rightarrow C \rightarrow D \rightarrow A$

$A \rightarrow B \rightarrow D \rightarrow B \rightarrow A$

$A \rightarrow B \rightarrow D \rightarrow C \rightarrow A$

By symmetry, we know that there are $7$ ways if the cricket's first hop was to leaf $C$ , and there are $7$ ways if the cricket's first hop was to leaf $D$ . So, there are $21$ ways in total 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

Solution 3 (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

@@ Line 6: / Line 6: @@
 <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>
+==Video Solution==
+https://www.youtube.com/watch?v=85A6av3oqRo
+~Mathematical Dexterity
 ==Solution 1 (Casework)==

2022 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 24	Followed by Last Problem
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