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2022 AMC 8 Problems/Problem 25

Revision as of 13:25, 27 February 2023 by Sujoyg (talk | contribs)

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 $A$ after $4$ hops is $\frac19+\frac{4}{27}=\boxed{\textbf{(E) }\frac{7}{27}}.$

~MRENTHUSIASM

Solution 2 (Casework)

We can label the leaves as shown below:

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 $B$:

  1. $A \rightarrow B \rightarrow A \rightarrow B \rightarrow A$

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

  3. $A \rightarrow B \rightarrow A \rightarrow D \rightarrow A$

  4. $A \rightarrow B \rightarrow C \rightarrow B \rightarrow A$

  5. $A \rightarrow B \rightarrow C \rightarrow D \rightarrow A$

  6. $A \rightarrow B \rightarrow D \rightarrow B \rightarrow A$

  7. $A \rightarrow B \rightarrow D \rightarrow C \rightarrow A$

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 (Complement)

There are always three possible leaves to jump to every time the cricket decides to jump, so there is a total number of $3^4$ routes. Let $A$ denote the leaf cricket starts at, and $B, C, D$ be the other leaves. If we want the cricket to move to leaf $A$ for its last jump, the cricket cannot jump to leaf $A$ for its third jump. Also, considering that the cricket starts at leaf $A$, he cannot jump to leaf $A$ for its first jump. Note that there are $3\cdot2=6$ paths if the cricket moves to leaf $A$ for its third jump. Therefore, we can conclude that the total number of possible paths for the cricket to return to leaf $A$ after four jumps is $3^3 - 6 = 21$, so the answer is $\frac{21}{3^4} = \frac{21}{81}=\boxed{\textbf{(E) }\frac{7}{27}}$.

~Bloggish

Solution 4 (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 $P_1=0$ (After one hop, the cricket can never be back to the target leaf.), we have \[P_2 = \frac13, P_3 = \frac29, P_4 = \frac7{27},\] so our answer is $\boxed{\textbf{(E) }\frac{7}{27}}$.

~wamofan

Solution 5 (Dynamic Programming)

Let $A$ denote the leaf cricket starts at, and $B, C, D$ be the other leaves, similar to Solution 2.

Let $A(n)$ be the probability the cricket lands on $A$ after $n$ hops, $B(n)$ be the probability the cricket lands on $B$ after crawling $n$ hops, and etc.

Note that $A(1)=0$ and $B(1)=C(1)=D(1)=\frac13.$ For $n\geq2,$ the probability that the cricket land on each leaf after $n$ hops is $\frac13$ the sum of the probability the cricket land on other leaves after $n-1$ hops. So, we have \begin{align*} A(n) &= \frac13 \cdot [B(n-1) + C(n-1) + D(n-1)], \\ B(n) &= \frac13 \cdot [A(n-1) + C(n-1) + D(n-1)], \\ C(n) &= \frac13 \cdot [A(n-1) + B(n-1) + D(n-1)], \\ D(n) &= \frac13 \cdot [A(n-1) + B(n-1) + C(n-1)]. \end{align*} It follows that $A(n) = B(n-1) = C(n-1) = D(n-1).$

We construct the following table: \[\begin{array}{c|cccc} & & & & \\ [-2ex] n & A(n) & B(n) & C(n) & D(n) \\ [1ex] \hline & & & & \\ [-1ex] 1 & 0 & \frac13 & \frac13 & \frac13 \\ & & & & \\ 2 & \frac13 & \frac29 & \frac29 & \frac29 \\ & & & & \\ 3 & \frac29 & \frac{7}{27} & \frac{7}{27} & \frac{7}{27} \\ & & & & \\ 4 & \frac{7}{27} & \frac{20}{81} & \frac{20}{81} & \frac{20}{81} \\ [1ex] \end{array}\] Therefore, the answer is $A(4)=\boxed{\textbf{(E) }\frac{7}{27}}$.

~isabelchen

Solution 6 (Generating Function)

Assign the leaves to $0, 1, 2,$ and $3$ modulo $4,$ and let $0$ be the starting leaf. We then use generating functions with relation to the change of leaves. For example, from $3$ to $1$ would be a change of $2,$ and from $1$ to $2$ would be a change of $1.$ This generating function is equal to $(x+x^2+x^3)^4.$ It is clear that we want the coefficients in the form of $x^{4n},$ where $n$ is a positive integer. One application of roots of unity filter gives us a successful case count of $\frac{81+1+1+1}{4} = 21.$

Therefore, the answer is $\frac{21}{3^4}=\boxed{\textbf{(E) }\frac{7}{27}}.$

~sigma

Solution 7 (Simple Enumeration)

- By sujoyg

There are only two paths for the moth go back to the original leaf on 4th hop:

1. 1st hop to some other leaf, 2nd hop to original leaf, 3rd hop to some other leaf, and 4th hop again back to original leaf. 2. 1st, 2nd and 3rd hops to some other leaf and back to the original leaf on the 4th hop back.

The probability of it being on some other leaf after being on the original leaf is 1. Otherwise, the probability of hopping to the original leaf is $\frac{1}{3}$ and some other leaf is $\frac{2}{3}$.

Therefore, the total probability is:

p(1st path) = p(2nd path) = 1 * $\frac{1}{3}$ * 1 * $\frac{1}{3}$ + 1 * $\frac{2}{3}$ * $\frac{2}{3}$ * $\frac{1}{3}$ = $\frac{1}{9}$ + $\frac{4}{27}$ = $\frac{7}{27}$


So 1 * 1/3 * 1 * 1/3 (since in this case the probability of it not going to original leaf on 1st and 3rd hop are 1) +1 * 2/3 * 2/3 * 1/3 (similar logic) = 1/9 + 4/27 = 7/27?

Remark

This problem is a reduced version of 1985 AIME Problem 12, changing $7$ steps into $4$ steps.

This problem is also similar to 2003 AIME II Problem 13.

~isabelchen

Video Solution

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https://youtu.be/kE15Sy0B2Pk?t=633

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See Also

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

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