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2022 AMC 8 Problems/Problem 25

Revision as of 21:38, 18 February 2024 by Cxsmi (talk | contribs) (→‎Solution 5 (Dynamic Programming))

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, 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 (Also Generating Functions)

Let the leaves be $(0,0), (0,1), (1,0),$ and $(1,1)$ on the coordinate plane, with the cricket starting at $(0,0)$. Then we write a generating function. We denote $x$ a change in the x-value of the cricket, and similarly for $y$. Then our generating function is $(x+y+xy)^4,$ and we wish to compute the number of terms in which the exponents of both x and y are even. To do this, we first square to get $(x^2 + y^2 + x^2y^2 + 2xy + 2x^2y + 2xy^2)^2$. Note that every term squared will give even powers for x and y, so that gives us $3 + 4\cdot3 = 15.$ Then every combination of $x^2, y^2,$ and $x^2y^2$ will also give us even powers for x and y, so that yields $6$ more terms, for a total of $21.$ Now in total there $3^4 = 81$ possible sequences, so $21/81$ gives us the answer of $\boxed{\textbf{(E) }\frac{7}{27}}.$

~littlefox_amc

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 by Math-X (First understand the problem!!!)

https://youtu.be/oUEa7AjMF2A?si=n9aPrcW_qLqFC8IF&t=5261

~Math-X

Video Solution(🚀Under 2 min🚀 Easy logic with all paths color-coded ✨)

https://youtu.be/YiI9szmMWX4

~Education, the Study of Everything

Video Solution

https://youtu.be/GNFG4cmYDgw

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Video Solution by OmegaLearn

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

~ pi_is_3.14

Video Solution

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

~Mathematical Dexterity

Video Solution

https://youtu.be/Ij9pAy6tQSg?t=2588

~Interstigation

Video Solution

https://www.youtube.com/watch?v=H1zxrkq6DKg

~David

Video Solution

https://youtu.be/0orAAUaLIO0?t=609

~STEMbreezy

Video Solution

https://youtu.be/9SKUdTut3l4

~savannahsolver

See Also

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

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