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

m (→‎Solution 3 (Recursion))
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~wamofan
 
~wamofan
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==Solution 4 (Dynamic Programming)==
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Let A denote the leaf cricket starts at. B, C, D be the other leaves, similar to solution <math>2</math>.
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Let <math>A(n)</math> be the probability the bug lands on vertex A after crawling <math>n</math> meters, <math>B(n)</math> be the probability the bug lands on vertex B after crawling <math>n</math> meters and etc.
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The probability that the bug land on each vertex after <math>n</math> meters is one-third the sum of the probability the bug land on other vertices after crawling <math>n-1</math> meters.
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<math>A(n) = \frac13 \cdot [B(n-1) + C(n-1) + D(n-1)]</math>
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<math>B(n) = \frac13 \cdot [A(n-1) + C(n-1) + D(n-1)]</math>
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<math>C(n) = \frac13 \cdot [A(n-1) + B(n-1) + D(n-1)]</math>
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<math>D(n) = \frac13 \cdot [A(n-1) + B(n-1) + C(n-1)]</math>
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<math>B(n) = C(n) = D(n)</math>, <math>A(n) = B(n-1) = C(n-1) = D(n-1)</math>
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<cmath>\[ \begin{array}[b]{ccccc}
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n & A(n) & B(n) & C(n) & D(n) \\
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&  &  &  & \\
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1 & 0 & \frac13 & \frac13 & \frac13 \\
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&  &  &  & \\
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2 & \frac13 & \frac29 & \frac29 & \frac29 \\
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&  &  &  & \\
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3 & \frac29 & \frac{7}{27} & \frac{7}{27} & \frac{7}{27} \\
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&  &  &  & \\
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4 & \frac{7}{27} & \frac{20}{81} & \frac{20}{81} & \frac{20}{81}\\
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\end{array} \]</cmath>
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<math>n = \boxed{\textbf{182}}</math>
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~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen]
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==Remark==
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This problem is a reduced version of [https://artofproblemsolving.com/wiki/index.php/1985_AIME_Problems/Problem_12 1985 AIME Problem 12], changing <math>7</math> steps into <math>4</math> steps. This problem is also similar to [https://artofproblemsolving.com/wiki/index.php/2003_AIME_II_Problems/Problem_13 2003 AIME II Problem 13].
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~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen]
  
 
==Video Solution==
 
==Video Solution==

Revision as of 10:01, 20 February 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 $A$ 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 (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 4 (Dynamic Programming)

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

Let $A(n)$ be the probability the bug lands on vertex A after crawling $n$ meters, $B(n)$ be the probability the bug lands on vertex B after crawling $n$ meters and etc.

The probability that the bug land on each vertex after $n$ meters is one-third the sum of the probability the bug land on other vertices after crawling $n-1$ meters.

$A(n) = \frac13 \cdot [B(n-1) + C(n-1) + D(n-1)]$

$B(n) = \frac13 \cdot [A(n-1) + C(n-1) + D(n-1)]$

$C(n) = \frac13 \cdot [A(n-1) + B(n-1) + D(n-1)]$

$D(n) = \frac13 \cdot [A(n-1) + B(n-1) + C(n-1)]$

$B(n) = C(n) = D(n)$, $A(n) = B(n-1) = C(n-1) = D(n-1)$

\[ \begin{array}[b]{ccccc} n & A(n) & B(n) & C(n) & D(n) \\ & & & & \\ 1 & 0 & \frac13 & \frac13 & \frac13 \\ & & & & \\ 2 & \frac13 & \frac29 & \frac29 & \frac29 \\ & & & & \\ 3 & \frac29 & \frac{7}{27} & \frac{7}{27} & \frac{7}{27} \\ & & & & \\ 4 & \frac{7}{27} & \frac{20}{81} & \frac{20}{81} & \frac{20}{81}\\ \end{array} \]

$n = \boxed{\textbf{182}}$

~isabelchen

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

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

~Mathematical Dexterity

Video Solution

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

~Interstigation

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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