Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Solution 7 (Simple Enumeration))
(→‎Video Solution)
 
(18 intermediate revisions by 10 users not shown)
Line 65: Line 65:
 
Let <math>A</math> denote the leaf cricket starts at, and <math>B, C, D</math> be the other leaves, similar to Solution 2.
 
Let <math>A</math> denote the leaf cricket starts at, and <math>B, C, D</math> be the other leaves, similar to Solution 2.
  
−
Let <math>A(n)</math> be the probability the cricket lands on <math>A</math> after <math>n</math> hops, <math>B(n)</math> be the probability the cricket lands on <math>B</math> after crawling <math>n</math> hops, and etc.
+
Let <math>A(n)</math> be the probability the cricket lands on <math>A</math> after <math>n</math> hops, <math>B(n)</math> be the probability the cricket lands on <math>B</math> after crawling <math>n</math> hops, etc.
  
 
Note that <math>A(1)=0</math> and <math>B(1)=C(1)=D(1)=\frac13.</math> For <math>n\geq2,</math> the probability that the cricket land on each leaf after <math>n</math> hops is <math>\frac13</math> the sum of the probability the cricket land on other leaves after <math>n-1</math> hops. So, we have
 
Note that <math>A(1)=0</math> and <math>B(1)=C(1)=D(1)=\frac13.</math> For <math>n\geq2,</math> the probability that the cricket land on each leaf after <math>n</math> hops is <math>\frac13</math> the sum of the probability the cricket land on other leaves after <math>n-1</math> hops. So, we have
Line 102: Line 102:
 
~sigma
 
~sigma
  
−
==Solution 7 (Simple Enumeration)==
+
==Solution 7 (Also Generating Functions)==
−
- By [https://artofproblemsolving.com/wiki/index.php/User:Sujoyg sujoyg]
 
  
−
There are only two paths for the moth go back to the original leaf on the 4th hop:
+
Let the leaves be <math>(0,0), (0,1), (1,0),</math> and <math>(1,1)</math> on the coordinate plane, with the cricket starting at <math>(0,0)</math>. Then we write a generating function. We denote <math>x</math> a change in the x-value of the cricket, and similarly for <math>y</math>. Then our generating function is <math>(x+y+xy)^4,</math> 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 <math>(x^2 + y^2 + x^2y^2 + 2xy + 2x^2y + 2xy^2)^2</math>. Note that every term squared will give even powers for x and y, so that gives us <math>3 + 4\cdot3 = 15.</math> Then every combination of <math>x^2, y^2,</math> and <math>x^2y^2</math> will also give us even powers for x and y, so that yields <math>6</math> more terms, for a total of <math>21.</math> Now in total there <math>3^4 = 81</math> possible sequences, so <math>21/81</math> gives us the answer of <math>\boxed{\textbf{(E) }\frac{7}{27}}.</math>
  
−
<ol style="margin-left: 1.5em;">
+
~littlefox_amc
−
<li>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.</li>
 
−
<li>1st, 2nd and 3rd hops to some other leaf and back to the original leaf on the 4th hop back.</li>
 
−
</ol>
 
−
 
 
−
The probability of it being on some other leaf on the immediate next hop after being on the original leaf is 1. In all other cases, the probability of hopping to the original leaf is <math>\frac{1}{3}</math> and some other leaf is <math>\frac{2}{3}</math>.
 
−
 
 
−
Therefore, the total probability is:
 
−
<cmath>P(Path_1) = P(Path_2)</cmath>
 
−
<cmath>= (1 * \frac{1}{3} * 1 * \frac{1}{3}) + (1 * \frac{2}{3} * \frac{2}{3} * \frac{1}{3})</cmath>
 
−
<cmath>= \frac{1}{9} + \frac{4}{27}</cmath>
 
−
<cmath>= \frac{7}{27}</cmath>
 
  
 
==Remark==
 
==Remark==
Line 127: Line 115:
  
 
~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen]
 
~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen 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
 +
 +
<i>~Education, the Study of Everything</i>
 +
 
==Video Solution==
 
==Video Solution==
 
https://youtu.be/GNFG4cmYDgw
 
https://youtu.be/GNFG4cmYDgw
Line 147: Line 146:
 
==Video Solution==
 
==Video Solution==
 
https://www.youtube.com/watch?v=H1zxrkq6DKg
 
https://www.youtube.com/watch?v=H1zxrkq6DKg
 +
 +
~David
  
 
==Video Solution==
 
==Video Solution==
Line 157: Line 158:
  
 
~savannahsolver
 
~savannahsolver
 +
 +
== Video Solution Using States by SpreadTheMathLove ==
 +
https://www.youtube.com/watch?v=740Z355PtWs&t=777s
  
 
==See Also==  
 
==See Also==  
 
{{AMC8 box|year=2022|num-b=24|after=Last Problem}}
 
{{AMC8 box|year=2022|num-b=24|after=Last Problem}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 20:43, 2 June 2024

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

Please like and subscribe!

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

Video Solution Using States by SpreadTheMathLove

https://www.youtube.com/watch?v=740Z355PtWs&t=777s

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2022_AMC_8_Problems/Problem_25&oldid=219999"