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Difference between revisions of "2019 AIME I Problems/Problem 5"

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==Problem 5==
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==Problem==
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A moving particle starts at the point <math>(4,4)</math> and moves until it hits one of the coordinate axes for the first time. When the particle is at the point <math>(a,b)</math>, it moves at random to one of the points <math>(a-1,b)</math>, <math>(a,b-1)</math>, or <math>(a-1,b-1)</math>, each with probability <math>\frac{1}{3}</math>, independently of its previous moves. The probability that it will hit the coordinate axes at <math>(0,0)</math> is <math>\frac{m}{3^n}</math>, where <math>m</math> and <math>n</math> are positive integers. Find <math>m + n</math>.
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A moving particle starts at the point <math>(4,4)</math> and moves until it hits one of the coordinate axes for the first time. When the particle is at the point <math>(a,b)</math>, it moves at random to one of the points <math>(a-1,b)</math>, <math>(a,b-1)</math>, or <math>(a-1,b-1)</math>, each with probability <math>\frac{1}{3}</math>, independently of its previous moves. The probability that it will hit the coordinate axes at <math>(0,0)</math> is <math>\frac{m}{3^n}</math>, where <math>m</math> and <math>n</math> are positive integers such that <math>m</math> is not divisible by <math>3</math>. Find <math>m + n</math>.
  
 
==Solution 1==
 
==Solution 1==
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Obviously, the only way to reach (0,0) is to get to (1,1) and then have a <math>\frac{1}{3}</math> chance to get to (0,0). Let x denote a move left 1 unit, y denote a move down 1 unit, and z denote a move left and down one unit each. The possible cases for these moves are <math>(x,y,z)=(0,0,3),(1,1,2),(2,2,1)</math> and <math>(3,3,0)</math>. This gives a probability of <math>1 \cdot \frac{1}{27} + \frac{4!}{2!} \cdot \frac{1}{81} + \frac{5!}{2! \cdot 2!} \cdot \frac{1}{243} +\frac{6!}{3! \cdot 3!} \cdot \frac{1}{729}=\frac{245}{729}</math> to get to <math>(1,1)</math>. The probability of reaching <math>(0,0)</math> is <math>\frac{245}{3^7}</math>. This gives <math>245+7=\boxed{252}</math>.  
 
Obviously, the only way to reach (0,0) is to get to (1,1) and then have a <math>\frac{1}{3}</math> chance to get to (0,0). Let x denote a move left 1 unit, y denote a move down 1 unit, and z denote a move left and down one unit each. The possible cases for these moves are <math>(x,y,z)=(0,0,3),(1,1,2),(2,2,1)</math> and <math>(3,3,0)</math>. This gives a probability of <math>1 \cdot \frac{1}{27} + \frac{4!}{2!} \cdot \frac{1}{81} + \frac{5!}{2! \cdot 2!} \cdot \frac{1}{243} +\frac{6!}{3! \cdot 3!} \cdot \frac{1}{729}=\frac{245}{729}</math> to get to <math>(1,1)</math>. The probability of reaching <math>(0,0)</math> is <math>\frac{245}{3^7}</math>. This gives <math>245+7=\boxed{252}</math>.  
  
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==Solution 3==
  
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*The MAA should have specified that <math>m</math> is not divisible by <math>3</math> or the greatest common divisor of <math>m</math> and <math>3^{n}</math> was 1.
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Since the particle stops at one of the axes, we know that the particle most pass through <math>(1,1)</math>. Thus, it suffices to consider the probability our particle will reach <math>(1,1)</math>. Then the only ways to get to <math>(1,1)</math> from <math>(4,4)</math> are the following:
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(1) 3 moves diagonally   
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(2) 2 moves diagonally, 1 move left, 1 move down
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(3) 1 move diagonally, 2 moves left and 2 moves down.
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(4) 3 moves left, 3 moves down.
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The probability of (1) is <math>\frac{1}{3^3}</math>. The probability of (2) is <math>\frac{\frac{4!}{2!}}{3^4} = \frac{12}{3^4}</math>. The probability of (3) is <math>\frac{\frac{5!}{2!2!}}{3^5} = \frac{30}{3^5}</math>. The probability of (4) is <math>\frac{\frac{6!}{3!3!}}{3^6} = \frac{20}{3^6}</math>. Adding all of these together, we obtain a total probability of <math>\frac{245}{3^6}</math> that our particle will hit <math>(1,1)</math>. Trivially, there is a <math>\frac{1}{3}</math> chance our particle will hit <math>(0,0)</math> from <math>(1,1)</math>. So our final probability will be <math>\frac{245}{3^6} \cdot \frac{1}{3} = \frac{245}{3^7} \implies m = 245, n = 7 \implies \boxed{252}</math>
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~NotSoTrivial
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==Solution 4 (Official MAA)==
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All paths that first hit the axes at the origin must pass through the point <math>(1,1)</math>. There are <math>63</math> paths from the point <math>(4,4)</math> to the point <math>(1,1)</math>: <math>\tbinom{6}{3}=20</math> that take <math>3</math> steps left and <math>3</math> steps down, <math>\tbinom{5}{2\,2\,1}=30</math> that take <math>2</math> steps left, <math>2</math> steps down, and <math>1</math> diagonal step, <math>\tbinom{4}{1\,1\,2}=12</math> steps that take <math>1</math> step left, <math>1</math> steps down, and <math>2</math> diagonal steps, and <math>1</math> that takes <math>3</math> diagonal steps. The total probability of moving from <math>(4,4)</math> to <math>(1,1)</math> is therefore <cmath>\frac{1}{3^6}\cdot20+\frac{1}{3^5}\cdot30+\frac{1}{3^4}\cdot12+\frac{1}{3^3}\cdot1=\frac{245}{3^6}.</cmath> Multiplying by <math>\tfrac13</math> gives <math>\tfrac{245}{3^7},</math> the probability that the path first reaches the axes at the origin. The requested sum is <math>245+7=252</math>.
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==Video Solution #1(A nice visual explanation)==
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https://youtu.be/JQdad7APQG8?t=1340
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==Video Solution==
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Unique solution: https://youtu.be/I-8xZGhoDUY
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~Shreyas S
  
 
==See Also==
 
==See Also==
 
{{AIME box|year=2019|n=I|num-b=4|num-a=6}}
 
{{AIME box|year=2019|n=I|num-b=4|num-a=6}}
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[[Category:Intermediate Probability Problems]]
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 14:51, 25 February 2021

Problem

A moving particle starts at the point $(4,4)$ and moves until it hits one of the coordinate axes for the first time. When the particle is at the point $(a,b)$, it moves at random to one of the points $(a-1,b)$, $(a,b-1)$, or $(a-1,b-1)$, each with probability $\frac{1}{3}$, independently of its previous moves. The probability that it will hit the coordinate axes at $(0,0)$ is $\frac{m}{3^n}$, where $m$ and $n$ are positive integers such that $m$ is not divisible by $3$. Find $m + n$.

Solution 1

One could recursively compute the probabilities of reaching $(0,0)$ as the first axes point from any point $(x,y)$ as \[P(x,y) = \frac{1}{3} P(x-1,y) + \frac{1}{3} P(x,y-1) + \frac{1}{3} P(x-1,y-1)\] for $x,y \geq 1,$ and the base cases are $P(0,0) = 1, P(x,0) = P(y,0) = 0$ for any $x,y$ not equal to zero. We then recursively find $P(4,4) = \frac{245}{2187}$ so the answer is $245 + 7 = \boxed{252}$.


If this algebra seems intimidating, you can watch a nice pictorial explanation of this by On The Spot Stem. https://www.youtube.com/watch?v=XBRuy3_TM9w

Solution 2

Obviously, the only way to reach (0,0) is to get to (1,1) and then have a $\frac{1}{3}$ chance to get to (0,0). Let x denote a move left 1 unit, y denote a move down 1 unit, and z denote a move left and down one unit each. The possible cases for these moves are $(x,y,z)=(0,0,3),(1,1,2),(2,2,1)$ and $(3,3,0)$. This gives a probability of $1 \cdot \frac{1}{27} + \frac{4!}{2!} \cdot \frac{1}{81} + \frac{5!}{2! \cdot 2!} \cdot \frac{1}{243} +\frac{6!}{3! \cdot 3!} \cdot \frac{1}{729}=\frac{245}{729}$ to get to $(1,1)$. The probability of reaching $(0,0)$ is $\frac{245}{3^7}$. This gives $245+7=\boxed{252}$.

Solution 3

Since the particle stops at one of the axes, we know that the particle most pass through $(1,1)$. Thus, it suffices to consider the probability our particle will reach $(1,1)$. Then the only ways to get to $(1,1)$ from $(4,4)$ are the following:

(1) 3 moves diagonally

(2) 2 moves diagonally, 1 move left, 1 move down

(3) 1 move diagonally, 2 moves left and 2 moves down.

(4) 3 moves left, 3 moves down.

The probability of (1) is $\frac{1}{3^3}$. The probability of (2) is $\frac{\frac{4!}{2!}}{3^4} = \frac{12}{3^4}$. The probability of (3) is $\frac{\frac{5!}{2!2!}}{3^5} = \frac{30}{3^5}$. The probability of (4) is $\frac{\frac{6!}{3!3!}}{3^6} = \frac{20}{3^6}$. Adding all of these together, we obtain a total probability of $\frac{245}{3^6}$ that our particle will hit $(1,1)$. Trivially, there is a $\frac{1}{3}$ chance our particle will hit $(0,0)$ from $(1,1)$. So our final probability will be $\frac{245}{3^6} \cdot \frac{1}{3} = \frac{245}{3^7} \implies m = 245, n = 7 \implies \boxed{252}$

~NotSoTrivial

Solution 4 (Official MAA)

All paths that first hit the axes at the origin must pass through the point $(1,1)$. There are $63$ paths from the point $(4,4)$ to the point $(1,1)$: $\tbinom{6}{3}=20$ that take $3$ steps left and $3$ steps down, $\tbinom{5}{2\,2\,1}=30$ that take $2$ steps left, $2$ steps down, and $1$ diagonal step, $\tbinom{4}{1\,1\,2}=12$ steps that take $1$ step left, $1$ steps down, and $2$ diagonal steps, and $1$ that takes $3$ diagonal steps. The total probability of moving from $(4,4)$ to $(1,1)$ is therefore \[\frac{1}{3^6}\cdot20+\frac{1}{3^5}\cdot30+\frac{1}{3^4}\cdot12+\frac{1}{3^3}\cdot1=\frac{245}{3^6}.\] Multiplying by $\tfrac13$ gives $\tfrac{245}{3^7},$ the probability that the path first reaches the axes at the origin. The requested sum is $245+7=252$.

Video Solution #1(A nice visual explanation)

https://youtu.be/JQdad7APQG8?t=1340

Video Solution

Unique solution: https://youtu.be/I-8xZGhoDUY

~Shreyas S

See Also

2019 AIME I (Problems • Answer Key • Resources)
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