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

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The 2019 AIME I takes place on March 13, 2019..
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==Problem 5==
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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>.
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==Solution 1==
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One could recursively compute the probabilities of reaching <math>(0,0)</math> as the first axes point from any point <math>(x,y)</math> as <cmath>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)</cmath> for <math>x,y \geq 1,</math> and the base cases are
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<math>P(0,0) = 1, P(x,0) = P(y,0) = 0</math> for any <math>x,y</math> not equal to zero.
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We then recursively find <math>P(4,4) = \frac{245}{2187}</math> so the answer is <math>245 + 7 = \boxed{252}</math>.
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If this algebra seems intimidating, you can watch a nice pictorial explanation of this by On The Spot Stem.
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https://www.youtube.com/watch?v=XBRuy3_TM9w
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==Solution 2==
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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>.  
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==Video Solution==
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Unique solution: https://youtu.be/I-8xZGhoDUY
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~Shreyas S
  
==Problem 5==
 
==Solution==
 
 
==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 14:37, 16 October 2020

Problem 5

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 $245+7=\boxed{252}$.

Video Solution

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

~Shreyas S

See Also

2019 AIME I (ProblemsAnswer KeyResources)
Preceded by
Problem 4
Followed by
Problem 6
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions

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