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

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==Solution 2==
 
==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>.
+
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>.
 +
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.
  
 
==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}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 17:36, 10 March 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. Find $m + n$.

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Solution

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 one. 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 $245+7=\boxed{252}$. The MAA should have specified that $m$ is not divisible by $3$ or the greatest common divisor of $m$ and $3^{n}$ was 1.

See Also

2019 AIME I (Problems • Answer Key • Resources)
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Problem 4 		Followed by
Problem 6
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All AIME Problems and Solutions

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