# 2019 AIME I Problems/Problem 5

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

## 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 $(1,1)$. Denote a move to the left, down, diagonally, as X, Y, Z, respectively. Then the only ways to get to $(1,1)$ from $(4,4)$ are the following: $(1) 0X 0Y 3Z$ $(2) 1X 1Y 2Z$ $(3) 2X 2Y 1Z$ $(4) 3X 3Y 0Z$

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

## Video Solution

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

~Shreyas S

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