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. Find $m + n$ .

Solution

A move from $(a,b)$ to $(a,b-1)$ is labeled as down ( $D$ ), from $(a,b)$ to $(a-1,b)$ is labeled as left ( $L$ ), and from $(a,b)$ to $(a-1,b-1)$ is labeled as slant ( $S$ ). To arrive at $(0,0)$ without arriving at an axis first, the particle must first go to $(1,1)$ then do a slant move. The particle can arrive at $(1,1)$ through any permutation of the following 4 different cases: $SSS$ , $SSDL$ , $SDLDL$ , and $DLDLDL$ .

There is only $1$ permutation of $SSS$ . Including the last move, there are $4$ possible moves, making the probability of this move $\frac{1}{3^4}$ .

There are $\frac{4!}{2!} = 12$ permutations of $SSDL$ , as the ordering of the two slants do not matter. There are $5$ possible moves, making the probability of this move $\frac{12}{3^5}$ .

There are $\frac{5!}{2! \cdot 2!} = 30$ permutations of $SDLDL$ , as the ordering of the two downs and two lefts do not matter. There are $6$ possible moves, making the probability of this move $\frac{30}{3^6}$ .

There are $\frac{6!}{3! \cdot 3!} = 30$ permutations of $DLDLDL$ , as the ordering of the three downs and three lefts do not matter. There are $7$ possible moves, making the probability of this move $\frac{20}{3^7}$ .

Adding these, the total probability is $\frac{1}{3^4} + \frac{12}{3^5} + \frac{30}{3^6} + \frac{20}{3^7} = \frac{245}{3^7}$ . Therefore, the answer is $245 + 7 = \boxed{252}$ .

Solution by Zaxter22

Solution 2

Alternatively, one could recursively compute the probabilities of reaching $(0,0)$ as the first axes point from any point $(x,y)$ as $P(x,y) = \frac{1}{3} P(x-1,y) + \frac{1}{3} P(x,y) + \frac{1}{3} P(x-1,y-1)$ for $x,y \geq 1,$ and the base cases are $P(0,0) = 1, P(x,0) = P(y,0)$ for any $x,y.$ We then recursively find $P(4,4) = \frac{245}{2187}$ so the answer is $245 + 7 = \boxed{252}$ .

2019 AIME I (Problems • Answer Key • Resources)
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

Page

Toolbox

Search

2019 AIME I Problems/Problem 5

Contents

Problem 5

Solution

Solution 2

See Also