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

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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 one.
 
<math>P(0,0) = 1, P(x,0) = P(y,0) = 0</math> for any <math>x,y</math> not equal to one.
 
We then recursively find <math>P(4,4) = \frac{245}{2187}</math> so the answer is <math>245 + 7 = \boxed{252}</math>.
 
We then recursively find <math>P(4,4) = \frac{245}{2187}</math> so the answer is <math>245 + 7 = \boxed{252}</math>.
 +
 +
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
  
 
==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 00:39, 4 October 2019

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 $\frac{6!}{3! \cdot 3!} = 20$ 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 $P(0,0) = 1, P(x,0) = P(y,0) = 0$ 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

See Also

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

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