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

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There are <math>\frac{4!}{2!} = 12</math> permutations of <math>SSDL</math>, as the ordering of the two slants do not matter. There are <math>5</math> possible moves, making the probability of this move <math>\frac{12}{3^5}</math>.  
 
There are <math>\frac{4!}{2!} = 12</math> permutations of <math>SSDL</math>, as the ordering of the two slants do not matter. There are <math>5</math> possible moves, making the probability of this move <math>\frac{12}{3^5}</math>.  
  
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There are <math>\frac{5!}{2! \cdot 2!} = 30</math> permutations of <math>SDLDL</math>, as the ordering of the pairs of  do not matter. There are <math>6</math> possible moves, making the probability of this move <math>\frac{30}{3^6}</math>.  
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There are <math>\frac{5!}{2! \cdot 2!} = 30</math> permutations of <math>SDLDL</math>, as the ordering of the two downs and two lefts do not matter. There are <math>6</math> possible moves, making the probability of this move <math>\frac{30}{3^6}</math>.  
  
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There are <math>\frac{6!}{3! \cdot 3!} = 30</math> permutations of <math>DLDLDL</math>, as the ordering of the two slants do not matter. There are <math>7</math> possible moves, making the probability of this move <math>\frac{20}{3^7}</math>.  
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There are <math>\frac{6!}{3! \cdot 3!} = 30</math> permutations of <math>DLDLDL</math>, as the ordering of the three downs and three lefts do not matter. There are <math>7</math> possible moves, making the probability of this move <math>\frac{20}{3^7}</math>.  
  
 
Adding these, we get the total probability as <math>\frac{1}{3^4} + \frac{12}{3^5} + \frac{30}{3^6} + \frac{20}{3^7} = \frac{245}{3^7}</math>. Therefore, the answer is <math>245 + 7 = 252</math>.
 
Adding these, we get the total probability as <math>\frac{1}{3^4} + \frac{12}{3^5} + \frac{30}{3^6} + \frac{20}{3^7} = \frac{245}{3^7}</math>. Therefore, the answer is <math>245 + 7 = 252</math>.

Revision as of 22:52, 14 March 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

We label a move from $(a,b)$ to $(a,b-1)$ as down ($D$), from $(a,b)$ to $(a-1,b)$ as left ($L$), and from $(a,b)$ to $(a-1,b-1)$ 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 can be done 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, we get the total probability as $\frac{1}{3^4} + \frac{12}{3^5} + \frac{30}{3^6} + \frac{20}{3^7} = \frac{245}{3^7}$. Therefore, the answer is $245 + 7 = 252$.

Solution by Zaxter22

See Also

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

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IMO 1999

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