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

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A moving particle starts at the point <math>(4,4)</math> and moves until it hits one of the coordinate axes for the first time. When the particle is at the point <math>(a,b)</math>, it moves at random to one of the points <math>(a-1,b)</math>, <math>(a,b-1)</math>, or <math>(a-1,b-1)</math>, each with probability <math>\frac{1}{3}</math>, independently of its previous moves. The probability that it will hit the coordinate axes at <math>(0,0)</math> is <math>\frac{m}{3^n}</math>, where <math>m</math> and <math>n</math> are positive integers. Find <math>m + n</math>.
 
A moving particle starts at the point <math>(4,4)</math> and moves until it hits one of the coordinate axes for the first time. When the particle is at the point <math>(a,b)</math>, it moves at random to one of the points <math>(a-1,b)</math>, <math>(a,b-1)</math>, or <math>(a-1,b-1)</math>, each with probability <math>\frac{1}{3}</math>, independently of its previous moves. The probability that it will hit the coordinate axes at <math>(0,0)</math> is <math>\frac{m}{3^n}</math>, where <math>m</math> and <math>n</math> are positive integers. Find <math>m + n</math>.
  
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==Solution==
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jaskdfjskdjfksdfj
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A move from <math>(a,b)</math> to <math>(a,b-1)</math> is labeled as down (<math>D</math>), from <math>(a,b)</math> to <math>(a-1,b)</math> is labeled as left (<math>L</math>), and from <math>(a,b)</math> to <math>(a-1,b-1)</math> is labeled as slant (<math>S</math>). To arrive at <math>(0,0)</math> without arriving at an axis first, the particle must first go to <math>(1,1)</math> then do a slant move. The particle can arrive at <math>(1,1)</math> through any permutation of the following 4 different cases: <math>SSS</math>, <math>SSDL</math>, <math>SDLDL</math>, and <math>DLDLDL</math>.
 
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There is only <math>1</math> permutation of <math>SSS</math>. Including the last move, there are <math>4</math> possible moves, making the probability of this move <math>\frac{1}{3^4}</math>.
 
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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>.
 
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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!} = 20</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>.
 
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Adding these, the total probability is <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 = \boxed{252}</math>.
 
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Solution by Zaxter22
 
  
 
==Solution 2==
 
==Solution 2==

Revision as of 14:32, 7 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 $m + n$.

jaskdfjskdjfksdfj

Solution 2

Alternatively, one could recursively compute the probabilities of reaching $(0,0)$ as the first axes point from any point $(x,y)$ as \[P(x,y) = \frac{1}{3} P(x-1,y) + \frac{1}{3} P(x,y-1) + \frac{1}{3} P(x-1,y-1)\] 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

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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