2019 AIME I Problems/Problem 5
A moving particle starts at the point and moves until it hits one of the coordinate axes for the first time. When the particle is at the point , it moves at random to one of the points , , or , each with probability , independently of its previous moves. The probability that it will hit the coordinate axes at is , where and are positive integers such that is not divisible by . Find .
One could recursively compute the probabilities of reaching as the first axes point from any point as for and the base cases are for any not equal to zero. We then recursively find so the answer is .
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
Obviously, the only way to reach (0,0) is to get to (1,1) and then have a 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 and . This gives a probability of to get to . The probability of reaching is . This gives .
Since the particle stops at one of the axes, we know that the particle most pass through . Thus, it suffices to consider the probability our particle will reach . Denote a move to the left, down, diagonally, as X,Y,Z, respectively. Then the only ways to get to from are the following:
$(1) 0X 0Y 3Z \$ (Error compiling LaTeX. ! Missing $ inserted.)(2) 1X 1Y 2Z \$ $(3) 2X 2Y 1Z \$ (Error compiling LaTeX. ! Missing $ inserted.)(4) 3X 3Y 0Z \$
The probability of (1) is . The probability of (2) is . The probability of (3) is . The probability of (4) is . Adding all of these together, we obtain a total probability of that our particle will hit . Trivially, there is a chance our particle will hit from . So our final probability will be
Unique solution: https://youtu.be/I-8xZGhoDUY
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