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2020 AMC 10A Problem 13

Problem 13 of 25IntermediateAlgebraProbability & StatisticsProblem-Solving Techniques

A frog sitting at the point (1,2)(1, 2) begins a sequence of jumps, where each jump is parallel to one of the coordinate axes and has length 1,1, and the direction of each jump (up, down, right, or left) is chosen independently at random. The sequence ends when the frog reaches a side of the square with vertices (0,0),(0, 0), (0,4),(0, 4), (4,4),(4, 4), and (4,0).(4, 0). What is the probability that the sequence of jumps ends on a vertical side of the square?

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Solution

Let p(x,y)p(x,y) be the probability of eventually hitting a vertical side first from point (x,y)(x,y). By symmetry, set a=p(1,1)=p(1,3)a=p(1,1)=p(1,3), b=p(2,1)=p(2,3)b=p(2,1)=p(2,3), c=p(1,2)c=p(1,2), and d=p(2,2)d=p(2,2). The averaging equations are a=1+b+c4a=\dfrac{1+b+c}{4}, b=2a+d4b=\dfrac{2a+d}{4}, c=1+2a+d4c=\dfrac{1+2a+d}{4}, and d=b+c2d=\dfrac{b+c}{2}. Solving gives c=58c=\dfrac58, which is the desired probability from (1,2)(1,2). Thus, B is the correct answer.
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Tagged: random walk · system of equations · symmetry

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