2020 AMC 10A Problem 13

Attempt Problem 13 of the 2020 AMC 10A below, then check your answer against the video solution and professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2020 AMC 10A solutions, or check the answer key.

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13.

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?

12\displaystyle \frac{1}{2}

58\displaystyle \frac{5}{8}

23\displaystyle \frac{2}{3}

34\displaystyle \frac{3}{4}

78\displaystyle \frac{7}{8}

Answer: B
Concepts:random walksystem of equationssymmetry
Difficulty rating: 1950
Video solution:
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Written 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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