2020 AMC 10A Problem 13

Question

A frog sitting at the point

(1, 2)

begins a sequence of jumps, where each jump is parallel to one of the coordinate axes and has length

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, 4),

(4, 4),

and

(4, 0).

What is the probability that the sequence of jumps ends on a vertical side of the square?

\displaystyle \frac{1}{2}

\displaystyle \frac{5}{8}

\displaystyle \frac{2}{3}

\displaystyle \frac{3}{4}

\displaystyle \frac{7}{8}

Answer: B

Concepts:random walk system of equations symmetry

Difficulty rating: 1950

Small hint:

Let each interior lattice point store the probability of hitting a vertical side first

Big hint:

Use symmetry to reduce the random-walk equations to four unknowns

Video solution:

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Written solution:

Let

p(x,y)

be the probability of eventually hitting a vertical side first from point

(x,y)

. By symmetry, set

a=p(1,1)=p(1,3)

,

b=p(2,1)=p(2,3)

,

c=p(1,2)

, and

d=p(2,2)

.

The averaging equations are

a=\dfrac{1+b+c}{4}

,

b=\dfrac{2a+d}{4}

,

c=\dfrac{1+2a+d}{4}

, and

d=\dfrac{b+c}{2}

. Solving gives

c=\dfrac58

, which is the desired probability from

(1,2)

. Thus, B is the correct answer.