2020 AMC 12A Problem 11

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?

\dfrac{1}{2}

\dfrac{5}{8}

\dfrac{2}{3}

\dfrac{3}{4}

\dfrac{7}{8}

Answer: B

Concepts:random walk system of equations symmetry

Difficulty rating: 1630

Small hint:

Let

P(x, y)

be the probability of ending on a vertical side; it equals the average of

P

at the four neighbors

Big hint:

Use symmetry:

P(2, 2) = \tfrac12,

and the interior points reduce to a small system of equations

Solution:

Let

P(x, y)

be the probability of ending on a vertical side. On a vertical side

P = 1,

on a horizontal side

P = 0,

and at an interior point

P

is the average of its four neighbors.

By left-right symmetry

P(2, 2) = \tfrac12.

Let

a = P(1, 2),

b = P(1, 1) = P(1, 3),

and

c = P(2, 1) = P(2, 3).

Then

a = \tfrac14\left(1 + \tfrac12 + 2b\right),

\;b = \tfrac14(1 + c + a),

and

c = \tfrac14\left(2b + \tfrac12\right).

Substituting gives

b = \tfrac12,

hence

a = \tfrac38 + \tfrac12 b = \tfrac58.

Thus, B is the correct answer.