2023 AMC 10B Problem 19

Question

Sonya the frog chooses a point uniformly at random lying within the square

[0, 6] \times [0, 6]

in the coordinate plane and hops to that point. She then chooses a distance uniformly at random from

[0, 1]

and a direction uniformly at random from

\{\text{north}, \text{south}, \text{east}, \text{west}\}.

All her choices are independent. She now hops the distance in the chosen direction. What is the probability that she lands outside the square?

\dfrac{1}{6}

\dfrac{1}{12}

\dfrac{1}{4}

\dfrac{1}{10}

\dfrac{1}{9}

By symmetry, condition on one direction, say east; only the coordinate toward that edge matters

For a fixed hop distance

d,

the landing is outside with probability

\frac{d}{6};

average over

d

The four directions behave the same by symmetry, so say she hops east. She lands outside exactly when her

x

-coordinate plus the hop distance

d

tops

6.

Fix

d.

Her

x

-coordinate is uniform on

[0, 6],

so it beats

6 - d

with probability

\frac{d}{6}.

Now average over

d

uniform on

[0, 1]:

\frac{1}{6} \cdot \frac{1}{2} = \frac{1}{12}.

Thus, B is the correct answer.

Problem 19 in other years