2023 AMC 10B Problem 19

The square looks the same from all four sides, so the probability of exiting is the same for each direction; compute it for north.

Hopping north only changes the $y$-coordinate, which is uniform on $[0, 6]$ and independent of the hop distance $d$, uniform on $[0, 1]$. She lands outside exactly when $y + d > 6$, i.e. $y > 6 - d$.

Picture the pair $(d, y)$ as a uniformly random point in the rectangle $[0, 1] \times [0, 6]$, of area $6$. The exit condition $y > 6 - d$ cuts off the triangle with vertices $(0, 6)$, $(1, 6)$, $(1, 5)$, of area $\frac{1}{2}\cdot 1 \cdot 1 = \frac{1}{2}$. So
$$P(\text{exit}) = \frac{1/2}{6} = \frac{1}{12}.$$
Since every direction gives $\frac{1}{12}$, the overall probability is $\frac{1}{12}$.

The answer is $\boxed{\textbf{(B)}\ \frac{1}{12}}$.

Independent uniform choices make a uniform point in a product region, so probabilities become area ratios. Symmetry removes the direction entirely, and the irrelevant coordinate ($x$ for a north hop) drops out, leaving a two-variable picture whose bad region is a small triangle.

Condition on the distance: for a given $d$, the frog exits northward iff $y > 6 - d$, probability $\frac{d}{6}$. Averaging over $d$ uniform on $[0,1]$ gives $\frac{E[d]}{6} = \frac{1/2}{6} = \frac{1}{12}$.

• Assuming she always hops the full distance 1, which gives 1/6 instead of averaging over the uniform distance.
• Multiplying the $\frac{1}{12}$ by $\frac{1}{4}$ for the choice of direction; the four cases are averaged, not intersected.