2023 AMC 10B Problem 19

The square looks the same from all four sides, so the chance of hopping out is the same for every direction. Work out the chance for a hop north; that is the answer.

1. What matters for a north hop. Only her height $y$ (a random number from $0$ to $6$) and the hop length $d$ (a random number from $0$ to $1$). Her left-right position does not matter.
2. When does she leave? She starts at height $y$ and ends at height $y + d$. She is outside when
$$y + d > 6.$$
Since $d$ is at most $1$, this can only happen when she starts within $1$ of the top edge.
3. Draw a picture of all the possibilities. Plot $d$ across (from $0$ to $1$) and $y$ up (from $0$ to $6$). Every pair $(d, y)$ is equally likely, so the possibilities fill a $1 \times 6$ rectangle of area $6$.
4. Shade the "outside" pairs. The line $y + d = 6$ goes from $(0, 6)$ to $(1, 5)$. The pairs with $y + d > 6$ form the triangle with corners $(0, 6)$, $(1, 6)$, $(1, 5)$. Its legs are $1$ and $1$, so its area is $\frac{1}{2} \cdot 1 \cdot 1 = \frac{1}{2}$.
5. Probability. Shaded area over total area:
$$\frac{1/2}{6} = \frac{1}{12}.$$
6. Every direction is the same, so the overall probability is also $\frac{1}{12}$.

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

When two random numbers are each chosen evenly from a range, every point of the rectangle they make is equally likely, so a probability becomes (shaded area) divided by (total area). Symmetry removed the direction, and the left-right position did not matter, which left a simple two-number picture.

Fix the hop length first. If she hops a distance $d$ north, she leaves exactly when she starts in the top strip of height $d$, which has probability $\frac{d}{6}$. For example, a hop of $1$ gives $\frac{1}{6}$ and a hop of $\frac{1}{2}$ gives $\frac{1}{12}$. The chance $\frac{d}{6}$ grows steadily with $d$, and $d$ is spread evenly from $0$ to $1$, so the overall chance is the value at the average hop $d = \frac{1}{2}$, which is $\frac{1}{12}$.