2022 AMC 10B Problem 23

The times and the step lengths are chosen separately, so split the problem: first find how many steps she takes, then where she ends up.

1. She never stops after one step. Each $t_n$ is less than $1$, so after one step less than $1$ minute has passed.
2. Two steps or three? She stops after step 2 exactly when $t_1 + t_2 > 1$. Picture $(t_1, t_2)$ as a random point in a $1$-by-$1$ square. The line $t_1 + t_2 = 1$ cuts the square into two equal triangles, so the chance is $\frac12$. Otherwise (chance $\frac12$) she takes step 3, and then she always stops, since at most $3$ steps are allowed.
3. After two steps. Her position is $x_1 + x_2$. By the same square picture, the chance that $x_1 + x_2 > 1$ is $\frac12$.
4. After three steps. Her position is $x_1 + x_2 + x_3$, a random point in a $1$-by-$1$-by-$1$ cube. The points where the sum is at most $1$ form a corner pyramid with corners $(0,0,0)$, $(1,0,0)$, $(0,1,0)$, $(0,0,1)$. Its volume is $\frac13 \times \text{base} \times \text{height} = \frac13 \times \frac12 \times 1 = \frac16$. So the chance that the sum is more than $1$ is $1 - \frac16 = \frac56$.
5. Combine. Because the lengths do not depend on the times,
$$\frac12 \times \frac12 + \frac12 \times \frac56 = \frac14 + \frac{5}{12} = \frac{3}{12} + \frac{5}{12} = \frac{8}{12} = \frac23.$$

The answer is $\boxed{\textbf{(C)}\ \frac{2}{3}}$.

The story looks tangled, but the times only decide how many step lengths get added, and the lengths are chosen independently of the times. So the problem splits into a "how many steps" chance and two standard picture problems: half of a square, and a cube with its corner pyramid removed. When random choices are independent, sort out the simpler one first and treat each outcome as its own case.