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.

First determine how many steps she takes. Since $t_1 < 1$, she never stops after one step. She stops after the second step exactly when $t_1 + t_2 > 1$; in the unit square of $(t_1, t_2)$ this is the half above the line $t_1 + t_2 = 1$, probability $\frac12$. Otherwise she takes the third step (and then stops regardless), also with probability $\frac12$.

The increments $x_n$ are independent of the times, so condition on the number of steps.

• Two steps: her position is $x_1 + x_2$, and $P(x_1 + x_2 > 1) = \frac12$ by the same half-square argument.
• Three steps: her position is $x_1 + x_2 + x_3$. The complement $x_1 + x_2 + x_3 \le 1$ is the tetrahedron cut from the unit cube by the plane $x + y + z = 1$, of volume $\frac{1}{3} \cdot \frac12 \cdot 1 = \frac16$. So $P(\text{sum} > 1) = \frac56$.

Combine:
$$P = \frac12 \cdot \frac12 + \frac12 \cdot \frac56 = \frac14 + \frac{5}{12} = \frac{8}{12} = \frac23.$$

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