2022 AMC 10B Problem 23

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

The process looks tangled, but the times only decide how many increments get added, and the increments are independent of the times. That splits the problem into a stopping-rule probability and two standard geometric probabilities (area of half a square, volume of a corner tetrahedron). Whenever random quantities are independent, condition on the simpler one and the rest becomes routine.

• Thinking the third step is skipped or truncated when time runs out (it is completed), or coupling the step lengths to the times.
• Using $P(x_1 + x_2 + x_3 > 1) = \frac12$ or $\frac23$ instead of $\frac56$; the region $x + y + z \le 1$ is a tetrahedron of volume $\frac16$, not half the cube.