Question

Ant Amelia starts on the number line at $0$ and crawls in the following manner. For $n = 1, 2, 3,$ Amelia chooses a time duration $t_n$ and an increment $x_n$ independently and uniformly at random from the interval $(0, 1).$ During the $n$ th step of the process, Amelia moves $x_n$ units in the positive direction, using up $t_n$ minutes. If the total elapsed time has exceeded $1$ minute during the $n$ th step, she stops at the end of that step; otherwise, she continues with the next step, taking at most $3$ steps in all. What is the probability that Amelia's position when she stops will be greater than $1?$

$\dfrac13$

$\dfrac12$

$\dfrac23$

$\dfrac34$

$\dfrac56$

Solution:

Because each $t_n \lt 1,$ Amelia always completes at least two steps. She stops after exactly two steps when $t_1 + t_2 \gt 1,$ which happens with probability $\tfrac12;$ otherwise she takes all three steps.

The increments are independent of the times. If she takes two steps, her position is $x_1 + x_2,$ and $P(x_1 + x_2 \gt 1) = \tfrac12.$ If she takes three, her position is $x_1 + x_2 + x_3,$ and $P(x_1 + x_2 + x_3 \gt 1) = 1 - \tfrac16 = \tfrac56.$

The answer is $\tfrac12 \cdot \tfrac12 + \tfrac12 \cdot \tfrac56 = \tfrac14 + \tfrac{5}{12} = \tfrac23.$

Thus, the correct answer is C.

2022 AMC 12B Problem 22

Solution:

Problem 22 in Other Years