2022 AMC 12B Problem 22

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?

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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.