2022 AMC 10B Problem 23

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?

\dfrac 13

\dfrac 12

\dfrac 23

\dfrac 34

\dfrac 56

Answer: C

Concepts:geometric probability independent events casework

Difficulty rating: 2150

Small hint:

Use geometric probabilities for sums of two or three independent numbers in

(0,1)

Big hint:

The stopping time depends only on the variables

t_1,

t_2,

and

t_3

Solution:

The stopping time depends only on the time variables, while the final position depends only on the distance variables, so the corresponding probabilities multiply.

For two independent numbers in

(0,1),

the probability that their sum is less than

1

is the area of a right triangle, namely

\frac12.

For three independent numbers in

(0,1),

the probability that their sum is less than

1

is the volume of a tetrahedron with side intercepts

1,

namely

\frac16.

If

t_1+t_2>1,

Amelia stops after two steps. This has probability

\frac12,

and independently

x_1+x_2>1

has probability

\frac12,

contributing

\frac14.

If

t_1+t_2<1,

Amelia takes the third step. This has probability

\frac12,