Problem:
Ant Amelia starts on the number line at and crawls in the following manner. For , Amelia chooses a time duration and an increment independently and uniformly at random from the interval . During the th step of the process, Amelia moves units in the positive direction, using up minutes. If the total elapsed time has exceeded minute during the th step, she stops at the end of that step; otherwise, she continues with the next step, taking at most steps in all. What is the probability that Amelia's position when she stops will be greater than ?
Answer Choices:
A.
B.
C.
D.
E.
Solution:
The conditions of the problem guarantee that Amelia stops after either 2 steps or 3 steps. The probability that she will stop after 2 steps is the probability that . This can be computed as the area of the triangular region in the square in the -coordinate plane that lies above the line , which is . If Amelia stops after 2 steps, the probability that her position is greater than 1 is, by the same reasoning with and , equal to .
Otherwise, Amelia stops after 3 steps, and this happens with probability . In this case the probability that her position is greater than 1 is equal to the probability that . This can be computed as the volume of the region in the cube in the coordinate space that lies above the plane . The complement of this region is a pyramid with isosceles right triangular base with leg length 1 and altitude 1 , so its volume is . It follows that the probability that Amelia's position if she stops after the third step will be greater than 1 is .
Combining both cases yields an answer of .
The problems on this page are the property of the MAA's American Mathematics Competitions