Sign up ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Working on this:

Alice and Bob agree to meet in the Copper Kettle after their Saturday lectures. They arrive at times that are independent and uniformly distributed between 12:00 and 13:00. Each is prepared to wait s minutes before leaving. Find a minimal s such that the probability that they meet is at least 25%.

I honestly can't figure out how to approach this one... It seems super simple, but I just can't wrap my head around it right now. Any advice on how to start?

Thank you.

share|cite|improve this question
This type of question has been asked and solved in great detail repeatedly on this forum. Look around a bit. –  Dilip Sarwate May 1 '12 at 2:59
@gfppaste: I hope you locate it, for a picture really helps, and someone will have drawn one. Let's measure time of wait in hours, can go to minutes later. Let $X$ be Alice's arrival time, $Y$ Bob's. We want $P(|X-Y|\le w)\ge 0.25$. The joint density of $X$ and $Y$ is $1$ on the square with corners $(0,0)$, $(1,0)$, $(1,1)$, $(0,1)$. what is the area of the region $|x-y|\le w$? Hint: Draw the line $x-y=w$ and $x-y=-w$. –  André Nicolas May 1 '12 at 3:41

1 Answer 1

We can define Alice's arrival time by a and Bob's arrival time by b, with both a and b between uniformly distributed between 0 and 1. We can therefore define a set $C = \{(a,b): |a-b| \le \frac{s}{60}\}$. You should be able to graph this and see that it is a diagonal strip around the graph of y=x. The area outside of this strip is $(1-\frac{s}{60})^2$, so the area inside the strip is $1-(1-\frac{s}{60})^2$. We want this to be at least 0.25. From here, it is relatively straightforward to see what the minimum value of s should be.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.