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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.

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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

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.

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