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This problem comes from an old math contest which has the answer, but no solution.

Two friends get home at 2:30 P.M. They both go to bed 10 P.M. If both friends will go to a park twice for 15 minutes, what is the probability that they will be able to see each other at any random time between the two time intervals? Assume that the travel to and from the park takes no time.

The standard method to determine the probability of meeting if they go to the park once yields $\frac{30^2-29^2}{30^2} = \frac{59}{900}$ (by considering the area of $|y-x| \leq 1$ within the square $0 \leq x \leq 30, 0 \leq y \leq 30$). How would you generalize this geometric argument to handle going to the park twice?

The listed answer is $\frac{29}{225}$. I noticed that this is the same as $\frac{30^2-28^2}{30^2} = \frac{15^2-14^2}{15^2}$, which is the probability that they see each other assuming they go to the park just once, but staying for 30 minutes (ie like the first situation, but going back-to-back). It seems strange to me that the probability would work out to this; it seems like this is a lower bound on the probability, since they could also meet within a 15-minute interval, and go another time completely removed from this time interval. Is the listed answer correct, and why? Or is there a problem with the answer?

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  • $\begingroup$ Notice that the problem states "at any random time". This means, "what is the probability that they are meeting at one particular random time", and not "what is the probability of meeting at all" which is what you seem to be trying to solve. $\endgroup$ – Bobson Dugnutt Jan 22 '16 at 17:00
  • $\begingroup$ @ringo - This is what I would like as well - I'm not convinced that the listed answer is correct. $\endgroup$ – user291130 Jan 24 '16 at 5:08
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The problem might be simplified by calculating the probability that the two friends do not meet at all, then subtracting this from $1.$

Notice that if the two friends do not meet at all, then we just have to find four distinct $15-$minute time intervals in the $450$ minute interval (from 2:30PM to 10:00PM). Notice that for each set of $4$ distinct intervals, there are $\dbinom{4}{2} = 6$ ways to distribute them between the two friends.

Can you take it from here?

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