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We are given that no 1 person will wait for more than 15 minutes for the other person after arriving. I want the probability that they meet during that hour. They don't visit the meeting point again. My attempt: Let $X, Y$ denote their arrival time respectively. Then, $X \sim U(0,60)$ and $Y \sim U(0,60)$. We want $P(|X-Y|<15)$ $\large{P(|X-Y|<15)=P(X

I have seen the geometric method to solve this problem. My answer doesn't agree with that approach. Please tell me where and why, I'm wrong? Thanks

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    Your $P(X – 2017-02-11
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    @HagenvonEitzen Yes. Isnt the restriction by 60 needed? – 2017-02-11

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You aren't calculating the probabilities correctly. For instance, $$P(X < Y+15) = \int_0^{45}dy\int_{0}^{y+15}dx \frac{1}{60^2} + \int_{45}^{60}dy\int_0^{60}dx \frac{1}{60^2} $$

(Also you don't need to take integrals. Just sketch a square and sketch the region and find its area.)

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    Thank you so much. You're right, I forgot to include the part where X arrives between 45 and 60 minutes. I am learning double integrals so that's why I decided to see if I could do this problem without relying on geometry. – 2017-02-11