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Two people are meeting for lunch. Each arrives between 12PM and 1PM. All pairs of arrival times are equally likely. The first to arrive will wait for 15 minutes. What is the probability that they meet. I would like to know how to find the integral formula for the joint probability distribution as I've used simulation to determine the answer to be .25

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    $\begingroup$ Let $X$ be the arrival time, in hours after 12:00, of Alice, and let $Y$ be the arrival time of Bob. By independence, the joint density of their arrival times is $1$ over the square with corners $(0,0)$, $(1,0)$, $(1,1)$, and $(0,1)$, and $0$ elsewhere. We want the probability that $|X-Y|\le 1/4$. You can do this by integration, or more simply by geometry, by identifying the probability as the area of a suitable regon. $\endgroup$ – André Nicolas Feb 16 '15 at 19:55
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You have the arrival times $t_1,t_2$ independently and uniformly distributed on $[12,13]$.

You want to compute $p(A)$, where $A= \{ (t_1,t_2) | |t_1-t_2| \le {1 \over 4} \}$.

If you draw a picture, it is clear that the answer is ${7 \over 16}$. If you wish to integrate, you need to compute $p(A) = \int_{12}^{13} \int_{12}^{13} 1_A((t_1,t_2)) d t_1 d t_2$.

It is easy to see that $A$ is symmetric, that is, if $(t_1,t_2) \in A$, then $(t_2,t_1) \in A$ and it is clear that the probability of both arriving at exactly the same time is zero.

Hence \begin{eqnarray} p(A) &=& 2\int_{t_1=12}^{13} \int_{t_2=12}^{t_1} 1_A((t_1,t_2)) d t_1 d t_2 \\ &=& 2 \left( \int_{t_1=12}^{12{1 \over 4}} \int_{t_2=12}^{t_1} 1_A((t_1,t_2)) d t_1 d t_2 + \int_{t_1=12{1 \over 4}}^{13} \int_{t_2=12}^{t_1} 1_A((t_1,t_2)) d t_1 d t_2 \right) \\ &=& 2 \left( \int_{t_1=12}^{12{1 \over 4}} \int_{t_2=12}^{t_1} d t_1 d t_2 + \int_{t_1=12{1 \over 4}}^{13} \int_{t_2=t_1-{1 \over 4}}^{t_1} d t_1 d t_2 \right) \\ &=& {7 \over 16} \end{eqnarray}

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