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Two people, A and B meet in a bar between 6 pm and 7 pm. They arrive independently of each other, evenly distributed between 6 pm and 7 pm. The first person who arrived, do not wait more than 15 minutes. What is the probability that they meet?

I have a geometric argument correction for this exercice but I would like to try it with formula argument.

Let $T_A,T_B$ uniform distribution from 0 to 60.

$P(T_A-T_B<15)=\int_{-\infty}^{\infty}f_{T_B}(x)f_{-T_B}(15-x)dx$ by a proposition.

But I don't know how to write the next step, I would like to write $=\int_{-\infty}^{\infty}\frac{\mathbb{1_{[0,60]}}}{60}\frac{\mathbb{1_{[0,45]}}}{45}dx$ but I feel that it is wrong.

Waspe
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  • It is more handsome to go for $P(T_A<T_B<T_A+15)+P(T_B<T_A<T_B+15)$. By symmetry the probabilities are the same so it is enough to find $P(T_A<T_B<T_A+15)$. Use conditional distribution. – drhab Dec 15 '18 at 10:28
  • For a graphical approach see https://math.stackexchange.com/questions/103015/chance-of-meeting-in-a-bar and the posts linked to it. For a couple with a more analytical approach see https://math.stackexchange.com/questions/1275612/coffee-shop-meeting and https://math.stackexchange.com/questions/1168454/two-people-meeting-expected-time-of-waiting – Henry Dec 15 '18 at 10:29

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It is simpler to express time in hours rather minutes. The joint PDF is $f_{X,Y}(x,y)=1$. Now consider the three cases: $$\begin{cases}0\le x\le \frac14 \\ 0\le y\le x+\frac14\end{cases} \ \ \text{or} \ \ \begin{cases}\frac14\le x\le \frac34 \\ x-\frac14\le y\le x+\frac14\end{cases} \ \ \text{or} \ \ \begin{cases}\frac34\le x\le 1 \\ x-\frac14\le y\le 1\end{cases}$$ Integate on each domain: $$\int_0^{\frac14}\int_{0}^{x+\frac14} 1 \ dydx + \int_{\frac14}^{\frac34}\int_{x-\frac14}^{x+\frac14} 1 \ dydx + \int_{\frac34}^{1}\int_{x-\frac14}^{1} 1 \ dydx =\\ \left(\frac3{32}\right)+\left(\frac14\right)+\left(\frac3{32}\right)=\frac7{16}.$$

farruhota
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