# two people meeting, uniform distribution

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.

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: $$\int_0^{\frac14}\int_{0}^{x+\frac14} 1 \ dydx + \int_{\frac14}^{\frac34}\int_{x-\frac14}^{x+\frac14} 1 \ dydx + \int_{\frac34}^{1}\int_{x-\frac14}^{1} 1 \ dydx =\\ \left(\frac3{32}\right)+\left(\frac14\right)+\left(\frac3{32}\right)=\frac7{16}.$$