Relation between exponential random variables and their min

Question

If $$U=exponential(\lambda), \quad V = exponential(\mu)$$ and if they are independent, then $T = \min (U,V)$ is exponential $(\lambda + \mu)$.

How do we show that $P(T = U) = \lambda / (\lambda + \mu)$? Any hint would be much appreciated.

Answer 1

\begin{align} & \Pr(T=U) \\[10pt] = {} & \Pr(U<V) = \iint\limits_{u,v\,:\, u\,<\,v} e^{-\lambda u} e^{-\mu v} (\lambda\mu\,d(u,v)) \\[10pt] = {} & \int_0^\infty \left( e^{-\lambda u} \int_u^\infty e^{-\mu v} (\mu\,dv) \right) (\lambda \, du) \\[10pt] = {} & \frac\lambda{\lambda+\mu}. \end{align}