I have a theorem in my notes that says that
if X is an exponential RV, $X$ ~ $exp(\lambda_{1}+\lambda_{2})$, $$Pr(x_{1} < x_{2}) = \frac{\lambda_{1}}{\lambda_{1}+\lambda_{2}}$$
I was too far away from the board while trying to copy this and I think the proof I copied has several errors. I haven't been able to google it (I'm not sure of the name), so could someone post it for me please? I think it has something to do with the total probability theorem.
Feel free to edit the title/post if there's a better way to describe this.
Actually, it is that if $X_1\sim\mathcal{Exp}(\lambda_1), X_2\sim\mathcal{Exp}(\lambda_2)$, and $X_1, X_2$ are independent, then $\mathsf P(X_1\leq X_2) = \frac{\lambda_1}{\lambda_1+\lambda_2}$
Via integration:
$$\begin{align} \mathsf P(X_1\leq X_2) = & ~ \int_0^\infty\int_0^t f_{X_1}(s)~f_{X_2}(t)\operatorname d s\operatorname d t \\[1ex] & = ~ \lambda_1\lambda_2 \int_0^\infty \mathsf e^{-\lambda_2 t}\int_0^t\mathsf e^{-\lambda_1 s}\operatorname d s\operatorname d t \\[1ex] & = ~ \lambda_2 \int_0^\infty \mathsf e^{-\lambda_2 t}(1-e^{-\lambda_1 t})\operatorname d t \\[1ex] & = ~ \lambda_2 \int_0^\infty \mathsf e^{-\lambda_2 t}-e^{-(\lambda_1+\lambda_2) t}\operatorname d t \\[1ex] & = ~ 1 - \frac{\lambda_2}{\lambda_1+\lambda_2} \\[1ex] & = ~ \frac{\lambda_1}{\lambda_1+\lambda_2} \end{align}$$