How to Solve Multivariate Probability — Multivariate — min/max of independent lifetimes Questions on Exam P

The next event time is the minimum of two independent exponential waiting times. Competing-risk probabilities are proportional to source rates, and the minimum is exponential with the sum of those rates. Let \(T_{1}\sim\operatorname{Exp}(\lambda_{1}=1/2)\) and \(T_{2}\sim\operatorname{Exp}(\lambda_{2}=1/3)\) be independent waiting times. The minimum waiting time \(M=\min(T_{1}, T_{2})\) is exponential with total rate \(\lambda=\lambda_{1}+\lambda_{2}=\frac{5}{6}\). The competing-risk probability that source 2 occurs first is \(\Pr(T_{2}<T_{1})=\dfrac{\lambda_{2}}{\lambda_{1}+\lambda_{2}}=\dfrac{1/3}{1/2+1/3}=\frac{2}{5}\). Thus \(\Pr(\text{source 2 first})=\frac{2}{5}\).