How to Solve Multivariate Probability — Multivariate — min/max of independent lifetimes Questions on Exam P
Sample Practice Problem
ID: #H8G7T
Two independent event streams have exponential waiting times with means 2 and 3 days. Calculate the probability the second source produces the next event.
📖 Worked Solution & Strategy
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 \(\Pr(\text{source 2 first})=\frac{2}{5}\), which matches the correct option.