X and Y are waiting time between phone calls for company A and B respectively and they are independent from each other. X and Y are exponentially distributed with expected waiting time of 10 min and 8 min respectively. What is the probability that the next phone call from either X or Y is within 5 min? What is the expected waiting time until the next phone call from either one of them?
I am not really sure what probability I am asked to look for. I'm guessing its P((X<5)U(Y<5)). For the expected waiting time I have no idea how to find it. Is it just E(X)+E(Y)?
The next phone call is the minimum of $X$ and $Y$; write $Z=X\wedge Y$. The intuition behind this should be fairly obvious. Now, the distribution of the minimum of two independent exponentially distributed random variables is again exponential, with rate the sum of rates of the two. To see this, suppose $X\sim\mathsf{Exp}(\lambda)$ and $Y\sim\mathsf{Exp}(\mu)$, then for any $t>0$ we have \begin{align} \mathbb P(Z>t) &= \mathbb P(X>t, Y>t)\\ &= \mathbb P(X>t)\mathbb P(Y>t)\\ &= e^{-\lambda t}e^{-\mu t}\\ &= e^{-(\lambda+\mu)t}, \end{align} so that $Z\sim\mathsf{Exp}(\lambda+\mu)$. Plugging in $\lambda = \frac1{10}$, $\mu=\frac18$, and $t=5$, we have $$ \mathbb P(Z\leqslant 5) = 1 - e^{-\left(\frac1{10}+\frac18 \right)\cdot5} = 1-e^{-\frac98}. $$ The expected value of $Z$ is simply $$\left(\frac1{10}+\frac18 \right)^{-1}=\frac{40}9. $$