In this lecture video, the professor describes a situation where the arrival of a phone call is a Poisson process with rate $\lambda$. Each call duration is exponentially distributed with parameter $\mu$. Next, assume that time is be split into infinitesimal small time slots with a duration $\delta$. Then, at any small duration $\delta$ the probability of a call ending is $\mu \delta$. Can someone please explain why this is the case? I thought to find the probability of departure at any given time slot would be:
$$P(x \leq X \leq x + \delta) = P(X \leq x + \delta) - P(X \leq x) = 1 - e^{-\lambda(x + \delta)} - (1 - e^{-\lambda x}) = e^{-\lambda x} - e^{-\lambda x}e^{-\lambda \delta} = e^{-\lambda x}(1-e^{-\lambda \delta})$$