I have following problem to solve :
There are $N$ customers waiting in the checkout line, where $N$ has Poisson distribution with parameter $3$. If a service begins when $m$ customers are in the line, that service time has exponential distribution with parameter $\frac{1}{2m}$ . Find the probability that a customer just approaching the checkout will be served in less than a minute.
What I did :
If X is a time serving, then
$P(X<1)=F(1)=1-e^{-\frac{1}{2N}}$
according to CDF for exponential distribution. But now I don't know what to do, how to use fact that $N$ has Poisson distribution. I'm puzzled. How to connect these two things?
Use conditional probability. Let $$N\sim \text{Poisson}(3)\\ X|N\sim \text{Exponential}(1/(2N))$$
Then from total probability $$P(X<1)=\sum_{N=0}^\infty F(1|N)P(N)$$
$$\begin{split}\sum_{N=0}^\infty (1-e^{-\frac 1{2N}})\frac{e^{-3}3^N}{N!}&=\sum_{N=0}^\infty \frac{e^{-3}3^N}{N!}-e^{-3}\sum_{N=0}^\infty \frac{e^{-\frac 1{2N}}3^N}{N!}\\ &=1-???\end{split}$$