The question I am trying to answer is along the lines of a music store has on average 8 customers every hour ($N$ ~ Poisson(8)) and the probability of them buying a CD is 0.7. If the $i$th customer searches through $c_i$ cd's before they find one to purchase ($c_i$ ~ Geo(0.7)), what is T (the total number of cd's browsed through in an hour by N customers, excluding the purchased ones) in terms of $c_i$ and N? What is $P(T = 0)$?
So far I have identified the two distributions and since I know that each person will look through $c_i$ books before a success, I'm assuming:
$$ T = \sum_{i=1}^N 0.7(1 - 0.7)^{c_i}$$
And in order to find $P(T = 0)$, I know that T will only equal zero in the case where no customers enter the store ($i = 0$), or where N customers purchase the first CD they look at ($c_i = 0$).
$$ P(T = 0) = P(N = 0) + P(T = 0 | N = n)P(N = n) $$ $$ P(T = 0) = e^{-8} + \sum_{n=0}^{\infty} 0.7\frac{8^ne^{-8}}{n!} $$ $$ P(T = 0) = e^{-8} + 0.7$$ Am I correct? My knowledge of using/combining two distributions is non-existent so I've tried my best.
Let $N\sim Po(\lambda)$ be the number of consumers in an hour.
$P(C=0)$ is the probability that all consumers (if any) who enter the store buy the first CD they see. For each of them, they buy the first CD with probability $p=0.7$. The probability of all $N$ people buying the first CD they see is $p^N$. So, $$ P(C=0)=\sum_{k=0}^\infty e^{-\lambda}\frac{\lambda^k}{k!}\cdot p^k\\ =e^{-\lambda} \sum_{k=0}^\infty \frac{(\lambda p)^k}{k!} =e^{-\lambda+\lambda p}=e^{-\lambda(1-p)}\\ =e^{-8\times 0.3}=0.0907 \ldots $$ You are "almost" correct but you've made two mistakes. First you need not count $e^{-8}$ seperately since it is in the sum. Secondly you need to note that the probability of $n$ independent events all happening is $p^n$.