Usbrth

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 ($c_i = 0$).

$$ P(T = 0) = P(N = 0) + P(T = 0 | N = n)P(N = n) $$
$$ P(T = 0) = e^-8 + sum_n=0^infty 0.7frac8^ne^-8n! $$
$$ P(T = 0) = e^-8 + 0.7$$
Am I correct? My knowledge of using/combining two distributions is non-existent so I've tried my best.

Let $Nsim Po(lambda)$ 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,
$$
P(C=0)=sum_k=0^infty e^-lambdafraclambda^kk!cdot p^k\
=e^-lambda sum_k=0^infty frac(lambda p)^kk!
=e^-lambda+lambda p=e^-lambda(1-p)\
=e^-8times 0.3=0.0907 ldots
$$
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 $p^n$.