# $C$ is Geometrically distributed and $N$ is Poisson distributed. What is the sum of N $C$ events?

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.

• Could you edit the first paragraph? There seems to be some confusions. – Holding Arthur Apr 8 at 6:22
• Sorry, is that better? – whereswally Apr 8 at 6:30
• " probability of them buying a CD is 0.7. If the ith customer searches through ci cd's before they find one to purchase (ci ~ Geo(0.8))". Why are you having 0.7 and 0.8 at the same time? – Holding Arthur Apr 8 at 6:45

## 1 Answer

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 $$p^n$$.

• Thank you. Should my summation for the total number of CD's browsed just be N*c_i? – whereswally Apr 8 at 7:06
• @whereswally What do you mean? Are you trying to find the expected number of CDs browsed? – Holding Arthur Apr 8 at 8:06