Probability Question: Poisson and pmf

I'm working through some practice exercises for my probability course and I think there's a concept I'm not totally understanding.

The question I'm trying to work through is as follows:

The number of mail a post office receives in one day is defined by a Poisson random variable with parameter $\lambda = 5$. The post office fails to process a piece of mail with probability $0.04$ independently with every ball. Find the probability that the post office processes exactly 5 pieces of mail in a working day.

From what I understand, we can model the number of calls as $\dfrac {(e^5)5^{10}} {10!}$ but how do I use the rest of the information to determine the probability mass function?

-
"The post office fails to process a piece of mail with probability 0.04 independently with every ball." What does that mean, "with every ball"? – CodyBugstein Nov 11 '12 at 20:53
i think it's: "The post office fails to process a piece of mail with probability 0.04 independently with every mail received." – Jaynathan Leung Nov 12 '12 at 1:29

It will turn out that the number of letters processed has Poisson distribution with parameter $\lambda(1-0.04)$. There are better conceptual ways of seeing this, but instead we do a somewhat unpleasant computation.

The Post Office receives $n$ letters with probability $e^{-\lambda}\dfrac{\lambda^n}{n!}$. Given that it receives $n$ letters, the number of letters processed has Binomial distribution. Specifically, the (conditional) probability that it processes $k$ letters is $\binom{n}{k}p^kq^{n-k}$. (In our case, $p=0.96$ and $q=1-p=0.04$.)

So the probability that the Post Office processes exactly $k$ letters is given by $$\sum_{n=k}^\infty e^{-\lambda}\frac{\lambda^n}{n!}\binom{n}{k}p^kq^{n-k}.$$ We are interested specifically in the sum for $k=5$. But we continue with the general analysis.

Use $\dbinom{n}{k}=\dfrac{n!}{k!(n-k)!}$ and some algebra to rewrite the above expression as $$e^{-\lambda}\frac{(p\lambda)^k}{k!} \sum_{n=k}^\infty\frac{1}{(n-k)!} (\lambda q)^{n-k}.$$ The $\sum_{n=k}^\infty$ part, perhaps after putting $j=n-k$, can be seen to be the power series expansion of $e^x$, with $x=\lambda q$. But $e^{-\lambda}e^{\lambda q}=e^{-\lambda p}$, so our expression simplifies to $$e^{-\lambda p}\frac{(\lambda p)^k}{k!}.$$ It follows that the number of letters processed has Poisson distribution, parameter $\lambda p$. In our case, $\lambda p=4.8$. Now finding the probability that $k=5$ is easy.

-

I don't know where you got $10$. You need $$\frac{e^{-\lambda} \lambda^x}{x!} = \frac{e^{-5}5^5}{5!}\text{ since }\lambda=5\text{ and } x=5.$$

-