Suppose that number of mistakes on a page is a Poisson RV and independent. From $n$ pages, find the expected number with no mistakes?

Question

A textbook has $n$ pages. The number of mistakes on each page is a Poisson RV with parameter $\lambda$ and is independent of the number of mistakes on all other pages. What is the expected number of pages with no mistake?

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For any given page, since it is Poisson, the probability of no mistakes is $e^{-\lambda}$.

Let $Y:=\{$# of pages with no mistakes$\}$. The choice is that it has or does not have a mistake. Since each page's # of mistakes is independent from those of others, $Y - Bin(n,e^{-\lambda})$

Thus,

$E[Y]=\sum_{k=0}^n{kP(Y=k)}=\sum_{k=0}^n{k {n\choose k}e^{-\lambda k}(1-e^{-\lambda})^{n-k}}$

Here is my attempt to simplify it to find a sum,

$=\sum_{k=0}^n{k {n\choose k}}e^{-\lambda k}(\frac{e^{\lambda}-1}{e^\lambda})^n(\frac{e^\lambda}{e^\lambda-1})^k=\sum_{k=0}^n{k {n\choose k}}(\frac{e^{\lambda}-1}{e^\lambda})^n(\frac{1}{e^\lambda-1})^k=\sum_{k=0}^n{k {n\choose k}}(e^\lambda-1)^{n-k}e^{-\lambda n}=e^{-\lambda n}\sum_{k=0}^n{k {n\choose k}}(e^\lambda-1)^{n-k}$

What mistakes have I made? Where should I go from here?

Answer 1

$Y$ is binomial with $p=e^{-\lambda}$, hence its expected value equals $ne^{-\lambda}$.