Poisson distribution simple question

Question

I am missing something very simple here.

I am given that the number of printing errors on any random page of a book of $N$ pages follows a poisson variate with parameter $\lambda$. The number of pages in a random sample of $n$ pages (where $N>>n$ and $n\geq2$) which contain fewer than $2$ errors is denoted by $Y$. Then one has to show that $P(Y=k) = {n \choose k} p^{k}q^{n-k}$ where $p=(1+\lambda)e^{-\lambda}$.

I am quite oblivious to this result...

Answer 1

the probability of zero or one error on any page is $$e^{-\lambda}+\lambda e^{-\lambda}=p$$ Then the number of pages in this condition follows a Binomial distribution with parameters $n$ and $p$