How to answer this poisson distribution? 
Writer is writing a book and he is doing 2 mistakes per page

What is the probability that the 2nd mistake of the writer is in page 3?
What I tried to do is as follows:
$X$~$Poi(2)$ -> 2 mistakes per page hence
$P(X=1)=2e^{-2}$  this is the chance to do 1 mistake in a page
$Y$~$NB(2,2e^{-2})$  we are looking for the first occurrence of 2 mistakes
So we are looking for $P(Y=3)$ which is the probability of doing the 2nd mistake in page 3 which equals to $8e^{-4}-16e^{-6}$ which is WRONG.
What have I done wrong here?
EDIT:
X is the number of mistakes per page, Y is the index of the first time we get 2 mistakes total
 A: You don't need two mistakes on the third page; you just need the second mistake on the third page.
For this problem it is best to consider TWO different Poisson processes:
(1) One Poisson process for the number of mistakes on the first two pages, with a rate of 4 for every two pages, and
(2) One Poisson process for the number of mistakes on the third page, with a rate of 2 on the third page.
There are two independent possibilities to consider for the second mistake to be on the third page:
(1) There are no mistakes on the first two pages, but at least two on the third page, and
(2) There is one mistake on the first two pages, and at least one mistake on the third page.
Can you take it from there?
A: $3$ possible scenarios might happen:
1- one error in the first page, no error in the second page and at least one error in the third page. the probability would be:
$2*e^{-2}*e^{-2}*\sum_{i=1}^\mathbb{\infty}\frac{2^{i}*e^{-2}}{i!}$
2- No error in the first page, one error in the second page and at least one error in the third page. the probability would be the same as the previous one.
3- No error in the first and the second pages and at least two errors in the third page. The corresponding probability is:
$e^{-2}*e^{-2}*\sum_{i=2}^\mathbb{\infty}\frac{2^{i}*e^{-2}}{i!}$
Please note that we can multiply probabilities, because each time interval is assumed to be independent. Also, we can add the three possibilities to get the final result, because the events are disjoint.
