Suppose that a book with $n$ pages contains on the average $λ$ misprints per page. What is the probability that there will be at least $m$ pages which contain more than $k$ misprints ?
I am getting two answers by two approaches, which one is wrong ?
Method $1$ :
If there are $m$ pages with at least $k$ misprints then there are a total of at least $mk$ misprints. Now the number of total misprints in the book is a Poisson with $\lambda n$ rate. Hence the probability becomes
$$\sum_{t=mk}^{\infty}\frac{e^{-\lambda n}(\lambda n)^{t}}{t!}$$
Method 2:
On a single page the probaility that there are at least $k$ misprints is $p = \sum_{i=k+1}^{\infty}\frac{e^{-\lambda}\lambda^i}{i!}$. Therefore the proabibility of a page having less than or equal to $k$ misprints is $1-p$. Now there are total $n$ pages we can have at least $m$ pages with $k$ misprints, then total probability is $$\sum_{x = m}^{n} {n\choose x}p^x(1-p)^{n-x}$$