If $k= 1- \sum _{p\in \mathbf{P}} \frac{1}{p^{2}}$, then there are at least $kN^{2}$ pairs $(n_{1},n_{2})$ in $\mathbf{Z}^{2}$, with $1\le n_{1}, n_{2} \le N$ and $gcd(n_{1},n_{2})=1$. Also since $k>0$ it follows that $gcd = 1$ exists with a probability greater than 0.
I don't understand a couple of things with this problem: why is k set to $1- \sum _{p\in \mathbf{P}} \frac{1}{p^{2}}$? F.e. if we plug in p=2 , we had $k=1-\sum_{2}\frac{1}{4} = \frac{3}{4}$. So that would mean if N was 6 we would have $36*\frac{3}{4} = 27$ relatively prime pairs ? I just don't see how to continue.