Two numbers with no common factor

Idea. In the limit, the greatest common divisor equals $d$ with probability $\frac{q}{d^2}$, where $q$ is the probability of being coprime: both numbers must be multiples of $d$, and after dividing by $d$ they must be coprime. These probabilities add up to 1.

Step 1: scaling. Both numbers are multiples of $d$ with probability tending to $\frac{1}{d^2}$. Given that, the quotients are (in the limit) again two uniform random integers, coprime with probability $q$. So $$\mathbb{P}(\gcd =d)\to \frac{q}{d^2}.$$

Step 2: add up. The events $\{\gcd =d\}$ for $d=1,2,\dots$ cover everything, so $$1=\sum _{d\ge 1}\frac{q}{d^2}=q\cdot \frac{{\pi }^2}{6}.$$ (Passing to the limit inside the sum is justified because the tail ${\sum }_{d>D}\frac{1}{d^2}$ is small uniformly in $N$.)

Step 3. $$q=\frac{6}{{\pi }^2}\approx \mathrm{0.608.}$$

The product view. Two numbers are coprime when no prime divides both, so $$q=\prod _{p\text{ prime}}\left(1-\frac{1}{p^2}\right)=\frac{1}{\zeta (2)},$$ by Euler's product formula. The question needs the limit: there is no uniform distribution on all positive integers.