If you pick a pair of integers then you can be guaranteed that they are coprime with probability $\frac1{\zeta(2)}$.
However if you fix $a\in\Bbb N$ and pick a pair in $[a,2a]$ what is the probability that they will be coprime?
For general $k$-tuples instead of pairs it is asked in On number of coprimes in a bounded interval - II.
First we establish asymptotic number of coprime pairs $(a,b)$ with $1\leq a\leq c_1n,1\leq b\leq c_2n$ with $c_1,c_2$ positive constants (this is well-known for $c_1=c_2$, and the argument seen here generalizes, but I add it here for completeness). Note that $$\sum_{d\mid a\text{ and }d\mid b}\mu(d)=\begin{cases} 1 & \text{ if }\gcd(a,b)=1,\\ 0 & \text{ otherwise,}\end{cases}$$ so the number of pairs in question is $$\sum_{a\leq c_1n,b\leq c_2n,d\mid a,d\mid b}\mu(d)$$ $$=\sum_{d\leq N}\mu(d)\sum_{k\leq c_1n/d , l\leq c_2n/d}1$$ $$=\sum_{d\leq N}\mu(d)\left(\frac{c_1c_2n^2}{d^2} + O\left(\frac{n}{d}\right)\right)$$ $$=c_1c_2n^2\sum_{d\leq N}\frac{\mu(d)}{d^2}+O(n\log n)$$ $$=c_1c_2n^2\sum_{d=1}^\infty\frac{\mu(d)}{d^2}+O(n\log n)$$ $$=c_1c_2n^2\frac{1}{\zeta(2)}+O(n\log n)$$ where $N$ is taken to be the larger of $c_1n,c_2n$. Applying this with $(c_1,c_2)$ equal to, in order, $(1,2),(2,1)$, adding these and subtracting for $(1,1)$, this gives precisely the number of coprime pairs in $[a,2a]^2$, which is seen to be equal $n^2\frac{1}{\zeta(2)}+O(n\log n)$.