If only Euler could help...

Let the two numbers be $a$ and $b$ , and fix a prime number $p$ . Since $p$ divides any positive integer with probability $\frac{1}{p}$ , the probability that both $a$ and $b$ is divisible by $p$ is $(\frac{1}{p})^2$ . Therefore the probability that their greatest common divisor is not divisible by $p$ is $1-\frac{1}{p^2}$ .

Note that $a$ and $b$ are relatively prime exactly when $p \nmid \gcd(a,b)$ for every positive prime $p$ .

Therefore the probability $P$ of $a$ and $b$ being relative primes can be expressed as

$P = \displaystyle \prod_{p}^{} (1-\frac{1}{p^2})$

where the value of $p$ runs through all positive primes.

Upon taking the reciprocal of $P$ it can be observed that each term of the product equals to the value of an infinite geometric sum with first term $1$ and quotient $\frac{1}{p^2}$ . (Remark that $\frac{1}{1-x}=1+x+x^2+x^3+\ldots$ whenever $|x|<1$ ):

$\frac{1}{P} = \displaystyle \prod_{p}^{} \frac{1}{1-\frac{1}{p^2}} = \displaystyle \prod_{p}^{} (1+\frac{1}{p^2}+\frac{1}{p^4}+\frac{1}{p^6}+\ldots)$

Notice that if we expand the product above we get to the infinite sum $1+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\ldots$ as our result, which was proven by Euler to be equal to $\frac{\pi^2}{6}$ . Therefore $P=\frac{6}{\pi^2} \approx 0.608$ .