Actually we can't choose a number at random from the set of integers, so let's rephrase the question. Find the probability that two numbers, drawn from 1 to n, are relatively prime, and take the limit as n approaches infinity.
Two numbers have a common divisor of p with probability (1/p)2. In other words, each number is divisible by p with probability 1/p. This probability may not be exact, due to the remainder of n mod p, but as n grows large the probability approaches 1/p2. Taking the complement, our two numbers fail to have a common factor p with probability 11/p2. By the chinese remainder theorem, all primes can be treated as independent events. Thus the probability of being relatively prime is the product of 11/p2, as p runs through all the primes ≤ n. As n increases we keep bringing in more factors of p, and the probability decreases monotonically. We only need find the limit.
Consider the reciprocal of this product. Each factor now looks like p2/(p21). Rewrite this as the convergent series:
1 + 1/p2 + 1/p4 + 1/p6 + 1/p8 + …
As a check, multiply this series by p21, and get p2/(p21) back again.
We must take the product of these infinite series, one series per prime. Expand this into a sum of products, just as (a+b)×(x+y) becomes ax+ay+bx+by. Of course this is a bit more complicated, since we have an infinite product of infinite sums. Still, the distributive law holds, and fortunately, we only need consider finite products in our expansion. Each finite product produces 1/n2 for some integer n. For instance, 1 × 1/4 × 1/9 = 1/36. By the fundamental theorem of arithmetic, each inverse square is produced exactly once. The expanded product is now the sum of 1/n2, or ζ(2). By the previous theorem, ζ(2) = π2/6. Take the reciprocal, and two numbers are coprime with probability 6/π2, roughly 60%.
This generalizes to more than two numbers. Select k numbers at random, and they will have a greatest common divisor of 1 with probability 1/ζ(k). The proof is just like the above, but uses kth powers instead of squares.
The Cost of Rehab By the Numbers:
Los Angeles Rehab

Rehab Cost Per Patient

Estimates and Cost Bands for Rehabs