>NOW, how about three random integers K, L, M?
>
>By same reasoning, for each p there must be at most one of the three random
>integers that p divides. The probability of this is
>
>Prob(p divides none of K,L,M) + Prob(p divides just one of K,L,M) =
>
>(1-1/p)3 + 3(1/p)*(1-1/p)2 =
>
>1 - 3/(p2) + 2/(p3).
>
>Thus Prob(K,L,M) are relatively prime =
>
>Product over all prime p of (1 - 3/(p2) + 2/(p3)).
>
>I don't know how to sum this in closed form, but
>numerically it turns out that
>
>Prob( (K,L,M) = 1 ) ~ 0.1284
>
>(taking the product over the first 1000 primes).
Am I making some mistake here:
Prob(p divides all three of K, L, M)=1/p^3
Prob(p does not divide all three) = 1 - 1/p^3
giving the final probability as the product of these terms over all p,
which is 1/zeta(3) = 0.8319...
Gary McGuire