--- Gary McGuire <gmg@maths.may.ie> wrote:
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
1/zeta(3) in addition to being the probability that, as Rich pointed out, GCD(x,y,z)=1, is also the probability that a random integer is cube-free. Gene __________________________________ Do you Yahoo!? Yahoo! Mail - 250MB free storage. Do more. Manage less. http://info.mail.yahoo.com/mail_250