Recent content by dnbwise

I don't disagree with the math, perhaps your argument isn't cogent though.

So, Bayes' Theorem states P(H|D) = P(D|H) X P(H) / [P(D|H) X P(H) + P(D|~H) x P(~H)], where ~ = negation If my hypothesis consists of two events - S and O - would this be P(S,O|D) = P(D|S,O) X P(S,O) / [P(D|S,O) X P(S,O) + P(D|~S,~O) x P(~S,~O)] or this P(S,O|D) = P(D|S,O) X...