I was wrong, and jimrandomh was right.. I said:
P(X|A,B) = 1-(1-P(X|A))(1-P(X|B)) if P(A,B) = P(A)P(B)
But P(X|A,B) = 1-P(~X|A,B)
therefore P(~X|A,B) = (1-P(X|A))(1-P(X|B)) = P(~X|A)P(~X|B)
This is equivalent to claiming P(X|A,B) = P(X|A)P(X|B)
And this is wrong for most distributions.
I was wrong, and jimrandomh was right.. I said:
P(X|A,B) = 1-(1-P(X|A))(1-P(X|B)) if P(A,B) = P(A)P(B)
But P(X|A,B) = 1-P(~X|A,B)
therefore P(~X|A,B) = (1-P(X|A))(1-P(X|B)) = P(~X|A)P(~X|B)
This is equivalent to claiming P(X|A,B) = P(X|A)P(X|B)
And this is wrong for most distributions.