Functor: "Are you saying P(A|B) = P(A \intersect B) ? if so, that is incorrect"
Johnny: You seem to be contradicting yourself.
I do not think I have said anything incorrect in this thread. P(A|B) is NOT P(A \intersect B) -- you still need to divide by P(B), that is what I was referring to as being incorrect in your earlier statement. I still maintain that P(A|B) is defined to be P(A \intersect B) / P(B) provided that P(B) is non-zero, and that there is no real meaning (and if there is it would be non-standard) to write P( (A|B) , C).
Good people think in terms of categories and groups -- Confucius
