Daredevel7 Adrenaline Junky Premium Member Joined Aug 7, 2008 Messages 4,335 Reaction score 5 Jun 7, 2011 #161 StSapphire said: Hah, if Daredevel was here, I could have simply posted this and ended my argument: p(A|X) = (p(X|A)*p(A)) / (p(X|A)*p(A) + p(X|~A)*p(~A)) Click to expand... Wow.. you are so right. End of discussion. I bow to you.
StSapphire said: Hah, if Daredevel was here, I could have simply posted this and ended my argument: p(A|X) = (p(X|A)*p(A)) / (p(X|A)*p(A) + p(X|~A)*p(~A)) Click to expand... Wow.. you are so right. End of discussion. I bow to you.
