Recent content by Cognac

Thanks, that example about independence makes things make more sense now. Oh sorry about that. I forgot to divide by the conditional. But does Bayes Rule still work for the case I presented? Given two joint events, X&Y?

So say I have Pr(Z|X&Y) I'm guessing that it follows the standard Pr(A|B)=[Pr(B|A)Pr(A)]/Pr(B) So Pr(Z|X&Y)=[Pr(X&Y|Z)Pr(Z)]/Pr(X&Y)?Also, if X&Y are independent, then would I get Pr(X&Y|Z)=Pr(X|Z)Pr(Y|Z)?