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Cognac
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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)?
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)?
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