It's true. If you are concerned with proof, following the this belove according to conditional probability p(a|b)=p(a,b)/p(b)
(1) P(A,B|C)=P(A,B,C)/P(C) (2) P(A,B,C)=P(A,C)*P(B|A,C) (3) P(A,C)=P(C)*P(A|C) WITH (2) AND (3) WE GET (4) P(A,B,C)=P(C)*P(A|C)*P(B|A,C) TAKING (1) AND (4) WE GET P(A*B|C)=P(A|C)*P(B|A*C)? Hope this help Nathaniel Użytkownik "Matt Dobrin" <[EMAIL PROTECTED]> napisał w wiadomości 9uh8ge$5hv$[EMAIL PROTECTED]">news:9uh8ge$5hv$[EMAIL PROTECTED]... > Does P(A*B|C)=P(A|C)*P(B|A*C)? If not, what does it equal? Thanks in > advance. > -Matt > > ================================================================= Instructions for joining and leaving this list and remarks about the problem of INAPPROPRIATE MESSAGES are available at http://jse.stat.ncsu.edu/ =================================================================
