Bryn Posted June 10, 2004 Share Posted June 10, 2004 Does P(A|B) = 1 - P(A|B')? Link to comment Share on other sites More sharing options...
MandrakeRoot Posted June 15, 2004 Share Posted June 15, 2004 No clearly not ! What is true is the following P(A) = P(A | B)P(B) + P(A | B' ) P(B ') (wet of total likelyhood) and where B' denotes the complement of B in Omega Mandrake Link to comment Share on other sites More sharing options...
5614 Posted June 15, 2004 Share Posted June 15, 2004 ive never seen that equation before, or maybe i have with different letters or something, or maybe i learnt it a while ago, but anyway; like madrakeRoot said, clearly one does not equal another, it is mathematically impossible Link to comment Share on other sites More sharing options...
bloodhound Posted June 15, 2004 Share Posted June 15, 2004 ive never seen that equation before' date=' or maybe i have with different letters or something, or maybe i learnt it a while ago, but anyway; [/quote'] thats just the "theorem of total probability" which just states that Let [math]E_1,E_2,...[/math] be a partition of [math]\Omega[/math] and let F be the proper subset of [math]\Omega[/math]. Then [math]P(F)=\sum_i{P(F|E_i)P(E_i)}[/math] Link to comment Share on other sites More sharing options...
MandrakeRoot Posted June 16, 2004 Share Posted June 16, 2004 Yes i partinioned Omega into two subsets to keep it simple Mandrake Link to comment Share on other sites More sharing options...
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