Does P(A|B) = 1 - P(A|B')?

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

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

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;

 

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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]

Yes i partinioned Omega into two subsets to keep it simple

 

Mandrake