# a probability

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

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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

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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

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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 $E_1,E_2,...$ be a partition of $\Omega$ and let F be the proper subset of $\Omega$.

Then $P(F)=\sum_i{P(F|E_i)P(E_i)}$

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Yes i partinioned Omega into two subsets to keep it simple

Mandrake

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