Oscar_Cunningham comments on Looking for proof of conditional probability - Less Wrong

-1 Post author: DanielLC 28 July 2011 02:24AM

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Comment author: Oscar_Cunningham 29 July 2011 07:03:19AM 0 points [-]

Um, no?

Comment author: lucidfox 29 July 2011 08:16:48AM * 0 points [-]

...Oops, yes, said that without thinking. But this

Basically, P(A|B) = 0 when A and B are disjoint, and P(A|C)/P(B|C) = P(A)/P(B) when A and B are subsets of C?

is correct.