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

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

You are viewing a comment permalink. View the original post to see all comments and the full post content.

Comments (33)

Sort By: Best

You are viewing a single comment's thread. Show more comments above.

Comment author: lucidfox 29 July 2011 06:26:10AM 0 points [-]

and P(A|C)/P(B|C) = P(A)/P(B) when A and B are subsets of C?

When A is a subset of C, P(A|C) = P(A).

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.