Oscar_Cunningham comments on Looking for proof of conditional probability

Oscar_Cunningham 29 Jul 2011 7:03 UTC
0 points
Um, no?
- lucidfox 29 Jul 2011 8:16 UTC
  0 points
  Parent
  ...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.