From Osher Doctorow Ph.D. mdoctorow@comcast.net
For readers who want to know what happens when P(A-->B) = P(A) or P(B), equation P(A-->B) = P(B) when expanded using the rules of probability theory is equivalent to 1 - P(A) + P(AB) = P(B), or in other words P(B) + P(A) - P(AB) = 1. The equation P(A-->B) = P(A) when expanded is equivalent to 1 - P(A) + P(AB) = P(A), or in other words 1 + P(AB) = 2P(A). Notice that P(A-->B) = 1 + P(AB) - P(A) is easy to derive from the definition P(A-->B) = P(A' U B) and the laws of probability which say: 1) P(A U B) = P(A) + P(B) - P(AB) 2) P(A' ) = 1 - P(A) where A' or "not A" (technically the "complement" of A) is the part of the Universe outside A. Then we get P(A-->B) = P(A' U B) = P(A') + P(B) - P(A' B) = 1 - P(A) + P(B) - P(A' B), and we can show that P(B) - P(A' B) = P(AB) because from (1) when A and B do not intersect, that is P(AB) is 0, we get P(A U B) = P(A) + P(B), and so P(B) = P(AB) + P(A' B) since AB and A' B don't intersect. There is a more complicated definition of A-->B than A' U B, namely A-->B = (AB' )', the complement of the intersection of A and the complement of B, but from the laws of set theory this reduces to A' U B. Osher Doctorow