# Stats Proof.

• Oct 14th 2010, 08:43 AM
mathsandphysics
Stats Proof.
Can anyone give help with this proof question.

The question states:

Let A and B be events with 0 < P(A) < 1 and 0 < P(B) < 1. B is said to favor A if P(A|B) > P(A).

(i) Show that if B favours A, then A also favours B, i.e. if P(A|B) > P(A) then P(B|A) > P(B).

(ii) Show that if B favours A, then B complement favours A complement, i.e. show that if P(A|B) > P(A) then P(A complement | B complement) > P(A complement).
• Oct 14th 2010, 03:49 PM
yannyy
if $P(A|B)>P(A)$
since $P(A|B)=\frac{P(AuB)}{P(B)}$
thus $\frac{P(AuB)}{P(B)}>P(A)$
rearranging the equation gives you $\frac{P(AuB)}{P(A)}>P(B)$

As for question two here a hint $P(\bar(A)|\bar(B))=\frac{P(B)-P(AnB)}{P(B)}$