If ACAC and BCBC are the events complementary to AA and BB, and p(A|B)>p(A)p(A|B)>p(A) and p(BC)>0p(BC)>0, prove that p(AC|BC)>p(AC)p(AC|BC)>p(AC).
Been trying to get my head around this but kind of stuck and any help would be much appreciated.
I know that p(A|B)=p(A and B)/p(B)p(A|B)=p(A and B)/p(B), p(AC)=1-p(A)p(AC)=1-p(A)... therefore p(A)=1-p(AC)p(A)=1-p(AC)... I