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
the events complementary to AA and BB
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