Let $$P(A), P(B) > 0$$ Prove or give a counterexample: $$P(A|B) > P(A) \rightarrow P(A|B^c) < P(A)$$
I have been able to do neither. It seems like the statement should be true by intuition, but I'm not sure.
2026-03-25 12:13:10.1774440790
proof or counterexample -- conditional probability & two events
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Since $P(A|B)=P(AB)/P(B)>P(A)$, we have $$P(AB)>P(A)P(B)$$ Note that if $P(B)=1$ then $P(AB)>P(A)$ which is impossible. Hence, $P(B^c)>0$.
Since $P(AB)+P(AB^c)=P(A)$, we have $$P(A|B^c)=\frac{P(AB^c)}{P(B^c)}=\frac{P(A)-P(AB)}{1-P(B)}<\frac{P(A)-P(A)P(B)}{1-P(B)}=P(A).$$