# How to prove if $P(A|B)>P(A)$ then $P(B|A)>P(B)$ [closed]

How to prove that If $$P(A|B)>P(A)$$ then $$P(B|A)>P(B)$$

• Do you know the definition of conditional probability: $P(A\,|\,B)={P(A \cap B)\over P(B)}$? Jul 11, 2013 at 12:45
• @DavidMitra Shouldn't that be $P(A \cap B)$? Jul 11, 2013 at 12:48
• @AndrewD Of course, thanks. Jul 11, 2013 at 12:48

$$P(A|B)>P(A)$$
$$\frac{P(A \wedge B)}{P(B)}>P(A)$$ $$\frac{P(B|A)P(A)}{P(B)}>P(A)$$ Now, since $P(A)$ and $P(B)$ are positive. $$\frac{P(B|A)P(A)}{P(A)}>P(B)$$ $$P(B|A)>P(B)$$ $$\square$$ It should be noted this works with all other comparison operators as well.
• It used the definition of conditional probability. $P(B|A)=P(A \land B)÷P(A)$ Oct 3, 2017 at 15:52