Finding counterexample for conditional probability

Question

If there is $\Omega$ -a sample space, and $A,B \subset\Omega$ events such that $P(A|B)>P(A)$ I need to find counterexample such that $P(B)<P(B|A)$ is not necessarily true.

Answer 1

According to Bayes Theorem

$$P(A|B)=\frac{P(B|A)P(A)}{P(B)}$$ $$\frac{P(A|B)}{P(A)}=\frac{P(B|A)}{P(B)}$$

Now, if $P(A|B)>P(A) \Rightarrow \frac{P(A|B)}{P(A)}>1$

So using Bayes Theorem, this implies that

$$ \frac{P(B|A)}{P(B)}>1 \Rightarrow P(B|A)>P(B)$$

So it will be true.