What does $P(A)\cdot P(B)$ mean for non independent events? When two events $A$ and $B$ are independent, $P(A)\cdot P(B)$ equals the probability of both events occurring together.
$$P(A|B) = \frac{P(A\cap B)}{P(B)}.$$
If $A$ and $B$ are not independent,  $P(A\cap B)$ is either greater than or less than $P(A)\cdot P(B)$.


*

*I want to understand the intuitive meaning of $P(A)\cdot P(B)$ for non independent event. What does it signify?

*Also, what is intuition behind 
If $P(A|B) > P(A)$, then $P(B|A) > P(B)$.
I understand it is direct consequence of $P(A\cap B) > P(A)\cdot P(B)$, but it is not very intuitive to me.
 A: When $A,B$ are not independent, and if there is no more information, there's no natural interpretation of $P(A){\,\cdot\,}P(B)$.

The condition $P(A|B) > P(A)$ means $A$ is more likely to have occurred if it's given that $B$ has occurred, than if there is no information as to whether or not $B$ has occurred.

Equivalently, $A$ is more likely to have occurred if it's given that $B$ has occurred, than if it's given that $B$ did not occur (i.e., $B'$ occurred, where $B'$ is the complement of $B$).

The interpretation for $P(B|A) > P(B)$ is analogous.

As regards the question as to why $P(A|B) > P(A)$ implies $P(B|A) > P(B)$ . . .

Assume $P(A),P(B) > 0$.
\begin{align*}
\text{Then}\;\;&P(A|B) > P(A)\\[4pt]
\implies\;&\frac{P(A\cap B)}{P(B)} > P(A)\\[4pt]
\implies\;&P(A\cap B) > P(A)P(B)\\[4pt]
\implies\;&\frac{P(A\cap B)}{P(A)} > P(B)\\[4pt]
\implies\;&P(B|A) > P(B)\\[4pt]
\end{align*}
As regards the intuition . . .

Assume $P(A),P(B) > 0$.

Suppose we have $P(A|B) > P(A)$.

Thus, when $B$ happens, $A$ tends to happen more than when $B$ doesn't happen.

So if $A$ happens, it suggests that $B$ might have been "responsible", hence it's more likely that $B$ occurred than $B'\;(\text{not}\,B)$.

Thus, the occurrence of $A$ increases the likelihood of the occurrence of $B$.

Which means $P(B|A) > P(B)$.
A: For events $A,B$, the correlation is measured by $$\mathsf{Corr}(A,B)=\dfrac{ \mathsf P(A\cap B)-\mathsf P(A)\cdotp \mathsf P(B)}{\sqrt{\mathsf P(A)\cdotp\mathsf P(B)\cdotp(1-\mathsf P(A))\cdotp(1-\mathsf P(B))}}$$
Similarly the covariance is measured as: $$\mathsf{Cov}(A,B)= \mathsf P(A\cap B)-\mathsf P(A)\cdotp \mathsf P(B)$$
You might extract an interpretation for $\mathsf P(A)\cdotp\mathsf P(B)$ from that.
$\mathsf P(A\mid B)>\mathsf P(A)$ intuitively means that outcomes of $A$ are overrepresented among event $B$ in comparison to their proportion of the whole outcome space -- we say there is a positive correlation between the events.   Then it should feel right that outcomes of $B$ are likewise overrepressented among event $A$.   Indeed, if they were underrepresented, there would be a negative correlation, but it would be a contradiction for the correlation to be both positive and negative.
Ipso facto, $\mathsf P(A\mid B)>\mathsf P(A)\iff \mathsf P(B\mid A)>\mathsf P(B)$ (preassuming $A,B$ both have positive probability mass).
A: Consider the following events $M$={pass midterm exam}, $F$={pass final exam} and the probabilities:
$$P(M)=0.8, P(F)=0.7,P(M\cap F)=0.6.$$
Then:
$$P(M|F)=\frac{P(M\cap F)}{P(F)}=\frac{0.6}{0.7}=\frac67>0.8=P(M); \\
P(F|M)=\frac{P(F\cap M)}{P(M)}=\frac{0.6}{0.8}=\frac34>0.7=P(F).$$
Hence, passing one exam increases the probability of passing the other exam, which is a consequence of a positive correlation. 
