Is it possible that probability of occuring intersection of two events will be greater than probability of occuring each of them? If $A$ and $B$ are correlated events, Is it possible that we have :
$$P\left(A \cap B \right) \geq P\left( A\right)$$
and
$$P\left(A \cap B \right) \geq P\left( B\right)$$
Is it possible?
 A: Consider $A=B$, then $\mathbb{P}(A\cap B) = \mathbb{P}(A)=\mathbb{P}(B)$.
Now observe that we have $\mathbb{P}(A\cap B) = \mathbb{P}(A\setminus (B^c\cap A)) = \mathbb{P}(A) - \mathbb{P}(B^c \cap A)$ so your statement can only be true if $A\subseteq B$ almost surely. If you must satisfy the two, then you have also $B\subseteq A$ almost surely which means the two are equals almost surely.
A: In general $(A \cap B) \subseteq A$ so $P\left(A \cap B \right) \leq P\left( A\right)$
So you can only have $P\left(A \cap B \right) \geq P\left( A\right)$ when $P\left(A \cap B \right) = P\left( A\right)$ and that only happens when $P(A \cap B^c)=0$, an example of which is $A \subseteq B$ 
Similarly you can only have $P\left(A \cap B \right) \geq P\left( B\right)$ when $P\left(A \cap B \right) = P\left( B\right)$ and that only happens when $P(A^c \cap B)=0$, an example of which is $B \subseteq A$
So the simplest example is when $A=B$ and all examples will have the difference (exclusive or) between $A$ and $B$ has zero probability where you get $P\left(A \cap B \right) = P\left( A\right) = P\left( B\right)$ 
