Probability of $P(A \cap B)$ in general We are given events $A$ and $B$, $0 < P(A), P(B) < 1$. Can $P(A|B) > P(A)$ and $P(B|A) > P(B)$ happen?
So I came to this conclusion:
$P(A|B) = \frac{P(A \cap B)}{P(B)} > P(A)$,
$P(B|A) = \frac{P(A \cap B)}{P(A)} > P(B)$,
so we get: $P(A \cap B) > P(A) P(B)$. We know that if A and B are independent $P(A \cap B) = P(A) P(B)$, but I am wondering what holds in general. My gut is saying, that in general $P(A \cap B) \leq P(A) P(B)$ should be true, but I don't know how to prove it. Is my understanding of this correct?
Thank you.
 A: This is a variation of other questions dealt with in related answers here and here.  To answer this question most clearly, suppose we define the posterior-to-prior probability ratio of two events as:
$$\Delta(A|B) \equiv \frac{\mathbb{P}(A|B)}{\mathbb{P}(A)}.$$
With some simple algebra you can get the following alternative expression for Bayes' theorem:
$$\Delta(A|B) = \frac{\mathbb{P}(A|B)}{\mathbb{P}(A)} = \frac{\mathbb{P}(A \cap B)}{\mathbb{P}(A) \mathbb{P}(B)} = \frac{\mathbb{P}(B|A)}{\mathbb{P}(B)} = \Delta(B|A).$$
Thus, it is a law of probability (essentially a re-expression of Bayes' theorem) that the posterior-to-prior probability ratio of two events is symmetric.  Your question effectively asks whether it is possible to get $\Delta(A|B) > 1$ and $\Delta(B|A) > 1$.  Not only is this outcome possible, but if one of those outcomes hold then the other must also hold, since $\Delta(A|B) = \Delta(B|A)$.  (The more interesting fact is that it is not possible to get $\Delta(A|B) > 1$ and $\Delta(B|A) \leqslant 1$.)
A: I don't have enough reputation to make comment so... along with the answer from @MarsPlastic, suppose $A=B$ and $P(A)<1$. Then $P(A|B) = 1 > P(A)$ and $P(B|A) = 1 > P(B)$. Yay!
A: Hint: If $P(A\cap B)\le P(A)P(B)$ were true in general, what would that imply for $A=B$?
A: You gut is misleading you 


*

*in general $P(A \cap B)$ can be greater than or less than or equal to $P(A)P(B)$

*depending on which of those three cases you have, you correspondingly get $P(A \mid B)$ is greater than or less than or equal to $P(A)$ assuming $P(B) > 0$, and similarly comparing $P(B\mid A)$ with $P(B)$
Consider throwing two dice, one red and one blue.  


*

*If $A$ is the event the red die shows $6$ and $B$ is the event the sum of the two dice is $12$, then $P(A \cap B) =\frac1{36} \gt \frac1{216}=P(A)P(B)$

*If $A$ is the event the red die shows $1$ and $B$ is the event the sum of the two dice is $12$, then $P(A \cap B) =0 \lt \frac1{216}=P(A)P(B)$

*If $A$ is the event the red die shows $6$ and $B$ is the event the sum of the two dice is $7$, then $P(A \cap B) =\frac1{36} =P(A)P(B)$
