Why doesn't this conditional probability equation hold This is probably a very stupid question but I can't wrap my head around it.
$$
P(B \cap A) = P(A \cap B) =  P(B \mid A)\cdot P(A) + P(A \mid B)\cdot P(B)
$$
Can someone explain intuitively why the above isn't true?
What I am essentially saying here is that A and B both occurred is same as saying A occurred and then B occurred,following A  OR B occurred and then A occurred, following B.
So probability of A AND B happening = (Probability that A happened AND then probability that given A, B happened) OR (Probability that B happened AND then probability that given B, A happened)
Where am I going wrong?

You are counting too much. The probability of two events occuring is the probability that one event occurs, and then, given this event, what is the probability that the other one happens. Hence you are counting twice as many outcomes as you want. 
Take for instance an easy example: Consider rolling a die. $A=\mbox{even number}$, $B=\mbox{multiple of 3}$. It is clear that $P(A\cap B)=\frac{1}{6}$. 
Now given that the number is even, (your space is reduced to ${2,4,6}$) what is the probability that your number is a multiple of $3$, it is $\frac{1}{3}$, hence you have $\frac{3}{6}\frac{1}{3}=\frac{1}{6}$.
Given that the number is multiple of $3$, (your space is reduced to $\{3,6\})$ what is the probability that it is even, it is $\frac{1}{2}$. Hence, in this case you have $\frac{2}{6}\frac{1}{2}=\frac{1}{6}$. 
If you add this probabilities you get $\frac{1}{3}$ which is double what the answer ought to be. 
