If $A$ and $B$ are events such that $P(A\mid B)=P(B\mid A)$, then If $A$ and $B$ are events such that $P(A\mid B)=P(B\mid A)$, then
(A) $A \subset B$ but $A \neq B$
(B) $A=B$
(C) $A \cap B = \emptyset $
(D) $P(A)= P(B)$
My try
It is given that $P(A\mid B)=P(B\mid A)$
$ \implies \dfrac {P\left( A\cap B\right) }{P\left( B\right) } =\dfrac {P\left( B\cap A\right) }{P\left( A\right) }$
$\implies ( P\left( A\cap B\right))(P(A)-P(B))=0$
$\implies \  either \ P(A)=P(B) \ or \ P(A\cap B)=0$
$\implies \  either \ P(A)=P(B) \ or \ A\cap B=\emptyset$
But the correct answer is P(A)=P(B) 
Can anyone explain why $A\cap B=\emptyset$ is wrong?
 A: Your logic implies, either $\mathbb{P}[A] = \mathbb{P}[B]$ or $\mathbb{P}[A \cap B] = 0$. Certainly, if $A \cap B = \emptyset$ the second condition holds, but consider $A \cap B$ being some discrete event out of a continuous probability, so $A \cap B \ne \emptyset$ but $\mathbb{P}[A\cap B] = 0$...
Thus it is possible (i.e. sufficient) that $A\cap B = \emptyset$ but it is not really necessary, and your question's style clearly implies a necessary condition...
A: You can suposse that we have the space $\Omega = [-1, 1]\times [-1, 1]$ and the events $A = [0, 1]\times [0, 1]$ and $B = [-1, 0]\times[-1, 0]$. It's clear that $A\cap B = \{(0, 0)\}$. If you consider the geometric probability, then $P(A)=P(B)=1/4$ and $P(A\cap B) = 0$ and you can see that $A\cap B\neq \emptyset$. Good luck!
A: Let us consider an experiment:
Pick a number uniformly from the interval $[0,1]$.
Let the event 
$A =$ The picked number is $\leq 0.5$, and
$B =$ The picked number is $\geq 0.5$.
Here $P(A \cap B) = P$(the picked number is .5) = 0 (property of continious uniform random variable). Hope this example works.
A: Hello Girish Kumar chandora that's a good try but you just missed the point close
What you did is correct to an extent but you have taken 
 $P(A∩B)$as common but think about this
If it is given that $P(A∩B)/P(B)=P(B∩A)/P(A)$ 
Then the $P(A∩B)$ would be cancelled on both L.H.S and R.H.S as they both are same (basic rule in division)
So then u will be left behind with 
$P(A)=P(B)$
