Are P[A,B] Independent When P[A|B]=P[A|B^c]?

[Dragonfall]

Homework Statement

Given that P[A|B]=P[A|B^c], prove that they are independent.

The Attempt at a Solution

So we have that \frac{P(A\cap B)}{P(B)}=\frac{P(A\cap B^c)}{P(B^c)}. I would have to show that \frac{P(A\cap B^c)}{P(B^c)}=P(A). I can't make that happen.

 

[danago]

Try using the fact that <br /> P(A \cap \overline B ) = P(A) - P(A \cap B), i think.

 

[Dragonfall]

Got it. This was unusually convoluted for something so trivial.