Show that $P(A|B^c) \neq 1-P(A|B)$

Question

Starting from the identity $P(A|B)=P(A\cap B)/P(B)$ i have to show that $P(A|B^c) \neq 1-P(A|B)$

I have been working on both sides of equation but can't get nowhere. A previous result is that $P(A^c|B) = 1-P(A|B)$

Answer 1

One counterexample suffices to refute an equation.

Consider the experiment of throwing a fair die once and observing its value. Let $A$ be the event that the roll was even, $B$ be that you throw a $6$.

Then $P(A|B^\complement)=\frac{2}{5}$ while $1-P(A|B)=1-1=0$.

Answer 2

Let $A$ be any event with $P(A) \neq \frac 1 2$ and let $B$ be any event independent of $A$. Then the left side is $P(A)$, the right side is $1-P(A)$ and hence the two side are not equal.

Answer 3

Hint:

What happens if we take for $A$ the whole space?