Probability Proof. $P(A | B) = P(A | B^c)$

Question

Be A,B are independent envents, if and only if

$P(A|B) = P(A|B^c)$

I know that $P(A) = P(A∩B) + P(A∩B^c)$

and $P(B)= 1- P(B^c)$ So...

$\frac {P(A∩B)}{P(B)} = \frac {P(A∩B^c)}{P(B^c)}$ hence

$(1-P(B)) P(A∩B) = P(B)P(A∩B^c)$

$P(A∩B)=P(B)P(A|B^c)$ (this change is not very clear to me)

But i can't see how continue...

Thanks for the time :)

Answer 1

Some corrections for your working:

$$\color{red}(1-P(B) \color{red})(P(A∩B) = P(B)P(A∩B^c)$$

$$P(A∩B)=P(B)P(A|B)$$

Now let's work on the question:

$$(1-P(B))(P(A∩B) = P(B)P(A∩B^c)$$

$$P(A \cap B) - P(B)P(A\cap B)=P(B)P(A \cap B^C)$$

which is equivalent to

$$P(A \cap B) = P(B)(P(A\cap B)+P(A \cap B^C))$$

Try to simplify the last line.

Answer 2

You have done it correctly till :- $$(1−P(B))P(A∩B)=P(B)P(A∩B^c)$$ Just try to convince yourself by some Venn diagrams that $$P(A∩B^c) = P(A)-P(A∩B)$$ Replace this in your equation and you will get $P(A∩B) =P(A)P(B)$ which proves their independence.