a simple issue on Conditional Probability

Question

Here is my question: I know if A & B are independent, P(A|B) = P(A); But If A and B are independent, is it true that: P(A'|B) = P (A')???

Answer 1

Proof

$$P(A'|B) =\frac{P(A' \cap B)}{P(B)}$$ But $P(A' \cap B) = P(A) - P(A n B)$ Thus $$P(A'| B) = P(B) \frac{1 - P(A)}{P(B)} = P(A')$$ since $1 - P(A) = P(A')$.

Answer 2

I know this question might've looked weird; But believe me, I don't like this subject much. That's why I couldn't solve such a simple question. Anyway, as @A.Webb commented: $P(A'|B) + P(A|B) = 1$, which could be written like this: $P(A'|B) = 1 - P(A|B)$.

As we had in hypothesis, A and B are independent, so $P(A|B) = P(A)$. Hence: $$P(A'|B) = 1 - P(A)$$ And we know $A$ and $A'$ are complements and to sum these two up, equals to $1$. Then $P(A'|B)$ MUST be equal to $P(A')$.