Probability of conditional complement

Question

I know that $$P(A^{C}|B) = 1 - P(A|B)$$ so I was wondering if can we also say that $P(B^{C}|A) = 1 - P(B|A)$? Can we just swap the terms like that?

Thanks

Answer 1

If you know the first equality for all $A$ and $B$ you can surely interchange $A$ and $B$. Here is the actual proof: $P(A)=P(A\cap B)+P(A\setminus B)$. Dividing by $P(A)$ this gives $1=P(B|A) +P(B^{c}|A)$. Hence $P(B^{c}|A)=1-P(B|A)$. Of course, you have to assume that $P(A)>0$.