Relationship between conditional probability and conditional with complements

Question

Does P(A|B^c) = 1 - P(A|B)??? Where B^c denotes B complement.

Can someone show work to show why it is true or not true? I intuitively think they are not equal, but I'd like some confirmation.

Answer 1

It is false.

Flip two coins. The space of all possible outcomes is $\Omega = \{(H,H), (H,T), (T,H), (T,T)\}$. Let $A$ denote the event that the first flip was a head, $A=\{(H,H), (H,T)\}$ and let $B$ denote the event that both flips were tails, $B=\{(T,T)\}$. Then $$P(A|B) = 0$$ but it is not necessary that the first flip a head was if $B$ did not occur. Thus $$P(A|B^c) \neq 1.$$ To compute the probabilities, you can use Bayes' rule, $$P(A|B) = \frac{P(A\cap B)}{P(B)}.$$