Answer
$P(A|B = empty)$,$P(A) = \frac{1}{3}$, $P(C|B = empty) $
Work Step by Step
Suppose the doors are $A$,$ B$ and $C$. Without Loss of Generality, assume you choose door $A$, and the game-show host opened the door $B$ to reveal it does not contain the prize. The statement $P(A) = P(C) = \frac{1}{2}$ is wrong because it is based on the extra information that $B$ does not contain the prize. The accurate expression is $P(A) = \frac{1}{3}$, $P(C|B = empty) =P(A^c) =\frac{2}{3}$ and whether you should change your decision matters on $P(A)$ instead of $P(A|B = empty)$.
