How to solve conditional probability for 3 events when one event is not indepedent

Question

I want to solve P(A AND B | C), but C relies on either A or B, or both happening. Does this affect the traditional conditioning rule?

Answer 1

$$P(X \text{ and } Y \mid C) = \dfrac{P(X \text{ and } Y \text{ and } C)}{P(C)} $$ so long as $P(C)>0$. This is what I would call the traditional conditioning rule

If $C$ was independent of $X$ and $Y$, then $P(X \text{ and } Y \mid C) = P(X \text{ and } Y)$, but then the conditional probability would not be particularly interesting

If $C = X \cup Y$ then since $X \cap Y \subset X \cup Y$ you have $P(X \text{ and } Y \mid C) = \dfrac{P(X \text{ and } Y )}{P( X \text{ or } Y)} $