P(A or B | C) =? P(A|C)+P(B|C)

Question

I would like to calculate the probability of either $A$ or $B$ or both given $C$. If A and B are mutually exclusive ($P(A,B)=0$) then I believe it is true that $P(A or B | C) = P(A|C)+P(B|C)$ although I have not seen this 'officially stated' in a brief search.

Is this true when A and B can occur simultaneously? When they are dependent?

Answer 1

$P(A \text{ or }B \mid C) = P(A\mid C)+P(B\mid C)$ is true if and only if $P(A \text{ and }B \mid C) = 0$

In other words, conditioning on event $C$ happening, you want $A$ and $B$ to be mutually exclusive, or at least have zero probability of occuring

More generally, $P(A \text{ or }B \mid C) = P(A\mid C)+P(B\mid C) - P(A \text{ and }B \mid C)$ as an example of inclusion-exclusion

Answer 2

A proof of the Henry's formula $$ P(\text{$A$ or $B$}\mid C) = P(A\cup B\mid C) = \frac{P((A\cup B)\cap C)}{P(C)} = \frac{P((A\cap C)\cup(B\cap C))}{P(C)} $$ $$ = \frac{P(A\cap C)+P(B\cap C)- P((A\cap B) \cap C)}{P(C)} = P(A\mid C)+P(B\mid C) - P(A\cap B\mid C) $$