Probability if non disjoint then $P(A \cup C \mid B) = P(A \mid B) + P(C \mid B) - P(A \cap B \cap C)$?

Question

If this is true for conditional probability: $P(A \cup C \mid B) = P(A \mid B) + P(C \mid B)$ if $ABC = \{\}$.

Then is it true if $ABC \neq \emptyset$ Then $P(A \cup C \mid B) = P(A \mid B) + P(C \mid B) - P(A \cap B \cap C)$

Is this true?

Answer 1

The right statement should be

$$P(A \cup C|B)=P(A|B)+P(C|B)-P(A \cap C|B)$$

That is the last term is still a conditional probability.

To see why this is true, you just have to multiply $P(B)$ throughout and you get inclusion exclusion in terms pf $A \cap B$ and $A \cap C$.

In general, we do not have $P(A \cap C|B)=P(A \cap B\cap C)$.