Probability the addition rule depending on events I am trying to understand something about the addition rule. It might be very very basic. I tried to search for it but I might not know the correct words in english for this.
I am trying to understand if it is true that:
$$P(A\cup B|C)=P(A|C)+P(B|C)-P(A\cap B|C)$$
thank you for your help!
 A: We would like to show that $P(A \cup B \mid C)$ and 
$P(A \mid C) + P(B \mid C) - P(A \cap B \mid C)$ are equal.
This will be a proof by direct calculation.
Assume $P(C)\neq 0$ so that the conditional probabilities are well-defined.
Consider the two events $A \cap C$ and $B \cap C$.
Apply the formula for the (unconditioned) probability of the
union of these two events:
$$P((A \cap C) \cup (B \cap C)) = 
   P(A \cap C) + P(B \cap C) - P((A \cap C) \cap (B \cap C)). \tag{1}$$
By DeMorgan's Laws, $(A \cap C) \cup (B \cap C) = (A \cup B) \cap C.$
Moreover, by associativity and commutativity of intersection,
$(A \cap C) \cap (B \cap C) = (A \cap B) \cap (C \cap C) = (A \cap B) \cap C.$
So we can rewrite Equation $(1)$:
$$P((A \cup B) \cap C) = 
   P(A \cap C) + P(B \cap C) - P((A \cap B) \cap C).$$
Use the fact that $P(X \cap C) = P(X \mid C) P(C)$ for any event $X$
to rewrite this equation again:
$$P((A \cup B) \mid C) P(C) = 
   P(A \mid C) P(C) + P(B \mid C) P(C) - P((A \cap B) \mid C) P(C).$$
Now, using the assumption that $P(C)\neq 0$, we can divide both
sides of the equation by $P(C)$:
$$P((A \cup B) \mid C) = P(A \mid C) + P(B \mid C) - P((A \cap B) \mid C).$$
QED
A: Inclusion-exclusion principle for probabilities  is used to reason about probabilities of union of events and so on.
In your case, all the events, $A \cup B, A, B, A \cap B $ are all conditioned on C, but still are events with corresponding probabilities.
Thus, there is no reason that the law will not hold.
You can use the venn diagram to convince yourself.

