Question: Let $(\Omega, \mathscr F, \mathbb P)$ be a probability space. Prove or disprove that $P(A|B \cup C) = P(A|B) + P(A|C) ~~~\forall~~ A,B,C~ \in \mathscr F$ where $B \cap C = \emptyset$.
Attempt: Starting of with the LHS:
$$ P(A|B \cup C) = \frac{P(AB \cup AC)}{P(B \cup C)}\\ =\frac{P(AB) + P(AC) - P(ABC)}{P(B)+P(C)}\\ =\frac{P(AB)+P(AC)}{P(B)+P(C)} $$ Then, expanding the RHS: $$ P(A|B) + P(A|C) = \frac{P(AB)}{P(B)} + \frac{P(AC)}{P(C)}\\ =\frac{P(C)P(AB)+P(B)P(AC)}{P(B)P(C)} $$
Let $A=\Omega$ and $C=B^c$. Then, if the relationship is true, we have $$ 1=P(\Omega\vert\Omega)=P(\Omega\vert B)+P(\Omega\vert B^c)=2. $$