Probability Proof - Bayes law multiplicative rule

Question

I have been trying to prove that P(A∪B|C) = P(A|C)+ P(B|C) - P(A∩B|C)

I have gotten to P(C|A∪B)*[P(A) + P(B) - P(A∩B)] = P(C|A)*P(A)/P(C) + P(C|B)*P(B)/P(C) + P(C|A∩B)*P(A∩B)/P(C)

but I am completely clueless regarding how to approach from here...

Thanks for any help!

Answer 1

Notice that: $$\begin{align*} P(A \cup B \mid C) &= \frac{P((A \cup B) \cap C)}{P(C)} \\ &= \frac{P((A \cap C) \cup (B \cap C))}{P(C)} \\ &= \frac{P(A \cap C) + P(B \cap C) - P((A \cap C) \cap (B \cap C))}{P(C)} \\ &= \frac{P(A \cap C) + P(B \cap C) - P((A \cap B) \cap C)}{P(C)} \\ &= \frac{P(A \cap C)}{P(C)} + \frac{P(B \cap C)}{P(C)} - \frac{P((A \cap B) \cap C)}{P(C)} \\ &= P(A \mid C) + P(B \mid C) - P(A \cap B \mid C) \\ \end{align*}$$