Show that $P\left(A\cup B|C\right) = P\left(A|C\right)+P\left(B|C\right) - P\left(A \cap B |C \right)$

Question

Let $A, B, C$ be events with $P\left(C\right) >0$.

Prove that $P\left(A\cup B|C\right) = P\left(A|C\right)+P\left(B|C\right) - P\left(A \cap B |C \right)$.

My attempt:

Let $A \cup B=x$.

Then, $P\left(x|C\right)= \frac{P\left(x \cap C\right)}{P\left(C\right)}$

I am stuck at where to go next. Does anyone have any hints? Thanks.

Answer 1

Hint:

$$ \begin{aligned} P(A\cup B) &=P(A\cap B’)+P(A\cap B)+P(A’\cap B)\\ \\ P(A) &=P(A\cap B)+P(A\cap B’)\\ P(B) &=P(B\cap A)+ P(B\cap A’) \end{aligned} $$