Probability: Union and Conditional Union

Question

I have $P(A\cup C|B)$
Does it equal to $P(A|B)+P(C|B)-P(A\cap C |B)$
If A,C are mutually exclusive, is it same as $P(A|B)+P (C|B)$?

Answer 1

The short answer is "Yes." We have $$\begin{align} \Pr(A\cup C|B)&=\frac{\Pr((A\cup C)\cap B)}{\Pr(B)}\\ &=\frac{\Pr((A\cap B)\cup (C\cap B))}{\Pr(B)}\\ &=\frac{\Pr(A\cap B)+\Pr(C\cap B)-\Pr(A\cap C\cap B)}{\Pr(B)}\\ &=\Pr(A|B)+\Pr(C|B)-\Pr(A\cap C|B) \end{align}$$

Now if $A$ and $C$ are mutually exclusive, or mutually exclusive given $B$, this reduces to $\Pr(A|B)+\Pr(C|B)$.

Answer 2

Application $A \to {\rm Pr}(A|B)$ is a probability, so the answer is yes.

In France, we note ${\rm Pr}_B(A)$. Both notations have pros and cons...