Conditional probability property

Question

How do I prove the following property:

P(A ∪ B ∣ C) = P(A ∣ C) + P(B ∣ C) - P(A ∩ B ∣ C)

I tried writing it as 1 - P((A∪B)' ∣ C) and then use the (A ∪ B)' = A' ∩ B' but I don't know how to continue from here, I'm stuck right after using the definition formula of conditional probability on that last form.

Answer 1

A more intuitive explanation here. $P(A \cup B)$ is the chance of $A$ occurring or $B$ occurring of both $AB$ occurring. So, you sum the probabilities of the sample points in $A$ and $B$, but you've counted the the sample points common to $A,B$ i.e. $A \cap B$ twice, so you subtract them once.

Therefore,

$$P(A \cup B) = P(A) + P(B) - P(A \cap B)$$

You can condition on as many events as you want, as long as you are consistent.

So,

$P(A \cup B \vert C_1)=P(A|C_1) + P(B|C_1) - P(AB | C_1)$ $P(A \cup B \vert C_1 C_2)=P(A|C_1 C_2) + P(B|C_1 C_2) - P(AB | C_1 C_2)$

This holds simply because, when you condition on $C$, you are interested in finding the frequency with which the event occurs, restricting your attention to the trials where $C$ occurs.

In a frequentist sense, your claim literally means the below:

The frequency of observing $A$ or $B$, restricting attention to trials where $C$ occurs = (Frequency with which $A$ occurs + Frequency with which $B$ occurs - Frequency with which both $AB$ occurred) all restricted to trials where $C$ occurs.

Hence, it holds.

Answer 2

First show that $P(A\cup B)=P(A)+P(B)-P(A\cap B)$. This follows from the fact that $ A\cup B $ is the disjoint union of $A\cap B^c$ and $A\cap B$ and $B\cap A^c$ whence $$ P(A\cup B)+P(A\cap B)=P(A\cap B^c)+P(A\cap B)+P(B\cap A^c)+P(A\cap B)=P(A)+P(B). $$ Now use the fact that $Q(A)=P(A\mid C)$ is a probability measure in its own right from which the result follows.