Addition Law of Probability

Theorem
Let $$\Pr$$ be a probability measure on an event space $$\Sigma$$.

Let $$A, B \in \Sigma$$.

Then:
 * $$\Pr \left({A \cup B}\right) = \Pr \left({A}\right) + \Pr \left({B}\right) - \Pr \left({A \cap B}\right)$$

That is, the probability of either event occurring equals the sum of their individual probabilities less the probability of them both occurring.

This is known as the addition law of probability, or the sum rule.

Proof
By definition, a probability measure is a measure.

Hence, again by definition, it is a countably additive function.

By Measure is Finitely Additive Function, we have that $$\Pr$$ is an additive function.

So we can apply Additive Function on Union of Sets directly.

Alternative Proof
Alternatively, we can prove it directly, although it works out exactly the same:

From Set Difference Union Intersection we have that:
 * $$A = \left({A \setminus B}\right) \cup \left({A \cap B}\right)$$

and from Set Difference with Intersection we have that:
 * $$\left({A \setminus B}\right) \cap B = \varnothing$$

from which it immediately follows that:
 * $$\left({A \setminus B}\right) \cap \left({A \cap B}\right) = \varnothing$$

So $$A$$ is the union of the two disjoint sets $$A \setminus B$$ and $$A \cap B$$.

So, by the definition of probability measure:
 * $$\Pr \left({A}\right) = \Pr \left({A \setminus B}\right) + \Pr \left({A \cap B}\right)$$

The same thing applies to $$B$$:
 * $$\Pr \left({B}\right) = \Pr \left({B \setminus A}\right) + \Pr \left({A \cap B}\right)$$

Hence:

$$ $$ $$

Hence the result.