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.

$\blacksquare$

Alternative Proof

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

From Set Difference and Intersection form Partition, we have that:

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

So, by the definition of probability measure:

$\Pr \left({A}\right) = \Pr \left({A \setminus B}\right) + \Pr \left({A \cap B}\right)$;

$\Pr \left({B}\right) = \Pr \left({B \setminus A}\right) + \Pr \left({A \cap B}\right)$.

We also have from Set Difference Disjoint with Reverse that $\left({A \setminus B}\right) \cap \left({B \setminus A}\right) = \varnothing$.

Hence:

	\(\displaystyle \)	\(\displaystyle \Pr \left({A}\right) + \Pr \left({B}\right)\)	\(=\)	\(\displaystyle \Pr \left({A \setminus B}\right) + 2 \Pr \left({A \cap B}\right) + \Pr \left({B \setminus A}\right)\)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \Pr \left({\left({A \setminus B}\right) \cup \left({A \cap B}\right) \cup \left({B \setminus A}\right)}\right) + \Pr \left({A \cap B}\right)\)	\(\displaystyle \)	Set Difference and Intersection form Partition: Corollary
	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \Pr \left({A \cup B}\right) + \Pr \left({A \cap B}\right)\)	\(\displaystyle \)

Hence the result.

$\blacksquare$

Sources

• Geoffrey Grimmett and Dominic Welsh: Probability: An Introduction (1986): $\S 1.4 \ (15)$