Addition Law of Probability/Proof 2

Theorem

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

Let $A, B \in \Sigma$.

Then:

$\map \Pr {A \cup B} = \map \Pr A + \map \Pr B - \map \Pr {A \cap B}$

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.

Proof

From Set Difference and Intersection form Partition:

$A$ 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:

$\map \Pr A = \map \Pr {A \setminus B} + \map \Pr {A \cap B}$

$\map \Pr B = \map \Pr {B \setminus A} + \map \Pr {A \cap B}$

From Set Difference is Disjoint with Reverse:

$\paren {A \setminus B} \cap \paren {B \setminus A} = \O$

Hence:

Hence the result.

$\blacksquare$

Sources

• 1986: Geoffrey Grimmett and Dominic Welsh: Probability: An Introduction ... (previous) ... (next): $1$: Events and probabilities: $1.4$: Probability spaces: $(15)$