# Addition Law of Probability/Proof 2

## Theorem

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

## Proof

$A$ 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:

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

Hence:

 $\displaystyle \map \Pr A + \map \Pr B$ $=$ $\displaystyle \map \Pr {A \setminus B} + 2 \map \Pr {A \cap B} + \map \Pr {B \setminus A}$ $\displaystyle$ $=$ $\displaystyle \map \Pr {\paren {A \setminus B} \cup \paren {A \cap B} \cup \paren {B \setminus A} } + \map \Pr {A \cap B}$ Set Difference and Intersection form Partition: Corollary 1 $\displaystyle$ $=$ $\displaystyle \map \Pr {A \cup B} + \map \Pr {A \cap B}$

Hence the result.

$\blacksquare$