Addition Law of Probability/Proof 2

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Theorem

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


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 Disjoint with Reverse:

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


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.

$\blacksquare$


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