3.0.0.3 証明

$\displaystyle P(A\cup B\cup C)$	$\displaystyle =$	$\displaystyle P(A\cup (B\cup C))$
	$\displaystyle =$	$\displaystyle P(A)+P(B\cup C)-P(A\cap (B\cup C))$
	$\displaystyle =$	$\displaystyle P(A)+[P(B)+P(C)-P(B\cap C)]-P((A\cap B)\cup(A\cap C))$
	$\displaystyle =$	$\displaystyle P(A)+P(B)+P(C)-P(B\cap C)-[P(A\cap B)+P(A\cap C))$
		$\displaystyle \qquad -P((A\cap B)\cap (A\cap C)]$
	$\displaystyle =$	$\displaystyle P(A)+P(B)+P(C)-P(B\cap C)-P(C\cap A)-P(A\cap B)+P(A\cap B\cap C). \quad\qed$

$\begin{jtheorem}[余事象の定理]\upshape \begin{displaymath} P(\overline A)=1-P(A). \end{displaymath}\end{jtheorem}$