(related to Proposition: Probability of Event Union)
We first note that
\[A\cup B=(A\cap B)\cup(A\cap\overline B)\cup (\overline A\cap B).\]
Because the events \((A\cap B)\), \((A\cap\overline B)\) and \((\overline A\cap B)\) are mutually exclusive, it follows from the definition of probability that
\[\begin{array}{ccl} p(A\cup B)&=&p(A\cap B)+p(A\cap\overline B)+p(\overline A\cap B)\\ &=&p(A\cap(B\cup \overline B))+p(\overline A\cap B)\\ &=&p(A\cap\Omega)+p(\overline A\cap B)\\ &=&p(A)+p(\overline A\cap B)\\ \end{array}~~~~~~~~~~~( * ).\]
Because \(B=(A\cap B)\cup (\overline A\cap B)\), and because again, both events \((A\cap B)\) and \((\overline A\cap B)\) are mutually exclusive, we have \[p(\overline A\cap B)=p(B)-p(A\cap B).\]
Together with \(( * )\), we get
\[p(A\cup B)=p(A)+p(B)-p(A\cap B),\]
as required.
