(related to Proposition: Probability of Included Event)
We first note that
\[B=\Omega\cap B=(A\cup \overline A)\cap B=(A\cap B)\cup (\overline A\cap B).\]
By assumption of the inclusion \(A\subset B\), we have that \(A\cap B=A\). Therefore
\[B=A\cup (\overline A\cap B).\]
Because \(A \cap (\overline A\cap B)=\emptyset\), the events \(A\) and \((\overline A\cap B)\) are mutually exclusive. Therefore, it follows from the definition of probability that
\[p(B)=p(A)+p(\overline A\cap B)\]
Because \(p(\overline A\cap B)\ge 0\), we have \(p(A)\le p(B)\), as required.
