Likewise, bounds can be put on \(p(X \cup Y)\). \(p(X\cup Y)\) can’t be bigger than when \(X\) and \(Y\) are disjoint, so it is bounded above by \(p(X)+p(Y)\). It is also bounded above by \(1\), and thus by the minimum of those expressions. It is also bounded below by \(p(X)\) and by \(p(Y)\) and thus by their maximum. Putting this together,