Counting two-set regions

Fill the overlap first. Each set total includes that overlap, so subtract it to find an “only” region. Adding both totals counts the overlap twice: \(n(A\cup B)=n(A)+n(B)-n(A\cap B)\).

Example

Of \(30\) pupils, \(18\) play badminton, \(14\) play tennis and \(6\) play both. Let these sets be \(A\) and \(B\).

\(A\text{ only}:18-6=12\). \(B\text{ only}:14-6=8\).

\(\begin{aligned}n(A\cup B)&=18+14-6\\&=26.\end{aligned}\)

Neither: \(30-26=4\). Exactly one sport: \(12+8=20\). At least one sport: \(26\), including the six who play both.

Check: \(12+6+8+4=30\). Counts must be nonnegative whole numbers. If one pupil is chosen uniformly from these \(30\), the probability of neither sport is \(\dfrac4{30}=\dfrac2{15}\).