Mutually Exclusive Events and Exhaustive Events

• If events \(A\) and \(B\) cannot occur together, then the two events are mutually exclusive.

\(\to \mathrm{P}\left( A\cap B \right)=0\)

\(\to \mathrm{P}\left( A\cup B \right)=\mathrm{P}\left( A \right)+\mathrm{P}\left( B \right)\)

• Note that \(\mathrm{P}\left( A\cup B \right)=\mathrm{P}\left( A \right)+\mathrm{P}\left( B \right)-\mathrm{P}\left( A\cap B \right)\).

• If \(A\) and \(B\) are exhaustive events, their union must cover the entire sample space, i.e.\(\mathrm{P}\left( A\cup B \right)=1\).

• For any event \(A\), \(A\) and \(A'\) are mutually exclusive and exhaustive events.

\(\to \mathrm{P}\left( A\cap A' \right)=0\)

\(\to \mathrm{P}\left( A\cup A' \right)=\mathrm{P}\left( A \right)+\mathrm{P}\left( A' \right)=1\)

Example

A card is drawn at random from cards numbered \(1\) to \(10\). Let \(A\) be "even", \(B\) be "odd" and \(C\) be "a multiple of \(3\)".

\(A\) and \(B\) are mutually exclusive and exhaustive: \(\mathrm{P}(A\cup B)=\dfrac{5}{10}+\dfrac{5}{10}=1\).

\(A\) and \(C\) share the card \(6\), so they are not mutually exclusive: \(\mathrm{P}(A\cup C)=\dfrac{5}{10}+\dfrac{3}{10}-\dfrac{1}{10}=\dfrac{7}{10}\).