College Algebra, Chapter 10, 10.3, Section 10.3, Problem 44

A die is rolled, and the number showing is observed. Determine whether the events $E$ and $F$ are mutually exclusive. Then find the probability of the event $E \bigcup F$.

a.) $E:$ The number is greater than 3.

$F:$ The number is less than 5.

In this case, the two events are not mutually exclusive since there will be a number greater than 3 that is less than 5. So, the probability in this event is


$
\begin{equation}
\begin{aligned}

P(E \bigcup F) =& P(E) + P(F) - P(E \bigcap F)
\\
\\
=& \left( \frac{3}{6} \right) + \left( \frac{4}{6} \right) - \left( \frac{1}{6} \right)
\\
\\
=& \frac{6}{6}
\\
\\
=& 1

\end{aligned}
\end{equation}
$



b.) $E:$ The number is divisible by 3.

$F:$ The number is less than 3.

In this case, the two events are mutually exclusive since there will be no number less than 3 that is divisible by 3 as well. This gives $P(E \bigcap F) = 0$. Thus, we have


$
\begin{equation}
\begin{aligned}

P(E \bigcup F) =& P(E) + P(F) - P(E \bigcap F)
\\
\\
=& P(E) + P(F)
\\
\\
=& \frac{2}{6} + \frac{2}{6}
\\
\\
=& \frac{2}{3}


\end{aligned}
\end{equation}
$