## Precalculus (6th Edition) Blitzer

The required solution is $\frac{7}{8}$
We know that it is possible for the pointer to land on a number that is both odd and greater than 3. Two of the numbers 5, 7 are both odd and greater than 3. Let us assume the probability of getting an odd number is $P\left( A \right)$, the probability of getting a number greater than 3 is $P\left( B \right)$, and the probability of getting an odd number greater than 3 is $P\left( A\cap B \right)$. So, these events are not mutually exclusive. And the probability of landing on a number that is odd or greater than 3 is evaluated as given below, $P\left( A\cup B \right)=P\left( A \right)+P\left( B \right)-P\left( A\cap B \right)$ Four of the eight numbers $1,3,5,7$ are odd. So, the probability of getting an odd number is, $P\left( A \right)\text{ }=\frac{4}{8}$ Five of the eight numbers $4,5,6,7,8$ are greater than 3. Therefore, the probability of getting a number greater than 3 is, $P\left( B \right)\text{ }=\frac{5}{8}$ Two out of eight numbers $5,7$ are odd and greater than 3. Therefore, the probability of getting an odd number greater than 3 is, $P\left( A\cap B \right)=\frac{2}{8}$ Thus, \begin{align} & P\left( A\cup B \right)=\frac{4}{8}+\frac{5}{8}-\frac{2}{8} \\ & =\frac{7}{8} \end{align}