Answer
The required solution is $\frac{7}{8}$
Work Step by Step
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, $\begin{align}
& P\left( A\cup B \right)=\frac{4}{8}+\frac{5}{8}-\frac{2}{8} \\
& =\frac{7}{8}
\end{align}$
