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