A number is selected at random from the first $100$ whole numbers. Find the probability that the number is divisible by $8or12.$

Hint: Probability of any given event is equal to the ratio of the favourable outcomes with the total number of the reference outcomes or space.

Complete step-by-step answer:
Given that: A number can be selected at random from the first whole numbers and can be done in $100$ways and this is the possible outcome in the event space.
$\therefore n(s) = 100$
Let us suppose that Event A be the event that the selected number is divisible by $8$
And
Event B be the event that the selected number is divisible by $12$
Event AB be the event that the selected number is divisible by both $8or12.$i.e. divisible by $24$.
Let us find out the favourable outcomes-
For Event A favourable outcomes
 $\begin{array}{l}
n(A) = \dfrac{{100}}{8}\\
n(A) = 12
\end{array}$
For Event B favourable outcomes,
$\begin{array}{l}
n(B) = \dfrac{{100}}{{12}}\\
n(B) = 8
\end{array}$
For Event AB favourable outcomes,
$\begin{array}{l}
n(AB) = \dfrac{{100}}{{24}}\\
n(AB) = 4
\end{array}$
Thus, we can find the probability of the events,
$\begin{array}{l}
P(A) = \dfrac{{n(A)}}{{n(S)}}\\
P(A) = \dfrac{{12}}{{100}}
\end{array}$
Similarly,
$\begin{array}{l}
P(B) = \dfrac{{n(B)}}{{n(S)}}\\
P(B) = \dfrac{8}{{100}}
\end{array}$
And Similarly,
$\begin{array}{l}
P(AB) = \dfrac{{n(AB)}}{{n(S)}}\\
P(AB) = \dfrac{4}{{100}}
\end{array}$
Now, the required probability is
$\begin{array}{l}
 = P(A + B)\\
 = P(A) + P(B) - P(AB)\\
 = \dfrac{{12}}{{100}} + \dfrac{8}{{100}} - \dfrac{4}{{100}}\\
 = \dfrac{{16}}{{100}}\\
 = \dfrac{4}{{25}}
\end{array}$
Additional Information: Questions based on probability are an important part of the quantitative aptitude section for the most of the competitive exams.

Note: The probability of any event always ranges between $0$ and $1$. It can never be negative nor the number can be greater than one