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A bag contains $5$ red and $3$ blue balls. A ball is randomly chosen with the probability of getting one red ball.
A) $\dfrac{5}{8}$
B) $\dfrac{5}{3}$
C) $\dfrac{5}{6}$
D) $\dfrac{5}{9}$

Answer
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Hint: Here in this question, we deal with probability. In order to solve probability related questions, we first need to understand what is probability and some of its related terms such as random experiment, outcome, sample space, equally likely outcomes and event. There are three different types of probability:
-Theoretical Probability
-Experimental Probability
-Axiomatic Probability

Complete step-by-step solution:
Probability helps us to know how often an event is likely to happen.
A random experiment is a trial or an experiment whose outcomes cannot be predicted with surety or certainty. Outcome is the result of any random experiment conducted. Sample space is a set of all the possible outcomes for a random experiment. Event is a set of possible outcomes under a specified condition.
The formula of probability is the ratio of number of favorable outcomes to total number of events in the sample space.
$P\left( E \right) = \dfrac{{n\left( E \right)}}{{n\left( S \right)}}$
Where, $P\left( A \right)$depicts the probability of an event ‘E’,$n\left( E \right)$ depicts the number of favorable outcomes and $n\left( S \right)$ depicts the total number of events in the sample space.
Now, according to the question
Let E be the event in which a red ball is chosen.
So, we get, $n\left( E \right) = 5$ and $n\left( S \right) = 5 + 3 = 8$.
Using the probability formula, we get,
$ \Rightarrow P\left( E \right) = \dfrac{{n\left( E \right)}}{{n\left( S \right)}}$
$ \Rightarrow P\left( E \right) = \dfrac{5}{8}$

Hence, option (A) is the correct answer.

Note: For solving such types of questions, the concept of probability and related terms should be clear. The formula for calculating probability is very easy to remember and use if the concepts are crystal clear. Probability helps to predict future events and the events which take place accordingly.