Consider the probability of an event E in which if $P\left( E \right)=0.42$ then what is the value of $P\left( \text{not E} \right)$?

Hint: It is given that $P\left( E \right)=0.42$ and we know that the total probability is always 1 so addition of $P\left( E \right)\And P\left( \text{not E} \right)$ is equal to 1. When we subtract P (E) from 1 then we will get the value of $P\left( \text{not E} \right)$.

Complete step-by-step answer:
The probability of an event E is given by:
$P\left( E \right)=0.42$
We know that probability of an event E is equal to favorable outcomes divided by the total outcomes.
$P\left( E \right)=\dfrac{\text{Favorable Outcomes}}{\text{Total Outcomes}}$
We also know that, total probability is equal to 1. Total probability means sum of favorable outcomes and unfavorable outcomes divided by the total outcomes.
$\begin{align}
  & \text{Total Probability}=\dfrac{\left( \text{Favorable + Unfavorable} \right)\text{Outcomes}}{\text{Total outcomes}}=1 \\
 & \Rightarrow \text{Total Probability}=\dfrac{\text{Favorable Outcomes}}{\text{TotalOutcomes}}+\dfrac{\text{Unfavorable Outcomes}}{\text{Total Outcomes}}=1.....Eq.(1) \\
\end{align}$
We have shown above that:
$P\left( E \right)=\dfrac{\text{Favorable Outcomes}}{\text{Total Outcomes}}$
So, $P\left( \text{not E} \right)$ is equal to:
$P\left( \text{not E} \right)=\dfrac{\text{Unfavorable Outcomes}}{\text{Total Outcomes}}$
Substituting the expressions of $P\left( E \right)\And P\left( \text{not E} \right)$ that we have shown above in eq. (1) we get,
$\text{Total Probability}=P\left( E \right)+P\left( \text{not E} \right)=1$
Substituting the value of $P\left( E \right)=0.42$ in the above equation we get,
$0.42+P\left( \text{not E} \right)=1$
Subtracting 0.42 on both the sides of the above equation we get,
$\begin{align}
  & P\left( \text{not E} \right)=1-0.42 \\
 & \Rightarrow P\left( \text{not E} \right)=0.58 \\
\end{align}$
From the above solution, we have got the value of $P\left( \text{not E} \right)=0.58$.

Note: You can also face problems in which in place of “not E” $\left( \overline{\text{E}} \right)$ is given. For instance, you have to find the probability of $P\left( \overline{E} \right)$ so the manner of calculating the probability of $\left( \overline{\text{E}} \right)$ is same as the probability of $P\left( \text{not E} \right)$ which we have solved in this question.
Whenever you see $P\left( \text{not E} \right)$ or $P\left( \overline{E} \right)$ directly write the following expression:
$P\left( \text{not E} \right)=P\left( \overline{E} \right)=1-P\left( \text{E} \right)$
Memorizing this concept will save your lot of time in solving questions in the exam.