Question

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

Answer 1

Hint: It is given that $P\left(E\right)=0.42$ and we know that the total probability is always 1 so addition of $P\left(E\right)\mathrm{\&}P\left(\text{not E}\right)$ 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.
$P\left(E\right)={\displaystyle \frac{\text{Favorable Outcomes}}{\text{Total Outcomes}}}$
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.
$\begin{array}{rl}& \text{Total Probability}={\displaystyle \frac{\left(\text{Favorable + Unfavorable}\right)\text{Outcomes}}{\text{Total outcomes}}}=1\\ & \Rightarrow \text{Total Probability}={\displaystyle \frac{\text{Favorable Outcomes}}{\text{TotalOutcomes}}}+{\displaystyle \frac{\text{Unfavorable Outcomes}}{\text{Total Outcomes}}}=1.....Eq.(1)\end{array}$
We have shown above that:
$P\left(E\right)={\displaystyle \frac{\text{Favorable Outcomes}}{\text{Total Outcomes}}}$
So, $P\left(\text{not E}\right)$ is equal to:
$P\left(\text{not E}\right)={\displaystyle \frac{\text{Unfavorable Outcomes}}{\text{Total Outcomes}}}$
Substituting the expressions of $P\left(E\right)\mathrm{\&}P\left(\text{not E}\right)$ 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,
$\begin{array}{rl}& P\left(\text{not E}\right)=1-0.42\\ & \Rightarrow P\left(\text{not E}\right)=0.58\end{array}$
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” $\left(\bar{\text{E}}\right)$ is given. For instance, you have to find the probability of $P\left(\bar{E}\right)$ so the manner of calculating the probability of $\left(\bar{\text{E}}\right)$ 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 $P\left(\bar{E}\right)$ directly write the following expression:
$P\left(\text{not E}\right)=P\left(\bar{E}\right)=1-P\left(\text{E}\right)$
Memorizing this concept will save your lot of time in solving questions in the exam.