
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
499.2k+ views
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.
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.
Recently Updated Pages
Master Class 9 General Knowledge: Engaging Questions & Answers for Success

Master Class 9 English: Engaging Questions & Answers for Success

Master Class 9 Science: Engaging Questions & Answers for Success

Master Class 9 Social Science: Engaging Questions & Answers for Success

Master Class 9 Maths: Engaging Questions & Answers for Success

Class 9 Question and Answer - Your Ultimate Solutions Guide

Trending doubts
For Frost what do fire and ice stand for Here are some class 10 english CBSE

What did the military generals do How did their attitude class 10 english CBSE

What did being free mean to Mandela as a boy and as class 10 english CBSE

What did Valli find about the bus journey How did she class 10 english CBSE

Can you say how 10th May is an Autumn day in South class 10 english CBSE

Explain the Treaty of Vienna of 1815 class 10 social science CBSE
