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
628.5k+ 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 12 Business Studies: Engaging Questions & Answers for Success

Master Class 12 Biology: Engaging Questions & Answers for Success

Master Class 12 Chemistry: Engaging Questions & Answers for Success

Class 12 Question and Answer - Your Ultimate Solutions Guide

Master Class 11 Social Science: Engaging Questions & Answers for Success

Master Class 11 English: Engaging Questions & Answers for Success

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

Who Won 36 Oscar Awards? Record Holder Revealed

What is the median of the first 10 natural numbers class 10 maths CBSE

Why is it 530 pm in india when it is 1200 afternoon class 10 social science CBSE

What is the full form of POSCO class 10 social science CBSE

The highest temperature in Karnataka is recorded in class 10 social science CBSE

