
Find ‘\[n\left( {{A^c} \cap {B^c}} \right) = \]’, given that ‘\[n\left( U \right) = 700\]’ \[n\left( A \right) = 200\]’ \[n\left( B \right) = 300\]’ \[n\left( {A \cap B} \right) = 100\]’.
A. \[400\]
B. \[600\]
C. \[300\]
D. \[200\]
Answer
161.7k+ views
Hint:
Recall De Morgan’s law.
Formula Used:
De Morgan’s Law:
\[{A^c} \cap {B^c} = {\left( {A \cup B} \right)^c}\]
Complementary formula:
\[n\left( {{A^c}} \right) = n\left( U \right) - n\left( A \right)\]
The formula of the number of element of \[A \cup B\]:
\[n\left( {A \cup B} \right) = n\left( A \right) + n\left( B \right) - n\left( {A \cap B} \right)\]
Complete step-by-step answer:
We have to find the value of \[n\left( {{A^c} \cap {B^c}} \right)\]
Apply De Morgan’s Law.
\[{A^c} \cap {B^c} = {\left( {A \cup B} \right)^c}\]
Calculating the number of elements of the above relation
\[n\left( {{A^c} \cap {B^c}} \right) = n{\left( {A \cup B} \right)^c}\]
Applying complementary formula on RHS
\[n\left( {{A^c} \cap {B^c}} \right) = n\left( U \right) - n\left( {A \cup B} \right)\]
\[n\left( {{A^c} \cap {B^c}} \right) = n\left( U \right) - \left[ {n\left( A \right) + n\left( B \right) - n\left( {A \cap B} \right)} \right]\]
Simplify the above equation:
\[n\left( {{A^c} \cap {B^c}} \right) = n\left( U \right) - n\left( A \right) - n\left( B \right) + n\left( {A \cap B} \right)\]
Substitute \[n\left( U \right) = 700\], \[n\left( A \right) = 300\], \[n\left( B \right) = 200\] and \[n\left( {A \cap B} \right) = 100\] in the above equation, we get;
\[n\left( {{A^c} \cap {B^c}} \right) = 700 - 300 - 200 + 100\]
\[n\left( {{A^c} \cap {B^c}} \right) = 300\]
The correct answer is option C.
Note:
Students often make mistakes when they apply De Morgan’s formula. They used a wrong formula that is \[\left( {{A^c} \cap {B^c}} \right) = {\left( {A \cap B} \right)^c}\]. The correct formula is \[\left( {{A^c} \cap {B^c}} \right) = {\left( {A \cup B} \right)^c}\].
Recall De Morgan’s law.
Formula Used:
De Morgan’s Law:
\[{A^c} \cap {B^c} = {\left( {A \cup B} \right)^c}\]
Complementary formula:
\[n\left( {{A^c}} \right) = n\left( U \right) - n\left( A \right)\]
The formula of the number of element of \[A \cup B\]:
\[n\left( {A \cup B} \right) = n\left( A \right) + n\left( B \right) - n\left( {A \cap B} \right)\]
Complete step-by-step answer:
We have to find the value of \[n\left( {{A^c} \cap {B^c}} \right)\]
Apply De Morgan’s Law.
\[{A^c} \cap {B^c} = {\left( {A \cup B} \right)^c}\]
Calculating the number of elements of the above relation
\[n\left( {{A^c} \cap {B^c}} \right) = n{\left( {A \cup B} \right)^c}\]
Applying complementary formula on RHS
\[n\left( {{A^c} \cap {B^c}} \right) = n\left( U \right) - n\left( {A \cup B} \right)\]
\[n\left( {{A^c} \cap {B^c}} \right) = n\left( U \right) - \left[ {n\left( A \right) + n\left( B \right) - n\left( {A \cap B} \right)} \right]\]
Simplify the above equation:
\[n\left( {{A^c} \cap {B^c}} \right) = n\left( U \right) - n\left( A \right) - n\left( B \right) + n\left( {A \cap B} \right)\]
Substitute \[n\left( U \right) = 700\], \[n\left( A \right) = 300\], \[n\left( B \right) = 200\] and \[n\left( {A \cap B} \right) = 100\] in the above equation, we get;
\[n\left( {{A^c} \cap {B^c}} \right) = 700 - 300 - 200 + 100\]
\[n\left( {{A^c} \cap {B^c}} \right) = 300\]
The correct answer is option C.
Note:
Students often make mistakes when they apply De Morgan’s formula. They used a wrong formula that is \[\left( {{A^c} \cap {B^c}} \right) = {\left( {A \cap B} \right)^c}\]. The correct formula is \[\left( {{A^c} \cap {B^c}} \right) = {\left( {A \cup B} \right)^c}\].
Recently Updated Pages
If tan 1y tan 1x + tan 1left frac2x1 x2 right where x frac1sqrt 3 Then the value of y is

Geometry of Complex Numbers – Topics, Reception, Audience and Related Readings

JEE Main 2021 July 25 Shift 1 Question Paper with Answer Key

JEE Main 2021 July 22 Shift 2 Question Paper with Answer Key

JEE Electricity and Magnetism Important Concepts and Tips for Exam Preparation

JEE Energetics Important Concepts and Tips for Exam Preparation

Trending doubts
JEE Main Eligibility Criteria 2025

NIT Delhi Cut-Off 2025 - Check Expected and Previous Year Cut-Offs

JEE Main Seat Allotment 2025: How to Check, Documents Required and Fees Structure

JEE Mains 2025 Cut-Off GFIT: Check All Rounds Cutoff Ranks

NIT Durgapur JEE Main Cut-Off 2025 - Check Expected & Previous Year Cut-Offs

JEE Main 2024 Cut-off for NIT Surathkal

Other Pages
JEE Advanced 2025: Dates, Registration, Syllabus, Eligibility Criteria and More

Degree of Dissociation and Its Formula With Solved Example for JEE

Free Radical Substitution Mechanism of Alkanes for JEE Main 2025

JEE Advanced 2025 Notes

List of Fastest Century in IPL History

Verb Forms Guide: V1, V2, V3, V4, V5 Explained
