यदि (n(U)=100), (n\(A^c\)=40), (n\(B^c\)=55) और (n\(A^c\cap B^c\)=20) है, तो (n\(A\cap B\)) कितना होगा?
If (n(U)=100), (n\(A^c\)=40), (n\(B^c\)=55), and (n\(A^c\cap B^c\)=20), what is (n\(A\cap B\))?
Explanation opens after your attempt
A. (25)
Concept
The number in \(A^c\cup B^c\) is (40+55-20=75). By De Morgan, (A^c\cup B^c=\(A\cap B\)^c), so (n\(A\cap B\)=25).
Why this answer is correct
The correct answer is A. (25). The number in \(A^c\cup B^c\) is (40+55-20=75). By De Morgan, (A^c\cup B^c=\(A\cap B\)^c), so (n\(A\cap B\)=25).
Exam Tip
\(A^c\cup B^c\) की संख्या (40+55-20=75) है। डी मॉर्गन से (A^c\cup B^c=\(A\cap B\)^c), इसलिए (n\(A\cap B\)=25)।
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