यदि (n\(A\cap B^c\)=33), (n\(A^c\cap B\)=27), (n\(A\cap B\)=18) और (n(U)=110) है, तो (n\(A^c\cap B^c\)) कितना है?
If (n\(A\cap B^c\)=33), (n\(A^c\cap B\)=27), (n\(A\cap B\)=18), and (n(U)=110), what is (n\(A^c\cap B^c\))?
Explanation opens after your attempt
Correct Answer
A. (32)
Concept
The sum of the three inside regions is (33+27+18=78), so outside is (110-78=32). The four regions together form the whole (U).
Why this answer is correct
The correct answer is A. (32). The sum of the three inside regions is (33+27+18=78), so outside is (110-78=32). The four regions together form the whole (U).
Exam Tip
तीन अंदरूनी भागों का योग (33+27+18=78) है, इसलिए बाहर (110-78=32)। चार क्षेत्र मिलकर पूरा (U) बनाते हैं।
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