यदि (n\(A\cap B^c\)=41), (n\(A^c\cap B\)=34), (n\(A\cap B\)=26) और (n(U)=130) है, तो (n\(A^c\cap B^c\)) कितना है?
If (n\(A\cap B^c\)=41), (n\(A^c\cap B\)=34), (n\(A\cap B\)=26), and (n(U)=130), what is (n\(A^c\cap B^c\))?
Explanation opens after your attempt
A. (29)
Concept
The sum of the three inside regions is (41+34+26=101), so outside is (130-101=29). The four regions together form the whole (U).
Why this answer is correct
The correct answer is A. (29). The sum of the three inside regions is (41+34+26=101), so outside is (130-101=29). The four regions together form the whole (U).
Exam Tip
तीन अंदरूनी भागों का योग (41+34+26=101) है, इसलिए बाहर (130-101=29)। चार क्षेत्र मिलकर पूरा (U) बनाते हैं।
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