MHT CET · Maths · Probability
If \(n(X)=700, n(A)=200, n(B)=300,\) \(n(A \cap B)=100\), where \(X\) is universal set and \(A\) and \(\mathrm{B}\) are subsets of \(\mathrm{X}\), then \(\mathrm{n}\left(\mathrm{A}^{\prime} \cap \mathrm{B}^{\prime}\right)=\)
- A 300
- B 400
- C 340
- D 240
Answer & Solution
Correct Answer
(A) 300
Step-by-step Solution
Detailed explanation
\(\mathrm{n}(\mathrm{A} \cap \mathrm{B})=\mathrm{n}(\mathrm{A} \cup \mathrm{B})\)
\(=\mathrm{n}(\mathrm{u})-\mathrm{n}(\mathrm{A} \cup \mathrm{B})\)
\(=\mathrm{n}(\mathrm{u})-\{\mathrm{n}(\mathrm{A})+\mathrm{n}(\mathrm{B})-\mathrm{n}(\mathrm{A} \cap \mathrm{B})\}\)
\(=700-\{200+300-100\}=300\)
\(=\mathrm{n}(\mathrm{u})-\mathrm{n}(\mathrm{A} \cup \mathrm{B})\)
\(=\mathrm{n}(\mathrm{u})-\{\mathrm{n}(\mathrm{A})+\mathrm{n}(\mathrm{B})-\mathrm{n}(\mathrm{A} \cap \mathrm{B})\}\)
\(=700-\{200+300-100\}=300\)
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