A) 3200
B) 3000
C) 4200
D) 3300
Correct Answer: D
Solution :
[d] We have, \[n\,(A)=40%\]of \[10000=4000\] \[n\,(B)=20%\]of \[10000=2000\] \[n\,(C)=10%\]of \[10000=1000\] \[n\,(A\cap B)=5%\]of \[10000=500\] \[n\,(B\cap C)=3%\]of \[10000=300\] \[n\,(C\cap A)=4%\]of \[10000=400\] and \[n\,(A\cap B\cap C)=2%\]of \[10000=200\] We want to find \[n\,(A\cap {{B}^{c}}\cap {{C}^{c}})=n\,[A\cap {{(B\cup C)}^{c}}]\] \[=n\,(A)-n\,[A\cap (B\cup C)]=n\,(A)-n\,[(A\cap B)\cup (A\cap C)]\] \[=n\,(A)-[n\,(A\cap B)+n\,(A\cap C)-\,n\,(A\cap B\cap C]\] \[=4000-[500+400-200]\] \[=4000-700\] \[=3300\] |
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