In a town of 10,000 families it was found that 40% family buy newspaper A, 20% buy newspaper B and 10% families buy newspaper C, 5% families buy A and B, 3% buy B and C and 4% buy A and C. If 2% families buy all the three newspapers, then number of families which buy A only is

n(A)=40% of 10,000=4,000n(B)=20% of 10,000=2,000n(C)=10% of 10,000=1,000n(A∩B)=5% of 10,000=500n(B∩C) =3% of 10,000=300n(C ∩A)=4% of 10,000=400n(A∩B∩C)=2% of 10,000=200We want to find n(A∩Bc∩Cc)=n[A∩(B∪C)c] = n(A)– n[A∩(B∪C)]=n(A)–n[(A∩B)∪(A∩C)]=n(A)–[n(A∩B)+n(A∩C)–n(A∩B∩C)]=400-500+400-200=4000-700=3300.