In a town of 10,000 families, it was found that 40% families buy newspaper A, 20% buy newspaper B and 10% buy newspaper C. Also,5o families buy newspapers A and B, 3% buy newspapers B and C and 4% buy newspapers A and C. lf 2% families buy all the three newspapers, then number of families which buy newspaper 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=200Number of families which buy newspaper A only=nA∩Bc∩Cc=nA∩(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)]=4000−[500+400−200]=4000−700=3300