in a town of 10000 families it was found that 40%families buy newspaper a,20% buy newspaper b and 10%buy newspaper c.5% families buy a and b.3% buy b and c and 4% buy a and c.if 2% buy all newspapers.find the number of families that buy a only b only none of a,b and c

n(A) = 40% of 10,000 = 4,000

n(B) = 20% of 10,000 = 2,000

n(C) = 10% of 10,000 = 1,000

n (A  B) = 5% of 10,000 = 500

n (B  C) = 3% of 10,000 = 300

n(C  A) = 4% of 10,000 = 400,

n(A  B  C) = 2% of 10,000 = 200

We want to find n(A  only) = n(A) – [n(A  B) + n(A  C)] + n(A  B C)=

 n(A  only) = 4000 – [500 + 400] + 200 = 4000 – 700 = 3300

 

Similarly, n(B only) = n(B) – [n(A  B) + n(B C)] + n(A  B C)

n(B only)  = 2000 - [500+300]+200 = 1400

 

n(none of A, B and C) = 10,000 - [n(A)+n(B) + n(C) - n (A  B) - n(B C) - n (C A) + n(AB C)]

n(none of A, B and C) = 10,000 - [4000+2000+1000 - 500-300-400 + 200]

n(none of A, B and C) = 4000