Question

In a town of 10000 families, it was found that 40% families buy newspaper A, 20% families buy newspaper B, 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 newspaper. Find
(i) the number of familiar which buy newspaper A only.
(ii) the number of familiar which buy none of A , B and C.

Answer 1

Let A be the set of familiar which by uy newspaper A, B be the set of families which buy newspaper A, B be the set of familis which bry newspaper B and C be the set of families which buy neswapaper C.
Then, `n(U) = 10000, n(A) = 40% n(B) = 20%` and `n(C) = 10%`
`n(A nn B) = 5%`
`n(B nn C) = 3%`
`n(A nn C) = 4%`
`n(A nn B nn C) = 2%`
(i) Number of familiar which buy newspaper A only
`= n(A) - n(A nn B) - n (A nn C) + n (A nn B nn C)`
` = (40 - 5 - 4 + 2) % = 33%`
`10000 xx 33//100 = 3300`
(ii) Number of families which buy none of A , B and C
`= n(U) - n(A uu B uu C)`
`= n(U) - [n(A) + n(B) + n(A) - n(A nn B) - n(B nn C) - n(A nn C) + n(A nn B nn C)]`
`= 100 - [ 40 + 20 + 10 - 5 - 3 - 4 +2]`
`= 100 - 60% = 40%`
`= 10000 xx 40/100 = 4000`