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
Text Solution
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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 ∩ ∩ B c ∩ ∩ C c ) = 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)]
= 4000 – [500 + 400 – 200] = 4000 – 700 = 3300.
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