In a town of 10,000 families it was found that 40% families buy newspaper A, 20% families 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 news papers, then number of families which buy newspaper A only is
Text Solution
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n = 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
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 – [n (A ∩ B) + n (A ∩ C) – n (A ∩ B ∩ C)]
= 4000 – [500 + 400 – 200] = 4000 – 700 = 3300.
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