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  • In a town of 10,000 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 newspapers. Find

In a town of 10,000 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 newspapers. Find
(a) The number of families which buy newspaper A only.
(b) The number of families which buy none of A, B and C
 

Answers (1)

Let A be a set of families which buy newspaper A, B be the set of families which buy newspaper B and C be the set of families which buy newspaper C
n(U) = 10000, n(A) = 40\%, n(B) = 20\%,n(c) = 10\%, n(A\cap B) = 5\%, n(B\cap C) = 3\%, n(A\cap C) = 4\% and n(A\cap B\cap C) = 2\%

Number of families which buy newspaper A only = n(A) - n(A\cap B) - n(A\cap C) +n(A\cap B\cap C)

                                                                              = 10000 \times (\frac{40}{100} - \frac{5}{100} - \frac{4}{100} +\frac{2}{100}) = 10000\times \frac{33}{100} = 3300 \ \text{families}

Number of families which buy none of A, B and C =n(U) - n(A\cup B\cup C)

                                                                                \small =n(U) - [n(A) + n(B) + n(C) - n(A\cap B) - n(B\cap C) - n(A\cap C) + n (A\cap B\cap C)]

                                                                                \small = [100 - (40 + 20 + 10 - 5-3-4+2)]\times 10000

                                                                                \small = \frac{40}{100}\times 10000 = 4000 \ families

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