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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
Asked by | 25 Feb, 2013, 10:04: AM Expert Answer
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(A B 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
Answered by | 25 Feb, 2013, 06:28: PM

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