In a town of 10,000 families, it was found that 40% of the families buy
newspaper A, 20% buy newspaper B, 10% buy newspaper C, 5% buy A and
B;3% buy B and C and 4% buy A and C. If 2% buy all the three newspapers,
find the number of families which buy
(i) A only
(ii) B only
(iii) none of A, B and C.
Answers
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Number of families=10000
n(A)=
100
40×10000
=4000
n(B)=
100
20×10000
=2000
n(C)=
100
10×10000
=1000
n(A∩B)=
100
5×10000
=500
n(A∩C)=
100
4×10000
=400
n(B∩C)=
100
3×10000
=300
n(A∩B∩C)=
100
2×10000
=200
no. of people buy only A, only B & only C=1400+200+3300=4900
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