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Q17.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 thre
newspapers. Then the number
of families which buy none of
A, B and C is equal to
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Answer:
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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