Question

In a town of 10000 families it was found that 40%
family buy newspaper A, 20% 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 newspapers, then number of families
which buy A only is​

Answer 1

Answer:

3300

Step-by-step explanation:

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

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