Question

4 A company produces three kinds of products A, B and C. The company studied the preference of 1600 consumers for these three products. It was found that the product A was liked by 1250, the product B was liked by 930 and product C was liked by 1000. The products A, B and C were liked by 650, the products B and C were liked by 610 and the products C and A were liked by 700 consumers. None of the products was liked by 30 consumers. Find number of consumers who liked: i) All the three products ii) Only two of these products I want it step by step

Answer 1

Answer:

Step-by-step explanation:

Answer 2

i) The total of 350 consumers liked all three products.

ii) The total of 910 consumers liked at least two products.

Given,

n(X) = 1600

n(A) = 1250, n(B) = 930, n(C) = 1000

n(AnB) = 650, n(BnC) = 610, n(AnC) = 700

n(A'nB'nC') = 30.

To Find,

The number of consumers who liked:  i) All three products ii) Only two of these products.

Solution,

Firstly, we will find the number of consumers who liked at least one of the products,

So,

n(AuBuC)= n(x) - n(A'nB'nC')

= 1600 - 30

n(AuBuC) = 1570.

i) All the three products

n(AuBuC) = n(A) + n(B) + n(C) - n(AnB)-n(BnC)-n(AnC) + n(AnBnC)

1570 = 1250+930+1000-650-610-700+ n(AnBnC)

n(AnBnC) = 350.

ii)Only two of these,

n[AnBnC'] = n(AnB) - n[(AnB)nC]

= 650 - 350.

n[(AnBnC']  = 300.

n[A'nBnC] = n(BnC) - n[(AnB)nC]

= 610 - 350

n[A'nBnC] = 260.

n[AnB'nC] = n(CnA) - n[(AnB)nC]

= 700 - 350

n[AnB'nC] = 350.

⇒ n[(AnBnC'] + n[A'nBnC]  + n[AnB'nC]

= 300 + 260 + 350

= 910.

i) The total of 350 consumers liked all three products.

ii) The total of 910 consumers liked at least two products.

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