In a survey of 120 consumers conducted in a shopping mall, 80 consumers indicated that they buy Brand A of a certain product, 68 buy Brand B, and 42 buy both brands. How many consumers participating the survey buy:

a) At least one of these brands?
b) Exactly one of these brands?

For a):
Would this translate as n(A U B)

For b):
Would this translate as n(A U Bcomplement) and n(Acomplement U B)

For a , I know that it is 106.

n(A U B) = 80+68-42=106

But I am stuck on part b.

Would it be :
n(A U Bcomplement) = n(a) - n(A ∩ B)
and n (Acomplement U B) = n(b) - n(A ∩ B).
Or does DeMorgan's law apply to sets with complements?

Would it apply in the sense of n(AUBcomplement) = n(AcomplementUB)?

To find the number of consumers who buy at least one of the brands (a), you need to find the size of the union of the two sets (A and B), which represents the total number of consumers who buy either Brand A or Brand B or both.

To find the number of consumers who buy exactly one of the brands (b), you need to find the size of the union of two disjoint sets: (A - B) and (B - A). In other words, you need to determine the number of consumers who buy Brand A but not Brand B, and the number of consumers who buy Brand B but not Brand A, and then add them together.

Given the information you provided:

- The number of consumers who buy Brand A (n(A)) = 80
- The number of consumers who buy Brand B (n(B)) = 68
- The number of consumers who buy both brands (n(A ∩ B)) = 42

To find the number of consumers who buy at least one of these brands (a), you can use the formula:

n(A ∪ B) = n(A) + n(B) - n(A ∩ B)

Substituting the values:

n(A ∪ B) = 80 + 68 - 42 = 106

So, a total of 106 consumers participating in the survey buy at least one of these brands.

To find the number of consumers who buy exactly one of these brands (b), you can use the formula:

n((A - B) ∪ (B - A))

Substituting the values:

n((A - B) ∪ (B - A)) = (80 - 42) + (68 - 42) = 38 + 26 = 64

So, a total of 64 consumers participating in the survey buy exactly one of these brands.