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Given N(A)=8, N(B)=12, N(F)=12, N(A intersect F)=5, N(B intersect F)=6, N(only A)=2, N(A intersect B intersect F)=3, and total=20, find N(only B).

Mathematics

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Answers

2 parts
b(i)

4\text{4}

b(ii)

11\text{11}

Step-by-step solution

4 steps

Let A be the set of people who liked Ampese, B be the set of people who liked Banku, and F be the set of people who liked Fufu.
The total number of people is 20.

Given information:

  • N(A)=8N(A) = 8
  • N(B)=12N(B) = 12
  • N(F)=12N(F) = 12
  • N(A∩F)=5N(A \cap F) = 5
  • N(B∩F)=6N(B \cap F) = 6
  • N(only A)=2N(\text{only } A) = 2
  • N(A∩B∩F)=3N(A \cap B \cap F) = 3

We will fill the Venn diagram regions step-by-step.

  1. Fill the intersection of all three sets.

    The number of people who liked all three foods is given:

    N(A∩B∩F)=3N(A \cap B \cap F) = 3
  2. Fill the intersections of two sets only.

    • People who liked Ampese and Fufu only: N(only A∩F)=N(A∩F)−N(A∩B∩F)=5−3=2N(\text{only } A \cap F) = N(A \cap F) - N(A \cap B \cap F) = 5 - 3 = 2
    • People who liked Banku and Fufu only: N(only B∩F)=N(B∩F)−N(A∩B∩F)=6−3=3N(\text{only } B \cap F) = N(B \cap F) - N(A \cap B \cap F) = 6 - 3 = 3
    • To find people who liked Ampese and Banku only, we use N(A)N(A): We know N(A)=N(only A)+N(only A∩B)+N(only A∩F)+N(A∩B∩F)N(A) = N(\text{only } A) + N(\text{only } A \cap B) + N(\text{only } A \cap F) + N(A \cap B \cap F). 8=2+N(only A∩B)+2+38 = 2 + N(\text{only } A \cap B) + 2 + 3 8=7+N(only A∩B)8 = 7 + N(\text{only } A \cap B) N(only A∩B)=8−7=1N(\text{only } A \cap B) = 8 - 7 = 1
  3. Fill the regions for only one type of food.

    • People who liked Ampese only is given: N(only A)=2N(\text{only } A) = 2
    • People who liked Banku only: We know N(B)=N(only B)+N(only A∩B)+N(only B∩F)+N(A∩B∩F)N(B) = N(\text{only } B) + N(\text{only } A \cap B) + N(\text{only } B \cap F) + N(A \cap B \cap F). 12=N(only B)+1+3+312 = N(\text{only } B) + 1 + 3 + 3 12=N(only B)+712 = N(\text{only } B) + 7 N(only B)=12−7=5N(\text{only } B) = 12 - 7 = 5
    • People who liked Fufu only: We know N(F)=N(only F)+N(only A∩F)+N(only B∩F)+N(A∩B∩F)N(F) = N(\text{only } F) + N(\text{only } A \cap F) + N(\text{only } B \cap F) + N(A \cap B \cap F). 12=N(only F)+2+3+312 = N(\text{only } F) + 2 + 3 + 3 12=N(only F)+812 = N(\text{only } F) + 8 N(only F)=12−8=4N(\text{only } F) = 12 - 8 = 4
  4. Check the total number of people.

    Sum of all regions: 2(only A)+5(only B)+4(only F)+1(only A and B)+2(only A and F)+3(only B and F)+3(A and B and F)=202 (\text{only A}) + 5 (\text{only B}) + 4 (\text{only F}) + 1 (\text{only A and B}) + 2 (\text{only A and F}) + 3 (\text{only B and F}) + 3 (\text{A and B and F}) = 20.
    This matches the total number of people in the house, so no one liked none of the foods.

    a) Illustrate the information on a Venn diagram.
    A Venn diagram with three overlapping circles (Ampese, Banku, Fufu) would show the following values in each region:

    • Only Ampese: 2
    • Only Banku: 5
    • Only Fufu: 4
    • Only Ampese and Banku: 1
    • Only Ampese and Fufu: 2
    • Only Banku and Fufu: 3
    • Ampese, Banku, and Fufu: 3

    b) Find the number of people who liked:

    i) only fufu
    The number of people who liked only fufu is N(only F)N(\text{only } F).

    N(only F)=4N(\text{only } F) = 4

    The number of people who liked only fufu is 4\boxed{\text{4}}.

    ii) only one type of food
    This is the sum of people who liked only Ampese, only Banku, and only Fufu.

    N(only one type of food)=N(only A)+N(only B)+N(only F)N(\text{only one type of food}) = N(\text{only } A) + N(\text{only } B) + N(\text{only } F) N(only one type of food)=2+5+4=11N(\text{only one type of food}) = 2 + 5 + 4 = 11

    The number of people who liked only one type of food is 11\boxed{\text{11}}.

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