Question

there are 123 children in a locality. 42 likes ice cream, 36 likes chocolates and 20 like soft drinks. 15 children like ice cream and chocolates, 10 children like ice cream and soft drinks, 4 like chocolates and soft drinks but not ice cream, 11 like chocolates and ice cream but not soft drinks. how many children (i) like all three items, (ii) do not like anything and (iii) like at least two items?​

Answer 1

Answer:

i) 4

ii) 50

iii)21

Step-by-step explanation:

First of all let us assume 3 sets:

• set I containing the children who like ice-cream
• set C containing the children who like Chocolate
• set D containing the children who like Soft Drinks

Now let us see what is given in question

The statements in the question give the following information

1. n(I) = 42
2. n(C) = 36
3. n(D) = 20
4. n(I ∩ C) = 15
5. n(I ∩ D) = 10
6. n(C ∩ D) = 4
7. n(C ∩ I ∩ D') = 11
8. n(Universal) = 123

n(I ∩ C ∩ D) = n(I ∩ C) - n(C ∩ I ∩ D')

                    = 15 - 11 = 4

n(I ∩ D ∩ C') = n(I ∩ D) - n(I ∩ C ∩ D)

                    = 10 - 4 = 6

n(I' ∩ D ∩ C) = n(C ∩ D) - n(I ∩ C ∩ D)

                    = 4 - 4 = 0

n(C only) = n(C) - n(C ∩ I ∩ D') - n(C ∩ I ∩ D) - n(C ∩ D ∩ I')

              = 36 - 11 - 4 -0 = 36 - 15 = 21

n(I only) = n(I) - n(C ∩ I ∩ D') - n(C ∩ I ∩ D) - n(C' ∩ D ∩ I)

              = 42 - 11 - 4 -6 = 42 - 21 = 21

n(D only) = n(D) - n(C ∩ I' ∩ D) - n(C ∩ I ∩ D) - n(C' ∩ D ∩ I)

              = 20 - 0 - 4 - 6 = 20 - 10 = 10

now part i)

all 3 items = n(I ∩ C ∩ D) = 4

part ii)

do not like anything = 123 - n(I ∪ C ∪ D)

                                 = 123 - (21 + 21 + 10 + 11 + 4 + 6)

                                 = 123 - 73 = 50

part iii)

like at least 2 = n(C ∩ I) + n (I ∩ D) + n(D ∩ C) - 2*n(C ∩ I ∩ D)

                      = 15 + 10 + 4 - 8 = 29 - 8 = 21

For more detail consider the Venn Diagram