#### Question

In a survey of 60 people, it was found that 25 people read newspaper H, 26 read newspaper T, 26 read newspaper I, 9 read both H and I,11 read both H and T, 8 read both T and I, 3 read all three newspapers. Find:

(i) the number of people who read at least one of the newspapers.

(ii) the number of people who read exactly one newspaper.

#### Solution

Let A be the set of people who read newspaper H.

Let B be the set of people who read newspaper T.

Let C be the set of people who read newspaper I.

Accordingly, *n*(A) = 25, *n*(B) = 26, and *n*(C) = 26

*n*(A ∩ C) = 9, *n*(A ∩ B) = 11, and *n*(B ∩ C) = 8

*n*(A ∩ B ∩ C) = 3

Let U be the set of people who took part in the survey.

(i) Accordingly,

*n*(A ∪ B ∪ C) = *n*(A) + *n*(B) + *n*(C) – *n*(A ∩ B) – *n*(B ∩ C) – *n*(C ∩ A) + *n*(A ∩ B ∩ C)

= 25 + 26 + 26 – 11 – 8 – 9 + 3

= 52

Hence, 52 people read at least one of the newspapers.

(ii) Let *a* be the number of people who read newspapers H and T only.

Let *b* denote the number of people who read newspapers I and H only.

Let *c* denote the number of people who read newspapers T and I only.

Let *d* denote the number of people who read all three newspapers.

Accordingly, *d* = *n*(A ∩ B ∩ C) = 3

Now, *n*(A ∩ B) = *a* + *d*

*n*(B ∩ C) = *c* + *d*

*n*(C ∩ A) = *b* + *d*

∴ *a + d + c + d + b + d* = 11 + 8 + 9 = 28

⇒ *a + b + c + d* = 28 – 2*d* = 28 – 6 = 22

Hence, (52 – 22) = 30 people read exactly one newspaper.