**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 numbers of people who read at least one of the newspapers.

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

**Solution:**

Given :

$n(H)=25$

$n(T)=26$

$n(I)=26$

$n(H \cap I)=9$

$n(H \cap T)=11$

$n(T \cap I)=8$

$n(H \cap T \cap I)=3$

(i) We know:

$n(H \cup T \cup I)=n(H)+n(T)+n(I)-n(H \cap T)-n(T \cap I)-n(H \cap I)+n(H \cap T \cap I)$

$\Rightarrow n(H \cup T \cup I)=25+26+26-11-8-9+3=52$

Thus, 52 people can read at least one of the newspapers.

(ii) Now, we have to calculate the number of people who read exactly one newspaper.

We have:

$n(H)+n(T)+n(I)-2 n(H \cap T)-2 n(T \cap I)-2 n(H \cap I)+3 n(H \cap T \cap I)$

$=25+26+26-22-16-18+9$

$=30$

Thus, 30 people can read exactly one newspaper.