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.
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.