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 $H\text{and}I,11$ read both $H\text{and}T,8$ read both $T\text{and}I,3$ read all three newspapers. Find
$\left(i\right)$ The number of people who read at least one of the newspapers.
$\left(ii\right)$ The number of people who read exactly one newspaper.

Answer 1

Given data:
Number of people who read newspaper $H=n\left(H\right)=25$
Number of people who read newspaper $T=n\left(T\right)=26$
Number of people who read newspaper $I=n\left(I\right)=26$
Number of people who read both $H\text{and}I=n(H\cap I)=9$
Number of people who read both $H\text{and}T=n(H\cap T)=11$
Number of people who read both $T\text{and}I=n(T\cap I)=8$
Number of people who read all $H,T\text{and}I=n(H\cap T\cap I)=3$

$\left(i\right)$ Solve for $n(H\cup T\cup I)$
We know,
$n(H\cup T\cup I)=n\left(H\right)+n\left(T\right)+n\left(I\right)-n(H\cap T)-n(H\cap I)-n(T\cap I)+n(H\cup T\cup I)$
$=25+26+26-11-9-8+3$
$\therefore n(H\cup T\cup I)=52$

Final Answer:
Hence, $52$ people read at least one of the newspapers.

$\left(ii\right)$
Let us draw a Venn diagram
Let $a$ denote the number of people of who read newspaper $H\text{and}T\text{but not}I$.
Let $b$ denote the number of people who read newspaper $I\text{and}H\text{but not}T$.
Let $c$ denote the number of people who read newspaper $T\text{and}I\text{but not}H$.
Let $d$ denote the number of people who read all three newspapers.

Solve for people who read eactly one newspaper.
$d=n(H\cap T\cap I)=3$
$n(H\cap T)=a+d$
$n(I\cap T)=c+d$
$n(H\cap I)=b+d$

Adding all three equations,
$n(H\cap T)+n(I\cap T)+n(H\cap I)$
$=a+d+b+d+c+d$
$\Rightarrow 11+8+9=a+b+c+d+2d$
$\Rightarrow 11+8+9=a+b+c+d+2\times 3$
$\Rightarrow a+b+c+d=28-6=22$

$\therefore$ People who read exactly one newspaper
$=n(H\cup T\cup I)-(a+b+c+d)$
$=52-22$
$=30$