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 $H\&I,11$ read both $H\&T,8$ read both $T\&I,3$ read all three newspapers, then

• A. $52$ People read at least one newspaper
• B. $48$ People read at least one newspaper
• C. $8$ People read neither of these newspaper
• D. $12$ People read neither of these newspaper

Answer 1

Answer: (A), (C)

$8$ People read neither of these newspaper

Let: $H=$ Set of people who read newspaper $H$
$T=$ Set of people who read newspaper $T$
$I=$ Set of people who read newspaper $I$
$\Rightarrow n\left(U\right)=60;n\left(H\right)=25;n\left(T\right)=26;$
$n\left(I\right)=26;n(H\cap I)=9;n(H\cap T)=11;$ and $n(T\cap I)=8,n(H\cap T\cap I)=3$
Now, let's create a venn diagram for the same and label them as shown:

Now, according to the diagram and the data given we can write:
$g=n(H\cap T\cap I)=3$
$\Rightarrow g=3$
$n(H\cap T)=e+g=11\Rightarrow e=8$

$n(H\cap I)=d+g=9\Rightarrow d=6$

$n(T\cap I)=f+g=8\Rightarrow f=5$

Now, $n\left(H\right)=a+e+g+d=25$
$\Rightarrow a+8+3+6=25\Rightarrow a=8$

$n\left(T\right)=b+e+g+f=26$
$\Rightarrow b+8+3+5=26\Rightarrow b=10$

$n\left(I\right)=c+f+g+d=25$
$\Rightarrow c+5+3+6=26\Rightarrow c=12$

Thus, Number of people reading at least one newspaper
$=a+b+c+d+e+f+g=52$

Now, number of people who read neither of these newspaper $=n\left(U\right)-n(H\cup T\cup I)=60-52=8.$