Question

In a survey of 60 people, it was found that 25 people read newspaper H, 26 read newspaper I, 26 read newspaper T, 9 read both H and I, 11 read both H and T, 8 read both T and I, 8 read all three newspapers. Find
A. The number of people who read at least one of the newspapers.
B. The number of people who read exactly one newspaper.

Answer 1

Hint: If $ n\left( A \right) $ is the number of outcomes of event A, $ n\left( B \right) $ is the number of outcomes of event B and $ n\left( C \right) $ is the number of outcomes of event C, then $ n\left( A\cup B\cup C \right)=n\left( A \right)+n\left( B \right)+n\left( C \right)-n\left( A\cap C \right)-n\left( A\cap B \right)-n\left( B\cap C \right)+n\left( A\cap B\cap C \right) $ , where A, B and C are not mutually exclusive events.

Complete step-by-step answer:

In the question, we have been given three events as,

1) The people who read newspaper H

2) The people who read newspaper T

3) The people who read newspapers I.

We have also been given the number of people who read each of these newspapers or the outcomes of these events, n as follows,

    n( H)  =25 

 $ n\left( T \right)=26 $

 $ n\left( I \right)=26 $

We have also been given the following too,

The number of people who read both newspapers H and I, $ n\left( H\cap I \right)=9 $ .

The number of people who read both newspapers H and T, $ n\left( H\cap T \right)=11 $ .

The number of people who read both newspaper T and I, $ n\left( T\cap I \right)=8 $ .

The number of people who read all the three newspapers, $ n\left( H\cap I\cap T \right)=3 $ .

A. We have been asked to find the number of people who read at least one of the newspapers, which means the number of people who read newspaper H or T or I. So, we will use the formula to find the same. So, we have the formulas as,

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

We have the values of these already, so we will substitute the values in this equality and get,

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

 $ \Rightarrow n\left( H\cup I\cup T \right)=52 $

Therefore, 52 people read at least one of the three newspapers.

B. Now, we are asked to find the number of people who read exactly one newspaper. We will draw a Venn diagram of the given data as follows,

Now, let us denote the number of people who read newspaper H and T but not I as a.

Let us denote the number of people who read newspaper T and I but not H as b.

Let us denote the number of people who read newspaper H and I but not T as c.

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

So, then the people who read exactly one newspaper will be, $ =n\left( H\cup T\cup I \right)-a-b-c-d $ .

Now here we have the number of people who read all three newspapers $ d=n\left( H\cap T\cap I \right)=3 $ .

From the figure, we get the number of people reading both H and T as,

  $ n\left( H\cap T \right)=a+d $

 Similarly, we will get,

   n( T$\cap I $)=b+d 

$ n\left( H\cap I \right)=c+d $

We will now add these 3 equations above, so we have,

   n( H$\cap T $)+n( T$\cap I $)+n( H$\cap I $)=9+8+11

 $ \Rightarrow \left( a+d \right)+\left( b+d \right)+\left( c+d \right)=28 $

$ \Rightarrow a+b+c+3d=28 $

Which can be written as follows,

  a+b+c+d+2d=28 

Now, we know that d = 3, so 2d = 6. So, substituting this value, we will get,

 a+b+c+d+6=28

$ \Rightarrow a+b+c+d=28-6 $

 $ \Rightarrow a+b+c+d=22 $

Therefore, the number of people who read exactly one newspaper will be,

 $ =n\left( H\cup T\cup I \right)-\left( a+b+c+d \right) $

 $ \Rightarrow 52-22 $

 $ \Rightarrow 30 $

Thus 30 people read exactly one newspaper.

Note: The students should always be careful while deducing the values from Venn diagrams. They should also remember that $ \cup $ represents or and $ \cap $ represents and. Also, the most common mistake that the students make is that, while finding the number of people who read exactly one newspaper, they may take it as, $ n\left( H\cup T\cup I \right) $ instead of $ n\left( H\cup T\cup I \right)-a-b-c-d $ an it is wrong. Also, some students may make mistake in the formula, $ n\left( H\cup T\cup I \right)=n\left( H \right)+n\left( T \right)+n\left( I \right)-n\left( H\cap I \right)-n\left( H\cap T \right)-n\left( T\cap I \right)+n\left( H\cap T\cap I \right) $ and writ it as, $ n\left( H\cup T\cup I \right)=n\left( H \right)+n\left( T \right)+n\left( I \right)-n\left( H\cap I \right)-n\left( H\cap T \right)-n\left( T\cap I \right)-n\left( H\cap T\cap I \right) $ . So all these mistakes must be avoided in order to get the correct answers.