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 number of people who read at least one of the newspapers.
(ii) the number of people who read exactly one newspaper.
​

Answer 1

AnSwer:-

• 52 people read at least one of the newspaper.

• 30 people read exactly one newspaper.

GiVen:-

• n(A)=25,n(B)=26, and n(C)=26
• n(AC)=9,n(AB)=11, and (BC)=8
• n(ABC)=3

(i) According to the 1st case :-

$\sf \implies \: n(ABC)=n(A)+n(B)+n(C) - n(AB) - n(BC - n)(CA)+n(ABC)$

$\sf \implies \: 25+26+26 - 11 - 8 - 9+3 \\ \sf \implies \:52</p><p>$

Hence, 52 people read at least one of the newspaper.

(ii) According to the 2nd case :-

• Let a be the number of people who read newspapers H and T only.
• Let b denote the number of people who read newspapers I and H only.
• Let c denote the number of people who read newspaper T and I only.
• Let d denote the number of people who read all three newspaper.

$\implies\sf \: d=n(ABC)=3$

Now,

$\implies \sf \: n(AB)=a+d \\\implies \sf \: n(BC)=c+d \\\implies \sf \: n(CA)=b+d$

$\implies \sf \: a+d+c+d+b+d=11+8+9=28 \\ \implies \sf \:a+b+c+d=28 - 2d=28 - 6=22$

Hence, (52−22)=30 people read exactly one newspaper.

Answer 2

αηѕωєя

Let A be the set of people who read newspaper H.

Let A be the set of people who read newspaper H.Let A be the set of people who read newspaper H.Let B be the set of people who read newspaper T.

• Let A be the set of people who read newspaper H.Let B be the set of people who read newspaper T.Let C be the set of people who read newspaper I.

Given

• n(A)=25,n(B)=26, and n(C)=26
• n(A∩C)=9,n(A∩B)=11, and (B∩C)=8
• n(A∩B∩C)=3

Let U be the set of people who took part in the survey.

• (i) n(A∪B∪C)=n(A)+n(B)+n(C)−n(A∩B)−n(B∩C)−n(C∩A)+n(A∩B∩C)
• =25+26+26−11−8−9+3
• =52

Hence, 52 people read at least one of the newspaper.

• (ii) Let a be the number of people who read newspapers H and T only.

• Let b denote the number of people who read newspapers I and H only.

• Let c denote the number of people who read newspaper T and I only.

• Let d denote the number of people who read all three newspaper.

Accordingly,

$\begin{gathered}d=n(A∩B∩C)=3 \\ Now \: , n(A∩B)=a+d \\ n(B∩C)=c+d \\ n(C∩A)=b+d \\ ∴a+d+c+d+b+d=11+8+9=28 \\ ⇒a+b+c+d=28−2d=28−6=22 \\\end{gathered} </p><p>d=n(A∩B∩C)=3</p><p>Now,n(A∩B)=a+d</p><p>n(B∩C)=c+d</p><p>n(C∩A)=b+d</p><p>∴a+d+c+d+b+d=11+8+9=28</p><p>⇒a+b+c+d=28−2d=28−6=22</p><p> </p><p>$

Hence, (52−22)=30 people read exactly one newspaper.