We are given n(s) = 60, n(N) = 25, n(T) = 26, n(F) = 26, $n(N \cap F) = 9, n(N \cap T) = 11, n(T \cap F) = 8, n(\bar{N} \cap \bar{T} \cap \bar{F})=8$

Now, $n (N \cup T \cup F) = n(N) + n(T) + c(F) - n(N \cap T) - n(T \cap F) - n(N \cap F) + n(N \cap T \cap F)$

Now, $n(N \cup T \cup F) = n(s) - n(\bar{N} \cap \bar{T} \cap \bar{F})$ = 60 - 8 = 52

$\therefore$ 52 = 25 + 26 + 26 - 11 - 8 - 9 + n($N \cap T \cap F$)

$\therefore$ $n(N \cap T \cap F) = 3$

Alternatively:-

Number of persons who read N and F only = 9 - 3 = 6.

No. of persons who read N and T only = 11- 3 = 8

No. of persons who read T and f only = 8 - 3 = 5

Enter these figure in the diagram.

No. of persons who read T only = 26 – 8 – 3 – 5 = 10

No. of persons who read N only = 25 – 8 – 3 – 6 = 8

No. of persons who read F only = 26 – 6 – 3 – 5 = 12

Q. In a survey of 60 people, it was found that 25 reads Newsweek Magazine, 26 reads Times and 26 reads Fortune. Also 9 reads both Newsweek and Fortune, 11 reads both Newsweek and Times, 8 reads Times and Fortune and 8 reads no magazine at all. (i) Find the number of people who read all three magazines. (ii) Determine number of people who read exactly one magazine.