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 $1,3$ read all three newspapers. Find:
the number of people who read exactly one newspaper.
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 $1,3$ read all three newspapers. Find:
the number of people who read exactly one newspaper.
Let $A$ be the set of people who read newspaper $H.$
Let $B$ be the of people who read newspaper $T.$
Let $C$ be the set of people who read newspaper $I.$
Accordingly, $n(A)=25, n(B)=26,$ and $n(C)=26$
$n(A \cap C)=9, n(A \cap B)=11,$ and $n(B \cap C)=8$
$n(A \cap B \cap C)=3$
Let $U$ be the set of people who took part in the survey.
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 newspapers $T$ and $I$ only.
Let $d$ denote the number of people who read all three newspapers.
Accordingly, $d=n(A \cap B \cap C)=3$
Now, $n(A \cap B)=a+d$
$n(B \cap C)=c+d$
$n(B \cap C)=c+d$
$n(C \cap A)=b+d$
$\therefore a+d+c+d+b+d=11+8+9=28$
$\Rightarrow a+b+c+d=28-2 d=28-6=22$
Hence, $(52-22)=30$ people read exactly one newspaper.
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