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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 at least one of the newspapers.
$52$
$52$
$52$
$52$
Solution
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.
Accordingly,
$n(A \cup B \cup C)=n(A)+n(B)+n(C)-n(A \cap B)-n(B \cap C)-n(C \cap A)+n(A \cap B \cap C)$
$=25+26+26-11-8-9+3$
$=52$
Hence, $52$ people read at least one of the newspapers.
