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In a town of $10,000$ families it was found that $40\%$ family buy newspaper $A, 20\%$ buy newspaper $B$ and $10\%$ families buy newspaper $C, 5\%$ families buy $A$ and $B, 3\%$ buy $B$ and $C$ and $4\%$ buy $A$ and $C$. If $2\%$ families buy all the three newspapers, then number of families which buy $A$ only is
$3100$
$3300$
$2900$
$1400$
Solution
(b) $n(A) = 40\% \ of 10,000 = 4,000$
$n(B) = 20\% \ of\ 10,000 = 2,000$
$n(C) = 10\% \ of \ 10,000 = 1,000$
$n (A \cap B)$ $= 5\% \ of\ 10,000 = 500$
$n (B \cap C)$ $= 3\% \ of\ 10,000 = 300$
$n(C \cap A)$ $= 4\% \ of \ 10,000 = 400$
$n(A \cap B \cap C)$ $= 2\% \ of \ 10,000 = 200$
We want to find $n(A \cap B^c \cap C^c) = n[A \cap (B \cap C)^c]$
$= n(A) -n[A \cap (B \cup C)] = n(A) -n[(A \cap B) \cup (A \cap C)]$
$= n(A) -[n(A \cap B) + n(A \cap C) -n(A \cap B \cap C)]$
$= 4000 -[500 + 400 -200] = 4000 -700 = 3300.$
