Question

In a town of $10000$ families, it was found that $40\%$ families buy newspaper $A$, $20\%$ buy newspaper $B$ and $10\%$ 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 newspaper $A$ only is

Answer 1

$n\left(A\right)=40\%\text{of}10000=4000;$
$n\left(B\right)=2000;$
$n\left(C\right)=1000$
$n(A\cap B)=500;$
$n(B\cap C)=300;$
$n(C\cap A)=400;$
$n(A\cap B\cap C)=200$

$\therefore n(A\cap B^C\cap C^C)=n(A\cap (B\cup C{)}^C)=n(A)-n(A\cap (B\cup C))$
$=n\left(A\right)-n\left(\right(A\cap B)\cup (A\cap C\left)\right)$
$=n\left(A\right)-\left[n\right(A\cap B)+n(A\cap C)-n(A\cap B\cap C\left)\right]=3300$