Question

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

• A. 3100
• B. 3300
• C. 2900
• D. 1400

Answer 1

The correct option is B 3300

$n\left(A\right)=40\%\text{of}10,000=4,000$
$n\left(B\right)=20\%\text{of}10,000=2,000$
$n\left(C\right)=10\%\text{of}10,000=1,000$
$n(A\cap B)=5\%\text{of}10,000=500$
$n(B\cap C)=3\%\text{of}10,000=300$
$n(C\cap A)=4\%\text{of}10,000=400$
$n(A\cap B\cap C)=2\%\text{of}10,000=200$
We want to find $n(A\cap B^c\cap C^c)$
$\Rightarrow n(A\cap B^c\cap C^c)=n[A\cap (B\cup C{)}^c]$.
$=n\left(A\right)-n[A\cap (B\cup C\left)\right]=n\left(A\right)-n\left[\right(A\cap B)\cup (A\cap C\left)\right]$
$=n\left(A\right)-\left[n\right(A\cap B)+n(A\cap C)-n(A\cap B\cap C\left)\right]$
$=4000-[500+400-200]=4000-700=3300$.