Question

In a town of $10,000$ 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 three newspapers, find number of families which buy None of $A,B,C$.

Answer 1

Given $N=10,000,n\left(A\right)=4000,n\left(B\right)=2000,n\left(C\right)=1000,n(A\cap B)=500,$

$n(B\cap C)=300,n(C\cap A)=400,n(A\cap B\cap C)=200$
$n(A^{\prime }\cap B^{\prime }\cap C^{\prime })=n\left(\right(A\cup B\cup C{)}^{\prime })$
$=N-n(A\cup B\cup C)$
$=N-\left[n\right(A)+n(B)+n(C)-n(A\cap B)-n(B\cap C)-n(A\cap C)+n(A\cap B\cap C\left)\right]$
$=10000-[4000+2000+1000-500-300-400+200]=4000$