Question

In a town of 10,000 families it was found that 40% families buy newspaper A, 20% families buy newspaper B and 10% families buy newspaper C, 5% families buy A and B, 3 % buy B and C, 4% buy A and C and 2% families buy all the three news papers. If probability of families to buy only newspaper B is $\frac{\lambda }{100}$, then find the value of $\lambda$

Answer 1

$n\left(A\right)=\frac{40}{100},n\left(B\right)=\frac{20}{100},n\left(C\right)=\frac{10}{100}n(A\cap B)=\frac{5}{100},n(B\cap C)=\frac{3}{100},n(A\cap C)=\frac{4}{100}n(A\cap B\cap C)=\frac{2}{100}$
Then required probability is
$=n\left(B\right)-n(A\cap B)-n(B\cap C)+n(A\cap B\cap C)=\frac{20}{100}-\frac{5}{100}-\frac{3}{100}+\frac{2}{100}=\frac{14}{100}$