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 newspapers.
If probability of families to buy only newspaper $B$ is $\frac{\lambda }{100},$ then the value of $\lambda =$

Answer 1

Given,
$P\left(A\right)=\frac{40}{100},P\left(B\right)=\frac{20}{100},P\left(C\right)=\frac{10}{100},P(A\cap B)=\frac{5}{100},P(B\cap C)=\frac{3}{100},P(A\cap C)=\frac{4}{100},P(A\cap B\cap C)=\frac{2}{100}$
Then the required probability, $P\left(\text{only}B\right)$
$=P\left(B\right)-P(A\cap B)-P(B\cap C)+P(A\cap B\cap C)=\frac{20}{100}-\frac{5}{100}-\frac{3}{100}+\frac{2}{100}=\frac{14}{100}$
So, $\lambda =14$