Question

In a certain city only two newspapers $A$ and $B$ are published, it is known that $25\%$ of the city population reads $A$ and $20\%$ reads $B$, while $8\%$ reads both $A$ and $B$. It is also known that $30\%$ of those who read $A$ but not $B$ look into advertisements and $40\%$ of those who read $B$ but not $A$ look into advertisements while $50\%$ of those who read both $A$ and $B$ look into advertisements. What is the percentage of the population reads an advertisement?

Answer 1

Let $P(A)$ and $P(B)$ denote respectively the percentage of city population that reads newspapers $A$ and $B$. Then,
$ P(A)=\frac{25}{100}=\frac{1}{4},P(B)=\frac{20}{100}=\frac{1}{5},$
$P(A\cap B) =\frac{8}{100}=\frac{2}{25} , $
$P ( A \cap B) = P(A) - P ( A \cap B ) =\frac{1}{4}-\frac{2}{25}=\frac{17}{100},$
$P ( \overline{A} \cap B) = P(B) - P ( A \cap B ) =\frac{1}{5}-\frac{2}{25}=\frac{3}{25},$
Let $P(C)$ be the probability that the population who reads advertisements.
$\therefore P(C) 30\%$ of $P(A\cap \overline{B}) + 40\%$ of $P( \overline{A} \cap B)$
$ + 50\%$ of $P(A \cap B)$
[since, $A \cap \overline{B}, \overline{A} \cap B$ and $ A \cap B$ are all mutually exclusive]
$\Rightarrow P(C)=\frac{3}{10} \times \frac{17}{100} + \frac{2}{5} \times \frac{3}{25} +\frac{1}{2} \times \frac{2}{25} =\frac{139}{1000}=13.9\%$