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]
$\Rightarrow P(C)=\frac{3}{10}\times \frac{17}{100}+\frac{2}{5}\times \frac{3}{25}+\frac{1}{2}\times \frac{2}{25}=\frac{139}{1000}=13.9\%$