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. If a person is chosen at random from the population, what is the percentage probability that he/she reads advertisements?

• A. $\frac{139}{1000}$
• B. $\frac{77}{250}$
• C. $\frac{481}{2000}$
• D. $\frac{61}{500}$

Answer 1

The correct option is A $\frac{139}{1000}$

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