Answer

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**Hint:**From the given data we will write the given percentages as the probability of reading those newspapers. So, we will get the probabilities for reading newspaper $A$ and newspaper $B$ along with the probability of people reading both the newspapers. Now we will calculate the probabilities of visiting advertisements from the given conditions and obtained values. After getting the probabilities of visiting advertisements for different conditions, we will calculate the required probability by using the relation between the probability of visiting advertisements to the required probability.

**Complete step by step answer:**

Given that,

$25\%$ of the city population reads $A$

$\therefore $ Probability of reading paper $A$ is $P\left( A \right)=25\%=\dfrac{1}{4}$

$20\%$ of the population reads $B$

$\therefore $ Probability of reading paper $B$ is $P\left( B \right)=20\%=\dfrac{1}{5}$

$8\%$ of the population reads both $A$ and $B$.

$\therefore $ Probability of reading both the papers $A$ and $B$ is $P\left( A\cap B \right)=8\%=\dfrac{2}{25}$

Now we need to calculate the probability of the people who read newspaper $A$ but not $B$ and also the probability of the people who read newspaper $B$ but not $A$.

We know that $P\left( A\cap \bar{B} \right)=P\left( A \right)-P\left( A\cap B \right)$ where $\bar{B}$ is the complementary event for the event $B$.

$\therefore $ The probability of the people who read newspaper $A$ but not $B$ is given by

$\begin{align}

& P\left( A\cap \bar{B} \right)=P\left( A \right)-P\left( A\cap B \right) \\

& \Rightarrow P\left( A\cap \bar{B} \right)=\dfrac{1}{4}-\dfrac{2}{25} \\

& \Rightarrow P\left( A\cap \bar{B} \right)=\dfrac{25-2\times 4}{4\times 25} \\

& \Rightarrow P\left( A\cap \bar{B} \right)=\dfrac{17}{100} \\

\end{align}$

Now the probability of the people who read newspaper $B$ but not $A$ is given by

$\begin{align}

& P\left( \bar{A}\cap B \right)=P\left( B \right)-P\left( A\cap B \right) \\

& \Rightarrow P\left( \bar{A}\cap B \right)=\dfrac{1}{5}-\dfrac{2}{25} \\

& \Rightarrow P\left( \bar{A}\cap B \right)=\dfrac{1\times 5-2}{25} \\

& \Rightarrow P\left( \bar{A}\cap B \right)=\dfrac{3}{25} \\

\end{align}$

Let $C$ be the event that the people visit the advertisement in the newspaper, then the probability of the people who reads advertisements is given by $P\left( C \right)$.

In the problem they have mentioned that the probability of the people who reads advertisements is equal to ‘$30\%$ of those who read $A$ but not $B$ look into advertisements and $40\%$ of those who read $B$ but not $A$ look advertisements while $50\%$ of those who read both $A$ and $B$ look into advertisements’. Mathematically we can write

$P\left( C \right)=30\%\text{ of }P\left( A\cap \bar{B} \right)+40\%\text{ of }P\left( \bar{A}\cap B \right)+50\%\text{ of }P\left( A\cap B \right)$

Substituting the all the values we have, then we will get

$\begin{align}

& P\left( C \right)=30\%\times \dfrac{17}{100}+40\%\times \dfrac{3}{25}+50\%\times \dfrac{2}{25} \\

& \Rightarrow P\left( C \right)=\dfrac{30}{100}\times \dfrac{17}{100}+\dfrac{40}{100}\times \dfrac{3}{25}+\dfrac{50}{100}\times \dfrac{2}{25} \\

& \Rightarrow P\left( C \right)=\dfrac{510}{10000}+\dfrac{120}{2500}+\dfrac{1}{25} \\

& \Rightarrow P\left( C \right)=\dfrac{139}{1000} \\

\end{align}$

**So, the correct answer is “Option C”.**

**Note:**If you are not comfortable with the percentages, you can assume that $100$ peoples are in that city. Now we can get the number of people who reads newspaper $A$ and number of people who reads newspaper $B$ and number of people who reads both $A$ and $B$. From these values we can easily solve the problem.

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