Question

Two newspapers A and B are published in a city. It is known that 25% of the city population reads A and 20% reads B while 8% reads both A and B. Further, 30% of those who read A but not B look into advertisements and 40% of those who read B but not A also look into advertisements while 50% of those who read A and B look into advertisements. Find the percentage of the population who look into the advertisements?
(a) 12.8
(b) 13.5
(c) 13.9
(d) 13

Answer 1

Hint: We start solving the problem by finding the probability of the individual events. We then find the probability of the population such that they read only A but not B, the probability of the population such that they read only B but not A. We then use these values and calculate the probability of the population who look into advertisements and then convert them into percentages.

Complete step-by-step answer:
According to the problem we have two newspapers A and B that were published in a given city.
It is given that 25% of the city population reads newspaper A. So, we get a probability of population reading paper A as $P\left( A \right)=\dfrac{25}{100}$.
$P\left( A \right)=\dfrac{1}{4}$ ---(1).
It is given that 20% of the city population reads newspaper B. So, we get a probability of the population reading paper B as \[P\left( B \right)=\dfrac{20}{100}\].
$P\left( B \right)=\dfrac{1}{5}$ ---(2).
It is given that 8% of the city population reads both newspapers A and B. So, we get a probability of the population reading both newspaper A and B is \[P\left( A\cap B \right)=\dfrac{8}{100}\].
\[P\left( A\cap B \right)=\dfrac{2}{25}\] ---(3).
Now, we find the probability of the population who reads newspaper A but not B. We have to find the value of $P\left( A\cap {{B}^{c}} \right)$.
We know that $P\left( A\cap {{B}^{c}} \right)=P\left( A \right)-P\left( A\cap B \right)$.
From equations (1) and (3) we get,
$\Rightarrow P\left( A\cap {{B}^{c}} \right)=\dfrac{1}{4}-\dfrac{2}{25}$.
$\Rightarrow P\left( A\cap {{B}^{c}} \right)=\dfrac{25-\left( 4\times 2 \right)}{4\times 25}$.
$\Rightarrow P\left( A\cap {{B}^{c}} \right)=\dfrac{25-8}{100}$.
$\Rightarrow P\left( A\cap {{B}^{c}} \right)=\dfrac{17}{100}$ ---(4).
Now, we find the probability of the population who reads newspaper B but not A. We have to find the value of $P\left( {{A}^{c}}\cap B \right)$.
We know that $P\left( {{A}^{c}}\cap B \right)=P\left( B \right)-P\left( A\cap B \right)$.
From equations (2) and (3) we get,
$\Rightarrow P\left( {{A}^{c}}\cap B \right)=\dfrac{1}{5}-\dfrac{2}{25}$.
$\Rightarrow P\left( {{A}^{c}}\cap B \right)=\dfrac{5-2}{25}$.
$\Rightarrow P\left( {{A}^{c}}\cap B \right)=\dfrac{3}{25}$ ---(5).
According to the problem, it is given that 30% of those who read A but not B look into advertisements and 40% of those who read B but not A also look into advertisements while 50% of those who read A and B look into advertisements. We need to find the percentage of the population who look into the advertisements and let us represent this even with C.
So, we have $P\left( C \right)=\left( 30\%\text{ of }P\left( A\cap {{B}^{c}} \right) \right)+\left( 40\%\text{ of }P\left( A\cap {{B}^{c}} \right) \right)+\left( 50\%\text{ of }P\left( A\cap B \right) \right)$.
From equations (3), (4) and (5) we get,
$\Rightarrow P\left( C \right)=\left( 30\%\text{ of }\dfrac{17}{100} \right)+\left( 40\%\text{ of }\dfrac{3}{25} \right)+\left( 50\%\text{ of }\dfrac{2}{25} \right)$.
$\Rightarrow P\left( C \right)=\left( \dfrac{30}{100}\times \dfrac{17}{100} \right)+\left( \dfrac{40}{100}\times \dfrac{3}{25} \right)+\left( \dfrac{50}{100}\times \dfrac{2}{25} \right)$.
$\Rightarrow P\left( C \right)=\left( \dfrac{510}{10000} \right)+\left( \dfrac{120}{2500} \right)+\left( \dfrac{100}{2500} \right)$.
$\Rightarrow P\left( C \right)=\left( \dfrac{510+\left( 4\times 120 \right)+\left( 4\times 100 \right)}{10000} \right)$.
$\Rightarrow P\left( C \right)=\left( \dfrac{510+480+400}{10000} \right)$.
$\Rightarrow P\left( C \right)=\left( \dfrac{1390}{10000} \right)$.
$\Rightarrow P\left( C \right)=0.1390$.
$\Rightarrow P\left( C \right)=0.1390\times 100\%$.
$\Rightarrow P\left( C \right)=13.90\%$.
We have found the percentage of the population in the city who look into the advertisements as 13.90%.
∴ The percentage of the population in the city who look into the advertisements is 13.90%.
The correct option for the given problem is (c).

Note: We can also solve this problem by taking the whole population of the city as 100 and drawing venn diagrams. In a venn diagram we can visualize the given information in a better way. Whenever we get this type of problem, we should start by writing the individual events first and then start solving the problems. We can also expect problems to find the percentage of people who don't read newspapers and don't look into advertisements.