Question

$60\%$ students read Hindi newspapers, $40\%$ students read Tamil newspapers and $20\%$ students read both Hindi and Tamil newspapers. Find the probability that a student selected at random reads
(i) Tamil newspaper given that he has already read Hindi newspapers.
(ii) Hindi newspaper given that he has already read Tamil newspapers.
(iii) Neither Hindi nor Tamil newspaper.
What values are being promoted in this question?

Answer 1

Hint : This question is based on probability. The percentage of students given in the question is the value of the probability for that group of students because we are considering the total number of the students as $100$ .

Complete step-by-step answer :
Let us assume the number of students who read Hindi newspapers are $H$ and the number of students who read Tamil newspapers are $T$ .
Then, the probability of students who read Hindi newspaper
$\begin{array}{c} P\left( H \right) = 60\% \\\ = \dfrac{{60}}{{100}}\\\ = 0.6 \end{array}$
The probability of students who read Tamil newspaper
$\begin{array}{c} P\left( T \right) = 40\% \\\ = \dfrac{{40}}{{100}}\\\ = 0.4 \end{array}$
And, the probability of students who read both Hindi and Tamil newspaper
$\begin{array}{c} P\left( {H \cap T} \right){\rm{ or }}P\left( {T \cap H} \right) = 20\% \\\ = \dfrac{{20}}{{100}}\\\ = 0.2 \end{array}$
(i) The probability that a student selected at random reads Tamil newspaper given that he has already read Hindi newspaper is given by-
$P\left( {T/H} \right) = \dfrac{{P\left( {T \cap H} \right)}}{{P\left( H \right)}}$
Substituting the values, we get,
$\begin{array}{c} P\left( {T/H} \right) = \dfrac{{0.2}}{{0.6}}\\\ = \dfrac{1}{3} \end{array}$
Therefore, the probability that a student selected at random reads Tamil newspaper given that he has already read Hindi newspaper is $\dfrac{1}{3}$ .

(ii) The probability that a student selected at random reads Hindi newspaper given that he has already read Tamil newspaper is given by-
$P\left( {H/T} \right) = \dfrac{{P\left( {H \cap T} \right)}}{{P\left( T \right)}}$
Substituting the values, we get,
$\begin{array}{c} P\left( {H/T} \right) = \dfrac{{0.2}}{{0.4}}\\\ = \dfrac{1}{2} \end{array}$
Therefore, the probability that a student selected at random reads Hindi newspaper given that he has already read Tamil newspaper is $\dfrac{1}{2}$ .

(iii) The probability that a student selected at random reads Neither Hindi nor Tamil newspaper is given by-
$P\left( {H' \cap T'} \right) = 1 - P\left( {H \cap T} \right)$
Substituting the value, we get,
$\begin{array}{c} P\left( {H' \cap T'} \right) = 1 - 0.2\\\ = 0.8 \end{array}$
Therefore, the probability that a student selected at random reads Neither Hindi nor Tamil newspaper is $0.8$ .

Note : In the solution we have used the Conditional Probability. This type of probability is calculated when the occurrence of one event depends upon another event. For example, in a random experiment there are two events $A{\rm{ and }}B$ . If the event $B$ has already happened and now we have to calculate the probability of happening of event $A$ then it is a conditional probability of event $A$ given event $B$ .
The conditional probability of event $A$ given event $B$ is given by-
$P\left( {A/B} \right) = \dfrac{{P\left( {A \cap B} \right)}}{{P\left( B \right)}}$