Question

In a hostel, 60% of the students read Hindi newspapers, 40% read English newspapers and 20% read both Hindi and English newspapers. A student selected at random:
(a) Find the probability that she reads neither Hindi nor English newspapers.
(b) If she reads Hindi newspapers, find the probability that she reads English newspapers.
(c) If she reads English newspaper. Find the probability that she reads Hindi newspapers.

Answer 1

- Hint: In this question, from the given percentage we can write the probability for a particular event. Then by using the corresponding formulae we can find the probability for the conditions being asked in the question.
\[\begin{align}
  & P\left( A\cup B \right)=P\left( A \right)+P\left( B \right)-P\left( A\cap B \right) \\
 & P\left( \bar{A}\cap \bar{B} \right)=1-P\left( A\cup B \right) \\
\end{align}\]
\[P\left( A/B \right)=\dfrac{P\left( A\cap B \right)}{P\left( B \right)}\]


Complete step-by-step answer:
PROBABILITY:
If there are n elementary events associated with a random experiment and m of them are favourable to an event A, then the probability of happening or occurrence of A, denoted by P(A), is given by
\[P\left( A \right)=\dfrac{m}{n}=\dfrac{\text{Number of favourable outcomes}}{\text{Total number of possible outcomes }}\]
Let us now look at some of the required formulae
\[\begin{align}
  & P\left( A\cup B \right)=P\left( A \right)+P\left( B \right)-P\left( A\cap B \right) \\
 & P\left( \bar{A}\cap \bar{B} \right)=1-P\left( A\cup B \right) \\
\end{align}\]
Conditional Probability: Let A and B be two events associated with a random experiment. Then, the probability of occurrence of event A under the condition that B has already occurred is called the conditional probability.
\[P\left( A/B \right)=\dfrac{P\left( A\cap B \right)}{P\left( B \right)}\]
Let us now assume that the reading of Hindi newspapers as A and the event of reading English newspapers as B.

From this, we can write the probability for each event according to the given question as:
\[P\left( A \right)=\dfrac{60}{100}=0.6\]
\[\begin{align}
  & P\left( B \right)=\dfrac{40}{100}=0.4 \\
 & P\left( A\cap B \right)=\dfrac{20}{100}=0.2 \\
\end{align}\]
Now, the event that she reads neither Hindi nor English newspaper can be written as:
\[\Rightarrow P\left( \bar{A}\cap \bar{B} \right)=1-P\left( A\cup B \right)\]
Now, by again substituting the above formula in this we get,
\[\Rightarrow P\left( \bar{A}\cap \bar{B} \right)=1-\left( P\left( A \right)+P\left( B \right)-P\left( A\cap B \right) \right)\]
Let us now substitute the respective probabilities in the above equation.
\[\Rightarrow P\left( \bar{A}\cap \bar{B} \right)=1-\left( 0.6+0.4-0.2 \right)\]
Now, on further simplification we get,
\[\begin{align}
  & \Rightarrow P\left( \bar{A}\cap \bar{B} \right)=1-\left( 1-0.2 \right) \\
 & \therefore P\left( \bar{A}\cap \bar{B} \right)=0.2 \\
\end{align}\]
Now, the probability of reading English when she already read Hindi newspaper is
\[\Rightarrow P\left( B/A \right)=\dfrac{P\left( A\cap B \right)}{P\left( A \right)}\]
Now, by substituting the respective values in the above formula we get,
\[\Rightarrow P\left( B/A \right)=\dfrac{0.2}{0.6}\]
Now, on further simplification we get,
\[\therefore P\left( B/A \right)=\dfrac{1}{3}\]
Now, the probability of reading Hindi when she already read English newspaper is
\[\Rightarrow P\left( A/B \right)=\dfrac{P\left( A\cap B \right)}{P\left( B \right)}\]
Now, by substituting the respective values in the above formula we get,
\[\Rightarrow P\left( A/B \right)=\dfrac{0.4}{0.6}\]
Now, on further simplification we get,
\[\therefore P\left( A/B \right)=\dfrac{2}{3}\]

Note: It is important to note that the probability of her reading English newspaper after reading the Hindi newspaper should be done according to the conditional probability because it is not the case of reading both the Hindi and English newspaper.
It is noted that as the given conditions are for 100% we can find that the students who read both the newspapers is equal to the number of students who read neither of the newspapers.