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 is selected at random. Find the probability that she reads neither Hindi nor English newspapers.
A. $\dfrac{1}{5}$
B. $\dfrac{2}{5}$
C. $\dfrac{3}{5}$
D. $\dfrac{4}{5}$

Answer 1

We will first make H be the event which means students who read Hindi newspapers and E be such that they read English newspapers. Now, we need to find $P(H' \cap E')$ now using the fact that $P(H' \cap E') = 1 - P(H \cup E)$.

Complete step by step answer:
We are provided that 60% of the students read Hindi newspaper, 40% read English newspaper and 20% read both Hindi and English newspapers.
Since, we are given that 60% of the students read Hindi newspaper in a hostel.
So, P(H) = $\dfrac{{60}}{{100}}$
$\Rightarrow$ P(H) = 0.6 ……………..(1)
Since, we are also given that 40% of the students read English newspaper in a hostel.
So, P(E) = $\dfrac{{40}}{{100}}$
$\Rightarrow$ P(E) = 0.4 …………….(2)
Since, we are also given that 20% of the students read both Hindi newspaper English newspaper in a hostel.
So, $P(H \cap E)$= $\dfrac{{20}}{{100}}$
$\Rightarrow P(H \cap E) = 0.2$ ……………(3)
Now, we will use the formula: $P\left( {H \cup E} \right) = P\left( H \right) + P\left( E \right) - P\left( {H \cap E} \right)$
Putting the values in the above expression, we will then obtain:-
$\Rightarrow P\left( {H \cup E} \right) = 0.6 + 0.4 - 0.2$
Simplifying the values on the right hand side, we will then get:-
$\Rightarrow P\left( {H \cup E} \right) = 0.8$ ……………(4)
Now, we also know that by using De Morgan’s Rule, we will get:-
$\Rightarrow \left( {H \cup E} \right)' = H' \cap E'$
Taking the probability on both the sides, we will get:-
$\Rightarrow P\left( {\left( {H \cup E} \right)'} \right) = P\left( {H' \cap E'} \right)$
Now, we know that for any event A, we have: P(A’) = 1 – P(A).
$\Rightarrow 1 - P\left( {H \cup E} \right) = P\left( {H' \cap E'} \right)$
Putting the values using the equation (4), we will get:-
$\Rightarrow P\left( {H' \cap E'} \right) = 1 - 0.8$
Simplifying the values on the right hand side, we will get:-
$\Rightarrow P\left( {H' \cap E'} \right) = 0.2$
We can write it as:-
$\Rightarrow P\left( {H' \cap E'} \right) = \dfrac{2}{{10}} = \dfrac{1}{5}$

Hence, the correct option is (A).

Note: The students must notice that the percentage is converted in probability by dividing the percentage by 100 and thus we get the probabilities as 0.6, 0.4 and 0.2.
The students must note that they must commit to the memory the following formulas:-
For any event A, we have: P(A’) = 1 – P(A), where A’ represents A complement.
For any two event H and E, we have: $\left( {H \cup E} \right)' = H' \cap E'$
For any two event H and E, we have : $P\left( {H \cup E} \right) = P\left( H \right) + P\left( E \right) - P\left( {H \cap E} \right)$