Given:
The letters $H$ and $E$ stand for the number of students who read the Hindi newspaper and the English daily, respectively.
Hence, $\mathrm{P}(\mathrm{H})=$ Probability of students who read Hindi newspaper $=60 / 100=3 / 5$
$P(E)=$ Probability of students who read English newspaper $=40 / 100=2 / 5$
$P(H \cap E)=$ Probability of students who read Hindi and English both newspaper $=20 / 100$
$=1 / 5$
(a) Find the probability that she reads neither Hindi nor English newspapers.
$P$ (neither H nor E)
$P($ neither $H$ nor $E)=P\left(H^{\prime} \cap E^{\prime}\right)$
As, $\left{\mathrm{H}^{\prime} \cap \mathrm{E}^{\prime}=(\mathrm{H} \cup \mathrm{E})^{\prime}\right}$
\&
$\Rightarrow P$ (neither $A$ nor $B)=P\left((H \cup E)^{\prime}\right)$
$=1-P(H \cup E)$
$=1-[\mathrm{P}(\mathrm{H})+\mathrm{P}(\mathrm{E})-\mathrm{P}(\mathrm{H} \cap \mathrm{E})]$
$=1-\left[\frac{3}{5}+\frac{2}{5}-\frac{1}{5}\right]$
$=1-\left[\frac{4}{5}\right]=\frac{1}{5}$
(b) If she reads a Hindi newspaper, calculate the likelihood that she also reads an English daily..
$P(E \mid H)=$ Hindi newspaper reading has already occurred and the probability that she reads English newspaper is to find.
As we know $P(E \mid H)=\frac{P(H \cap E)}{P(H)}$
$\Rightarrow \mathrm{P}(\mathrm{E} \mid \mathrm{H})=\frac{\frac{1}{3}}{\frac{3}{5}}=\frac{1}{5} \times \frac{5}{3}$
$\Rightarrow \mathrm{P}(\mathrm{E} \mid \mathrm{H})=1 / 3$
(c) If she reads English newspaper, find the probability that she reads Hindi newspaper.
$P(\mathrm{H} \mid \mathrm{E})=$ English newspaper reading has already occurred and the probability that she reads Hindi newspaper is to find.
As we know $P(H \mid E)=\frac{P(H \cap E)}{P(E)}$
$\Rightarrow P(\mathrm{H} \mid \mathrm{E})=\frac{\frac{1}{2}}{\frac{5}{5}}=\frac{1}{5} \times \frac{5}{2}$
$\Rightarrow \mathrm{P}(\mathrm{H} \mid \mathrm{E})=1 / 2$