(a) Let $\mathrm{H}$ and $\mathrm{E}$ represent the event that a student reads
Hindi and English newspaper respectively $\mathrm{P}(\mathrm{H})=0.6, \mathrm{P}(\mathrm{E})=0.4, \mathrm{P}(\mathrm{H} \cap \mathrm{E})=0.2$
Probability that the student reads at least one paper $=\mathrm{P}(\mathrm{H} \cup \mathrm{E})$
Now $\mathrm{P}(\mathrm{H})=0 \cdot 6, \mathrm{P}(\mathrm{E})=0 \cdot 4, \mathrm{P}(\mathrm{H} \cap \mathrm{E})=0 \cdot 2$
$\therefore \quad \mathrm{P}(\mathrm{H} \cup \mathrm{E})=0 \cdot 6+0 \cdot 4-0 \cdot 2=1-0.2=0.8$
$\therefore$ Probability that a student reads neither Hindi nor English newspaper
$=1-\mathrm{P}(\mathrm{H} \cup \mathrm{E})=1-0 \cdot 8=0 \cdot 2$
(b) The probability that the student reads English newspaper if she reads Hindi
$$
=\mathrm{P}(\mathrm{E} / \mathrm{H})=\frac{\mathrm{P}(\mathrm{E} \cap \mathrm{H})}{\mathrm{P}(\mathrm{H})}
$$
Now $\mathrm{P}(\mathrm{E} \cap \mathrm{H})=0 \cdot 2, \mathrm{P}(\mathrm{H})=0 \cdot 6$
$$
\therefore \mathrm{P}(\mathrm{E} / \mathrm{H})=\frac{0 \cdot 2}{0 \cdot 6}=\frac{1}{3}
$$
(c) The probability that she reads Hindi newspaper if she reads English newspaper
$$
\begin{aligned}
&=\mathrm{P}(\mathrm{H} / \mathrm{E})=\mathrm{P}(\mathrm{H} \cap \mathrm{E})=0 \cdot 2, \mathrm{P}(\mathrm{E})=0.4 \\
&\therefore \mathrm{P}(\mathrm{H} / \mathrm{E})=\frac{\mathrm{P}(\mathrm{H} \cap \mathrm{E})}{\mathrm{P}(\mathrm{E})}=\frac{0 \cdot 2}{0 \cdot 4}=\frac{1}{2}
\end{aligned}
$$