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In a hostel, $60 \%$ of the students read Hindi newspaper, $40 \%$ read English newspaper and $20 \%$ read both Hindi and English newspapers. A student is selected at random. If she reads Hindi newspaper, find the probability that she reads English newspaper.
$\frac{1}{3}$
$\frac{1}{3}$
$\frac{1}{3}$
$\frac{1}{3}$
Solution
$\mathrm{P}(\mathrm{H} \cup \mathrm{E})^{\prime}=1-\mathrm{P}(\mathrm{H} \cup \mathrm{E})$
$=1-\{\mathrm{P}(\mathrm{H})+\mathrm{P}(\mathrm{E})-\mathrm{P}(\mathrm{H} \cap \mathrm{E})\}$
$=1-\left(\frac{3}{5}+\frac{2}{5}-\frac{1}{5}\right)$
$=1-\frac{4}{5}$
$=\frac{1}{5}$
Probability that a randomly chosen student reads English newspaper, if she reads Hindi newspaper, is given by $\mathrm{P}(\mathrm{E} | \mathrm{H})$
$\mathrm{P}(\mathrm{E} | \mathrm{H})=\frac{\mathrm{P}(\mathrm{E} \,\cap \,\mathrm{H})}{\mathrm{P}(\mathrm{H})}$
$=\frac{\frac{1}{5}}{\frac{3}{5}}$
$=\frac{1}{3}$
