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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 Find the probability that she reads neither Hindi nor English newspapers.
$\frac{1}{5}$
$\frac{1}{5}$
$\frac{1}{5}$
$\frac{1}{5}$
Solution
Let $H$ denote the students who read Hindi newspaper and $E$ denote the students who read English newspaper.
It is given that, $\mathrm P(H)=60 \%=\frac{60}{100}=\frac{3}{5}$
$\mathrm{P}(\mathrm{E})=40 \%=\frac{40}{100}=\frac{2}{5}$
$P(H \cap E)=20 \%=\frac{20}{100}=\frac{1}{5}$
Probability that a student reads Hindi and English newspaper is,
$\mathrm{P}(\mathrm{H} \cup \mathrm{E})^{\prime}=1-\mathrm{P}(\mathrm{H} \cup \mathrm{E})$
$=1-\{\mathrm{P}(\mathrm{H})+\mathrm{P}(\mathrm{E})-\mathrm{P}(\mathrm{H} \cap \mathrm{E})\}$
$=1-\left(\frac{3}{5}+\frac{2}{5}-\frac{1}{5}\right)$
$=1-\frac{4}{5}$
$=\frac{1}{5}$
