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.

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,

$\mathrm{P}(\mathrm{H} \cup \mathrm{E})^{\prime}=1-\mathrm{P}(\mathrm{H} \cup \mathrm{E})$

$=1-\{\mathrm{P}(\mathrm{H})+\mathrm{P}(\mathrm{E})-\mathrm{P}(\mathrm{H} \cap \mathrm{E})\}$

$=1-\left(\frac{3}{5}+\frac{2}{5}-\frac{1}{5}\right)$

$=1-\frac{4}{5}$

$=\frac{1}{5}$

$A$ and $B$ are two events such that $P(A)=0.54$, $P(B)=0.69$ and $P(A \cap B)=0.35.$ Find $P \left( A ^{\prime} \cap B ^{\prime}\right)$.

If $A$ and $B$ are two events such that $P\,(A \cup B) = P\,(A \cap B),$ then the true relation is

- [IIT 1998]

Given two independent events $A$ and $B$ such $P(A)=0.3,\,P(B)=0.6 .$ Find $P($ neither $A$or $B)$

If $P(A)=\frac{3}{5}$ and $P(B)=\frac{1}{5},$ find $P(A \cap B)$ if $A$ and $B$ are independent events

In a city $20\%$ persons read English newspaper, $40\%$ read Hindi newspaper and $5\%$ read both newspapers. The percentage of non-reader either paper is