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,
$\mathrm{P}(\mathrm{H} \cup \mathrm{E})^{\prime}=1-\mathrm{P}(\mathrm{H} \cup \mathrm{E})$
$=1-\{\mathrm{P}(\mathrm{H})+\mathrm{P}(\mathrm{E})-\mathrm{P}(\mathrm{H} \cap \mathrm{E})\}$
$=1-\left(\frac{3}{5}+\frac{2}{5}-\frac{1}{5}\right)$
$=1-\frac{4}{5}$
$=\frac{1}{5}$
If the odds in favour of an event be $3 : 5$, then the probability of non-occurrence of the event is
The probability that a leap year selected at random contains either $53$ Sundays or $53 $ Mondays, is
Fill in the blanks in following table :
$P(A)$ | $P(B)$ | $P(A \cap B)$ | $P (A \cup B)$ |
$0.5$ | $0.35$ | ......... | $0.7$ |
The probabilities of three mutually exclusive events are $\frac{2}{3} , \frac{1}{4}$ and $\frac{1}{6}$. The statement is
A bag contains $9$ discs of which $4$ are red, $3$ are blue and $2$ are yellow. The discs are similar in shape and size. A disc is drawn at random from the bag. Calculate the probability that it will be either red or blue.