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 English newspaper, find the probability that she reads Hindi newspaper.
$\mathrm{P}(\mathrm{H} \cup \mathrm{E})^{\prime}=1-\mathrm{P}(\mathrm{H} \cup \mathrm{E})$
$=1-\{\mathrm{P}(\mathrm{H})+\mathrm{P}(\mathrm{E})-\mathrm{P}(\mathrm{H} \cap \mathrm{E})\}$
$=1-\left(\frac{3}{5}+\frac{2}{5}-\frac{1}{5}\right)$
$=1-\frac{4}{5}$
$=\frac{1}{5}$
Probability that a random chosen student reads Hindi newspaper, if she reads English newspaper, is given by $\mathrm{P}(\mathrm{H} | \mathrm{E})$
$\mathrm{P}(\mathrm{H} | \mathrm{E})=\frac{\mathrm{P}(\mathrm{H} \cap E)}{\mathrm{P}(\mathrm{E})}$
$=\frac{\frac{1}{5}}{\frac{2}{5}}$
$=\frac{1}{2}$
Let two fair six-faced dice $A$ and $B$ be thrown simultaneously. If $E_1$ is the event that die $A$ shows up four, $E_2 $ is the event that die $B$ shows up two and $E_3$ is the event that the sum of numbers on both dice is odd, then which of the following statements is NOT true $?$
Fill in the blanks in following table :
$P(A)$ | $P(B)$ | $P(A \cap B)$ | $P (A \cup B)$ |
$\frac {1}{3}$ | $\frac {1}{5}$ | $\frac {1}{15}$ | ........ |
In two events $P(A \cup B) = 5/6$, $P({A^c}) = 5/6$, $P(B) = 2/3,$ then $A$ and $B$ are
Probability of solving specific problem independently by $A$ and $B$ are $\frac{1}{2}$ and $\frac{1}{3}$ respectively. If both try to solve the problem independently, find the probability that the problem is solved.
If $P(A) = P(B) = x$ and $P(A \cap B) = P(A' \cap B') = \frac{1}{3}$, then $x = $