If $P(A / B) = \frac{3}{10}$, $P(B / A) = \frac{4}{5}$ and $P(A \cup B) = K P(B)$, then $\frac{1}{K} =$

$A, B$ are the events in a random experiment. If $P(A)=\frac{1}{2}, P(B)=\frac{1}{3}, P(A \cap B)=\frac{1}{4}$, then $P\left(\frac{A^{c}}{B^{c}}\right)+P\left(\frac{A}{B}\right)=$

For a biased die,the probabilities for different faces to turn up are given below:

$Face$	$1$	$2$	$3$	$4$	$5$	$6$
$Probability$	$0.1$	$0.32$	$0.21$	$0.15$	$0.05$	$0.17$

The die is tossed and you are told that either face $1$ or $2$ has turned up. Then the probability that it is face $1$ is:

In a group of $14$ males and $6$ females, $8$ males and $3$ females are aged above $40 \text{ yr}$. The probability that a person selected at random from the group is aged above $40 \text{ yr}$, given that the selected person is a female, is:

Let $E_{1}$ and $E_{2}$ be two events such that the conditional probabilities $P(E_{1} \mid E_{2}) = \frac{1}{2}$,$P(E_{2} \mid E_{1}) = \frac{3}{4}$ and $P(E_{1} \cap E_{2}) = \frac{1}{8}$. Then:

$A$ fair die is rolled. Consider events $E=\{1,3,5\}, F=\{2,3\}$ and $G=\{2,3,4,5\}$. Find $P(E | F)$ and $P(F | E)$.