If two events $A$ and $B$ are such that $P(\overline{A}) = 0.3$,$P(B) = 0.4$,and $P(A \cap \overline{B}) = 0.5$,then $P(B | (A \cup \overline{B})) = $

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:

$A$ bag contains $4$ Red and $6$ Black balls. $A$ ball is drawn at random from the bag,its colour is observed and this ball along with $3$ additional balls of the same colour are returned to the bag. If now a ball is drawn at random from the bag,then the probability that this drawn ball is red is

$A$ bag contains $ 17 $ tickets numbered from $ 1 $ to $ 17 $. $A$ ticket is drawn at random,then another ticket is drawn without replacing the first one. The probability that both the tickets show even numbers is

If $P(A) = \frac{7}{13}$,$P(B) = \frac{9}{13}$ and $P(A \cap B) = \frac{4}{13}$,evaluate $P(A | B)$.

$A$ die is thrown three times and the sum of the numbers obtained is $15$. What is the probability that the number $4$ appears in the first throw?