$P(A / A \cap B) + P(B / A \cap B) =$

$A$ 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 $N$ balls out of which $3$ are white,$6$ are green,and the remaining $(N-9)$ balls are blue. Three balls are drawn randomly one after the other without replacement. Let $W_i, G_i$,and $B_i$ denote the events that the ball drawn in the $i^{\text{th}}$ draw is white,green,and blue,respectively. If $P(W_1 \cap G_2 \cap B_3) = \frac{2}{5N}$ and $P(B_3 \mid W_1 \cap G_2) = \frac{2}{9}$,then $N$ equals:

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:

$A$ black and a red dice are rolled. Find the conditional probability of obtaining the sum $8$,given that the red die resulted in a number less than $4$.

If $P(A)=0.8, P(B)=0.5$ and $P(B | A)=0.4,$ find $P(A | B).$