Bag $B_1$ contains $6$ white and $4$ blue balls,Bag $B_2$ contains $4$ white and $6$ blue balls,and Bag $B_3$ contains $5$ white and $5$ blue balls. One of the bags is selected at random and a ball is drawn from it. If the ball is white,then the probability that the ball is drawn from Bag $B_2$ is:

$A$ man is known to speak the truth $3$ out of $4$ times. He throws a die and reports that it is $6$. Then the probability that it is actually $6$ is:

Let $A, B,$ and $C$ be three mutually independent events. Consider the two statements $S_1$ and $S_2$:
$S_1 : A$ and $B \cup C$ are independent.
$S_2 : A$ and $B \cap C$ are independent.
Then:

There are $3$ bags which are known to contain $2$ white and $3$ black balls; $4$ white and $1$ black balls and $3$ white and $7$ black balls respectively. $A$ ball is drawn at random from one of the bags and found to be a black ball. Then the probability that it was drawn from the bag containing the most black balls is

From a certain population, the probability of choosing a colour blind man is $\frac{1}{20}$ and that of a colour blind woman is $\frac{1}{10}$. If a randomly chosen person is found to be colour blind, then the probability that the person is a man is

Recent studies suggest that $12\%$ of the world population is left-handed. Depending on parents' hand usage, the chances of having left-handed children are as follows: $A$. Both parents are left-handed, chances of having left-handed children = $24\%$. $B$. Both parents are right-handed, chance of having left-handed children = $9\%$. $C$. Father left-handed and mother right-handed, chances of having left-handed children = $17\%$. $D$. Father right-handed and mother left-handed, chances of having left-handed children = $22\%$. Given $P(A) = P(B) = P(C) = P(D) = 1/4$ and $L$ denotes the event that the child is left-handed. What is the probability $P(A|L)$?