$A$ box contains $4$ black,$2$ white,and $6$ red balls. Another box contains $3$ black and $5$ white balls. An unbiased die is thrown. If $1$ or $2$ appears on the die,a ball is drawn from the first box; otherwise,a ball is drawn from the second box. If the drawn ball is black,what is the probability that $2$ appeared on the die?

Bag $A$ contains $2$ white and $3$ red balls and bag $B$ contains $4$ white and $5$ red balls. If one ball is drawn at random from one of the bags and is found to be red,then the probability that it was drawn from the bag $B$ is

Three boxes $B_1$,$B_2$ and $B_3$ contain balls with different colors as follows:

Box	White,Black,Red
$B_1$	$2, 1, 2$
$B_2$	$3, 2, 4$
$B_3$	$4, 3, 2$

$A$ die is thrown. Box $B_1$ is chosen if either $1$ or $2$ turns up. Box $B_2$ is chosen if $3$ or $4$ turns up and box $B_3$ is chosen if $5$ or $6$ turns up. Having chosen a box in this way,a ball is drawn at random from that box. If the ball drawn is found to be Red,then the probability that it is drawn from box $B_2$ is

$A$ person is known to speak the truth in $3$ out of $4$ occasions. If he throws a die and reports that it is six,then the probability that it is actually six is

$A$ box $A$ contains $2$ white,$3$ red and $2$ black balls. Another box $B$ contains $4$ white,$2$ red and $3$ black balls. If two balls are drawn at random,without replacement,from a randomly selected box and one ball turns out to be white while the other ball turns out to be red,then the probability that both balls are drawn from box $B$ 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 $P(A|L)$?