$A$ bag contains four balls. Two balls are drawn randomly and found to be white. The probability that all the balls in the bag are white is

Bag $I$ contains $3$ red and $4$ black balls,while Bag $II$ contains $5$ red and $6$ black balls. One ball is drawn at random from one of the bags and it is found to be red. Find the probability that it was drawn from Bag $II$.

Let $U_1$ and $U_2$ be two urns such that $U_1$ contains $3$ white and $2$ red balls,and $U_2$ contains only $1$ white ball. $A$ fair coin is tossed. If head appears,then $1$ ball is drawn at random from $U_1$ and put into $U_2$. However,if tail appears,then $2$ balls are drawn at random from $U_1$ and put into $U_2$. Now $1$ ball is drawn at random from $U_2$.
$1.$ The probability of the drawn ball from $U_2$ being white is
$(A)$ $\frac{13}{30}$ $(B)$ $\frac{23}{30}$ $(C)$ $\frac{19}{30}$ $(D)$ $\frac{11}{30}$
$2.$ Given that the drawn ball from $U_2$ is white,the probability that head appeared on the coin is
$(A)$ $\frac{17}{23}$ $(B)$ $\frac{11}{23}$ $(C)$ $\frac{15}{23}$ $(D)$ $\frac{12}{23}$
Give the answer for question $1$ and $2.$

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

$A$ bag contains $(N+1)$ coins: $N$ fair coins, and one coin with 'Head' on both sides. $A$ coin is selected at random and tossed. If the probability of getting 'Head' is $\frac{9}{16}$, then $N$ is equal to:

An urn contains $5$ balls. Two balls are drawn at random and they are found to be white. The probability that all the balls in the urn are white is: