There are three bags $B_1$,$B_2$,and $B_3$ containing $2$ Red and $3$ White,$5$ Red and $5$ White,and $3$ Red and $2$ White balls respectively. $A$ ball is drawn from bag $B_1$ and placed in bag $B_2$,then a ball is drawn from bag $B_2$ and placed in bag $B_3$,then a ball is drawn from bag $B_3$. The number of ways in which this process can be completed,if the same colour balls are used in the first and second transfers (assume all balls to be distinct),is

$12$ balls are distributed among $3$ boxes. The probability that the first box will contain exactly $3$ balls is:

If $A$ and $B$ are two independent events such that $P(B)=\frac{2}{7}$ and $P\left(A \cup B^c\right)=0.8$, then $P(A \cup B)$ $=$

$A$ and $B$ throw a die alternatively till one of them gets a '$6$' and wins the game. Find their respective probabilities of winning,if $A$ starts first.

$A$ bag contains $4$ red and $5$ black balls. Another bag contains $3$ red and $6$ black balls. If one ball is drawn from the first bag and two balls are drawn from the second bag at random,the probability that out of the three balls,two are black and one is red,is

$A$ box $B_1$ contains $1$ white ball,$3$ red balls and $2$ black balls. Another box $B_2$ contains $2$ white balls,$3$ red balls and $4$ black balls. $A$ third box $B_3$ contains $3$ white balls,$4$ red balls and $5$ black balls.
$1.$ If $1$ ball is drawn from each of the boxes $B_1, B_2$ and $B_3$,the probability that all $3$ drawn balls are of the same colour is
$(A)$ $\frac{82}{648}$ $(B)$ $\frac{90}{648}$ $(C)$ $\frac{558}{648}$ $(D)$ $\frac{566}{648}$
$2.$ If $2$ balls are drawn (without replacement) from a randomly selected box and one of the balls is white and the other ball is red,the probability that these $2$ balls are drawn from box $B_2$ is
$(A)$ $\frac{116}{181}$ $(B)$ $\frac{126}{181}$ $(C)$ $\frac{65}{181}$ $(D)$ $\frac{55}{181}$
Choose the correct options for question $1$ and $2$.