The corners of regular tetrahedrons are numbered $1, 2, 3, 4$. Three tetrahedrons are tossed. The probability that the sum of the upward corners will be $5$ is

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

Let $A, B$ and $C$ be three events such that the probability that exactly one of $A$ and $B$ occurs is $(1-k)$,the probability that exactly one of $B$ and $C$ occurs is $(1-2k)$,the probability that exactly one of $C$ and $A$ occurs is $(1-k)$ and the probability that all $A, B$ and $C$ occur simultaneously is $k^2$,where $0 < k < 1$. Then the probability that at least one of $A, B$ and $C$ occurs is:

$A$ speaks truth in $75 \%$ of the cases and $B$ in $80 \%$ of the cases. Then,the probability that their statements about an incident do not match,is

$A$ and $B$ are two independent events. $P(A)=\frac{2}{5}, P(B)=\frac{1}{3}$. Match the following List-$I$ with List-$II$.

List-$I$	List-$II$
$(A) P(\overline{A} \cup B)$	$(I) \frac{2}{3}$
$(B) P(\frac{A}{\overline{B}})$	$(II) \frac{11}{15}$
$(C) P(A \cup B)$	$(III) \frac{3}{5}$

$A$ 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 $2$.