$A$ man throws a fair coin repeatedly. He gets $10$ points for each head he throws and $5$ points for each tail he throws. If the probability that he gets exactly $30$ points is $\frac{m}{n}$, where $\text{gcd}(m, n) = 1$, then $m + n$ is equal to:

Football teams $T_1$ and $T_2$ play two games against each other. The outcomes of the two games are independent. The probabilities of $T_1$ winning,drawing,and losing a game against $T_2$ are $\frac{1}{2}$,$\frac{1}{6}$,and $\frac{1}{3}$,respectively. Each team gets $3$ points for a win,$1$ point for a draw,and $0$ points for a loss. Let $X$ and $Y$ denote the total points scored by teams $T_1$ and $T_2$,respectively,after two games.
$(1)$ $P(X>Y)$ is
$(A)$ $\frac{1}{4}$ $(B)$ $\frac{5}{12}$ $(C)$ $\frac{1}{2}$ $(D)$ $\frac{7}{12}$
$(2)$ $P(X=Y)$ is
$(A)$ $\frac{11}{36}$ $(B)$ $\frac{1}{3}$ $(C)$ $\frac{13}{36}$ $(D)$ $\frac{1}{2}$

Three urns respectively contain $2$ white and $3$ black,$3$ white and $2$ black,and $1$ white and $4$ black balls. If one ball is drawn from each urn,then the probability that the selection contains $1$ black and $2$ white balls 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 $2$.

$A$ pair of $12$-sided fair dice with faces numbered $1, 2, 3, \ldots, 12$ is rolled. The probability that the sum of the numbers appearing has a remainder of $2$ when divided by $9$ is

Let $B_{i} (i=1, 2, 3)$ be three independent events in a sample space. The probability that only $B_{1}$ occurs is $\alpha$,only $B_{2}$ occurs is $\beta$,and only $B_{3}$ occurs is $\gamma$. Let $p$ be the probability that none of the events $B_{i}$ occurs,and these $4$ probabilities satisfy the equations $(\alpha - 2\beta)p = \alpha\beta$ and $(\beta - 3\gamma)p = 2\beta\gamma$ (All the probabilities are assumed to lie in the interval $(0, 1)$). Then $\frac{P(B_{1})}{P(B_{3})}$ is equal to ..........