Let $X$ and $Y$ be two events such that $P(X \mid Y)=\frac{1}{2}$,$P(Y \mid X)=\frac{1}{3}$,and $P(X \cap Y)=\frac{1}{6}$. Which of the following is (are) correct?
$(A)$ $P(X \cup Y)=\frac{2}{3}$
$(B)$ $X$ and $Y$ are independent
$(C)$ $X$ and $Y$ are not independent
$(D)$ $P(X^C \cap Y)=\frac{1}{3}$

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
$(A)$ $\frac{11}{36}$ $(B)$ $\frac{1}{3}$ $(C)$ $\frac{13}{36}$ $(D)$ $\frac{1}{2}$

$A$ fair die is thrown until $2$ appears. Then the probability that $2$ appears in an even number of throws 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 $2$.

Three numbers are selected at random from the set ${1, 2, 3, \dots, 8}$ without replacement. Given that the minimum of the selected numbers is $3$ and the maximum is $6$, what is the probability that the third number is $4$ or $5$?

There are four fair dice $D_1, D_2, D_3$ and $D_4$. Each has six faces numbered $1, 2, 3, 4, 5$ and $6$. They are rolled one by one. What is the probability that the number shown on $D_4$ is equal to at least one of the numbers shown on $D_1, D_2$ and $D_3$ (in $/216$)?