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
$(D)$ $P(X^C \cap Y)=\frac{1}{3}$

An ordinary cube has four blank faces,one face marked $2$,and another marked $3$. The probability of obtaining a total of exactly $12$ in $5$ throws is:

For three events $A, B$ and $C$,$P(\text{Exactly one of } A \text{ or } B \text{ occurs}) = P(\text{Exactly one of } B \text{ or } C \text{ occurs}) = P(\text{Exactly one of } C \text{ or } A \text{ occurs}) = \frac{1}{4}$ and $P(\text{All the three events occur simultaneously}) = \frac{1}{16}$. Then the probability that at least one of the events occurs is:

Two players $A$ and $B$ alternatively toss $3$ coins simultaneously. The player who gets $2$ heads and $1$ tail first, wins the game. If the game continues until someone wins and if $A$ begins the game, the probability that $B$ wins the game is

For independent events $A$ and $B$,$P(A \cup B) =$ . . . . . . .

Bag $A$ contains $3$ white and $4$ red balls,bag $B$ contains $4$ white and $5$ red balls,and bag $C$ contains $5$ white and $6$ red balls. If one ball is drawn at random from each of these three bags,then the probability of getting one white and two red balls is