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

$A$ player $X$ has a biased coin whose probability of showing heads is $p$ and a player $Y$ has a fair coin. They start playing a game with their own coins and play alternately. The player who throws a head first is a winner. If $X$ starts the game,and the probability of winning the game by both the players is equal,then the value of $p$ is

$A$ random variable $X$ has the following probability distribution:
| $X=x$ | $1$ | $2$ | $3$ | $4$ | $5$ | $6$ | $7$ | $8$ |
| :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- |
| $P(X=x)$ | $0.15$ | $0.23$ | $0.12$ | $0.20$ | $0.08$ | $0.10$ | $0.05$ | $0.07$ |
For the events $E = \{X \text{ is a prime number}\}$ and $F = \{X < 5\}$,find $P(E \cup F)$.

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

If three numbers are randomly selected from the set $\{1, 2, 3, \ldots, 50\}$, then the probability that they are in arithmetic progression is