If $E$ and $F$ are events with $P(E) \le P(F)$ and $P(E \cap F) > 0,$ then

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

In a game, two dice are thrown simultaneously by a person $A$ and two cards are drawn at random simultaneously from a pack of $52$ playing cards by a person $B$. They win the game if $A$ gets a prime score as the sum of the numbers appearing on both the dice and $B$ gets a face card and a card having a prime number. Then the probability that both $A$ and $B$ win 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:

Consider a system of equations $ax + by = 0$ and $cx + dy = 0$,where $a, b, c, d \in \{0, 1\}$.
Statement $-1$: The probability that the system of equations has a solution is $1$.
Statement $-2$: The probability that the system of equations has a unique solution is $\frac{3}{8}$.

$A$ bag contains $19$ red balls and $19$ black balls. Two balls are chosen at a time repeatedly and discarded if they are of the same colour,but if they are different,the black ball is discarded and the red ball is returned to the bag. The probability that this process will terminate with one red ball is