If $A, B$ and $C$ are three independent events such that $P(A)=P(B)=P(C)=P$, then $P$ (at least two of $A, B$ and $C$ occur) is equal to

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

Two persons $A$ and $B$ take turns in throwing a pair of dice. The first person to throw a sum of $9$ with both dice will win the game. If $A$ throws first,then the probability that $B$ wins the game is:

$A$ bag contains $2n$ coins,out of which $n-1$ are unfair with heads on both sides and the remaining are fair. One coin is picked from the bag at random and tossed. If the probability that a head appears in the toss is $\frac{41}{56}$,then the number of unfair coins in the bag is:

$A$ bag contains $20$ coins. If the probability that the bag contains exactly $4$ biased coins is $1/3$ and the probability that it contains exactly $5$ biased coins is $2/3$,then the probability that all the biased coins are sorted out from the bag in exactly $10$ draws is:

If the numbers appearing on the two throws of a fair six-faced die are $\alpha$ and $\beta$,then the probability that $x^{2}+\alpha x+\beta > 0$ for all $x \in R$ is: