If $A$ and $B$ are independent events of a random experiment such that $P(A \cap B)=\frac{1}{6}$ and $P(\bar{A} \cap \bar{B})=\frac{1}{3}$, then $P(A)$ is equal to (Here, $\bar{E}$ is the complement of the event $E$)

Let $X$ and $Y$ be events such that $P(X \cup Y) = P(X \cap Y).$
Statement-$1$: $P(X \cap Y) = P(X' \cap Y') = 0$
Statement-$2$: $P(X) + P(Y) = 2P(X \cap Y).$

$A$ and $B$ are independent events. The probability that both $A$ and $B$ occur is $\frac{1}{20}$ and the probability that neither of them occurs is $\frac{3}{5}$. The probability of occurrence of $A$ is

Three dice are thrown simultaneously and the sum of the numbers appeared on them is noted. If $A$ is the event of getting a sum greater than $14$ and $B$ is the event of getting a sum which is a multiple of $3$,then $P(A \cap \overline{B}) + P(\overline{A} \cap B) = $

$A$ 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

Two persons $A$ and $B$ are throwing an unbiased six-faced dice alternatively,with the condition that the person who throws $3$ first wins the game. If $A$ starts the game,then the probabilities of $A$ and $B$ winning the game are,respectively: