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 $E$)

$A$ and $B$ are independent events. If $P(A \cup B)=0.5$ and $P(A)=0.2$,then $P(B) = $ . . . . . . . (in $/8$)

Three students $S_1, S_2$ and $S_3$ are given a problem to solve. Consider the following events:
$U:$ At least one of $S_1, S_2$ and $S_3$ can solve the problem,
$V: S_1$ can solve the problem,given that neither $S_2$ nor $S_3$ can solve the problem,
$W: S_2$ can solve the problem and $S_3$ cannot solve the problem,
$T: S_3$ can solve the problem.
For any event $E$,let $P(E)$ denote the probability of $E$.
If $P(U)=\frac{1}{2}, P(V)=\frac{1}{10}$ and $P(W)=\frac{1}{12}$,then $P(T)$ is equal to

Two players $A$ and $B$ are alternately throwing a coin and a die together. $A$ player who first throws a head and a $6$ wins the game. If $A$ starts the game, then the probability that $B$ wins the game is:

In a game,two players $A$ and $B$ take turns throwing a pair of fair dice,starting with player $A$. The total score on the two dice in each throw is noted. $A$ wins the game if he throws a total of $6$ before $B$ throws a total of $7$,and $B$ wins the game if he throws a total of $7$ before $A$ throws a total of $6$. The game stops as soon as either of the players wins. The probability of $A$ winning the game is:

There are three bags $B_1$,$B_2$,and $B_3$ containing $2$ Red and $3$ White,$5$ Red and $5$ White,and $3$ Red and $2$ White balls respectively. $A$ ball is drawn from bag $B_1$ and placed in bag $B_2$,then a ball is drawn from bag $B_2$ and placed in bag $B_3$,then a ball is drawn from bag $B_3$. The number of ways in which this process can be completed,if the same colour balls are used in the first and second transfers (assume all balls to be distinct),is