If $A$ and $B$ are independent events and $P(A) = p, P(B) = 2p$,and $P(\text{exactly one from } A \text{ and } B) = \frac{5}{9}$,then find the value of $p$.

$A$ and $B$ alternately throw a pair of dice. $A$ wins if he throws a sum of $5$ before $B$ throws a sum of $8$,and $B$ wins if he throws a sum of $8$ before $A$ throws a sum of $5$. The probability that $A$ wins,if $A$ makes the first throw,is

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

$A$ and $B$ each select one number at random from the distinct numbers $1, 2, 3, \ldots, n$. The probability that the number selected by $A$ is less than the number selected by $B$ is $\frac{1009}{2019}$. The probability that the number selected by $B$ is the number immediately next to the number selected by $A$ is:

For independent events $A$ and $B$,if $P(A) = \frac{1}{2}$ and $P(A \cup B) = \frac{3}{5}$,then $P(B) =$ . . . . . . .

There are four machines and it is known that exactly two of them are faulty. They are tested one by one, in a random order until both the faulty machines are identified. The probability that only two tests are needed is: