For independent events $A$ and $B$,$P(A \cup B) =$ . . . . . . .

Box-$I$ contains $3$ cards bearing numbers $1, 2, 3$; Box-$II$ contains $5$ cards bearing numbers $1, 2, 3, 4, 5$ and Box-$III$ contains $7$ cards bearing numbers $1, 2, 3, 4, 5, 6, 7$. One card is drawn at random from each of the boxes. If $x_i$ is the number on the card drawn from the $i^{\text{th}}$ box,$i=1, 2, 3$,then the probability that $x_1+x_2+x_3$ is odd is equal to

If $A$ and $B$ are two independent events such that $P(\bar{A})=0.75$,$P(A \cup B)=0.65$ and $P(B)=x$,then find the value of $x$.

Fifteen football players of a club-team are given $15$ $T$-shirts with their names written on the backside. If the players pick up the $T$-shirts randomly,then the probability that at least $3$ players pick the correct $T$-shirt is

Three numbers are selected at random from the set ${1, 2, 3, \dots, 8}$ without replacement. Given that the minimum of the selected numbers is $3$ and the maximum is $6$, what is the probability that the third number is $4$ or $5$?

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