A computer producing factory has only two plants $${T_1}$$ and $${T_2}.$$ Plant $${T_1}$$ produces $$20$$% and plant $${T_2}$$ produces $$80$$% of the total computers produced. $$7$$% of computers produced in the factory turn out to be defective. It is known that $$P$$ (computer turns out to be defective given that it is produced in plant $${T_1}$$)
$$ = 10P$$ (computer turns out to be defective given that it is produced in plant $${T_2}$$),
where $$P(E)$$ denotes the probability of an event $$E$$. A computer produced in the factory is randomly selected and it does not turn out to be defective. Then the probability that it is produced in plant $${T_2}$$ is

Box $$1$$ contains three cards bearing numbers $$1,2,3;$$ box $$2$$ contains five cards bearing numbers $$1,2,3,4,5;$$ and box $$3$$ contains seven cards bearing numbers $$1,2,3,4,5,6,7.$$ A card is drawn from each of the boxes. Let $${x_i}$$ be number on the card drawn from the $${i^{th}}$$ box, $$i=1,2,3.$$

The probability that $${x_1},$$, $${x_2},$$ $${x_3}$$ are in an arithmetic progression, is

Three boys and two girls stand in a queue. The probability, that the number of boys ahead of every girl is at least one more than the number of girls ahead of her, is

Box $$1$$ contains three cards bearing numbers $$1,2,3;$$ box $$2$$ contains five cards bearing numbers $$1,2,3,4,5;$$ and box $$3$$ contains seven cards bearing numbers $$1,2,3,4,5,6,7.$$ A card is drawn from each of the boxes. Let $${x_i}$$ be number on the card drawn from the $${i^{th}}$$ box, $$i=1,2,3.$$

The probability that $${x_1} + {x_2} + {x_3}$$ is odd, is