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

A box $${B_1}$$ contains $$1$$ white ball, $$3$$ red balls and $$2$$ black balls. Another box $${B_2}$$ contains $$2$$ white balls, $$3$$ red balls and $$4$$ black balls. A third box $${B_3}$$ contains $$3$$ white balls, $$4$$ red balls and $$5$$ black balls.

If $$2$$ balls are drawn (without replacement) from a randomly selected box and one of the balls is white and the other ball is red, the probability that these $$2$$ balls are drawn from box $${B_2}$$ is