Football teams $${T_1}$$ and $${T_2}$$ have to play two games against each other. It is assumed that the outcomes of the two games are independent. The probabilities of $${T_1}$$ winning, drawing and losing a game against $${T_2}$$ are $${1 \over 2},{1 \over 6}$$ and $${1 \over 3}$$ respectively. Each team gets $$3$$ points for a win, $$1$$ point for a draw and $$0$$ point for a loss in a game. Let $$X$$ and $$Y$$ denote the total points scored by teams $${T_1}$$ and $${T_2}$$ respectively after two games.

$$P\,\left( {X = Y} \right)$$ is

Football teams $${T_1}$$ and $${T_2}$$ have to play two games against each other. It is assumed that the outcomes of the two games are independent. The probabilities of $${T_1}$$ winning, drawing and losing a game against $${T_2}$$ are $${1 \over 2},{1 \over 6}$$ and $${1 \over 3}$$ respectively. Each team gets $$3$$ points for a win, $$1$$ point for a draw and $$0$$ point for a loss in a game. Let $$X$$ and $$Y$$ denote the total points scored by teams $${T_1}$$ and $${T_2}$$ respectively after two games.

$$\,\,\,\,P\,\left( {X > Y} \right)$$ is

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 $${x_1} + {x_2} + {x_3}$$ is odd, is