Four numbers are chosen at random (without replacement) from the set $$\left\{ {1,2,3,....20} \right\}.$$

Statement - 1: The probability that the chosen numbers when arranged in some order will form an AP is $${1 \over {85}}.$$

Statement - 2: If the four chosen numbers form an AP, then the set of all possible values of common difference is $$\left( { \pm 1, \pm 2, \pm 3, \pm 4, \pm 5} \right).$$

In a binomial distribution $$B\left( {n,p = {1 \over 4}} \right),$$ if the probability of at least one success is greater than or equal to $${9 \over {10}},$$ then $$n$$ is greater than :

One ticket is selected at random from $$50$$ tickets numbered $$00, 01, 02, ...., 49.$$ Then the probability that the sum of the digits on the selected ticket is $$8$$, given that the product of these digits is zer, equals :

A die is thrown. Let $$A$$ be the event that the number obtained is greater than $$3.$$ Let $$B$$ be the event that the number obtained is less than $$5.$$ Then $$P\left( {A \cup B} \right)$$ is :