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 :

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 :

It is given that the events $$A$$ and $$B$$ are such that
$$P\left( A \right) = {1 \over 4},P\left( {A|B} \right) = {1 \over 2}$$ and $$P\left( {B|A} \right) = {2 \over 3}.$$ Then $$P(B)$$ is :

Two aeroplanes $${\rm I}$$ and $${\rm I}$$$${\rm I}$$ bomb a target in succession. The probabilities of $${\rm I}$$ and $${\rm I}$$$${\rm I}$$ scoring a hit correctly are $$0.3$$ and $$0.2,$$ respectively. The second plane will bomb only if the first misses the target. The probability that the target is hit by the second plane is :