A random variable $$X$$ has Poisson distribution with mean $$2$$.
Then $$P\left( {X > 1.5} \right)$$ equals :

Three houses are available in a locality. Three persons apply for the houses. Each applies for one house without consulting others. The probability that all the three apply for the same house is :

Let $$A$$ and $$B$$ two events such that $$P\left( {\overline {A \cup B} } \right) = {1 \over 6},$$ $$P\left( {A \cap B} \right) = {1 \over 4}$$ and $$P\left( {\overline A } \right) = {1 \over 4},$$ where $${\overline A }$$ stands for complement of event $$A$$. Then events $$A$$ and $$B$$ are :

The mean and the variance of a binomial distribution are $$4$$ and $$2$$ respectively. Then the probability of $$2$$ successes is :