At a telephone enquiry system the number of phone cells regarding relevant enquiry follow Poisson distribution with an average of $$5$$ phone calls during $$10$$ minute time intervals. The probability that there is at the most one phone call during a $$10$$-minute time period is :

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 :

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

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 :