There are four machines and it is known that exactly two of them are faulty. They are tested, one by one, in a random order till both the faulty machines are identified. Then the probability that only two tests are needed is

If $$E$$ and $$F$$ are events with $$P\left( E \right) \le P\left( F \right)$$ and $$P\left( {E \cap F} \right) > 0,$$ then

For the three events $$A, B,$$ and $$C,P$$ (exactly one of the events $$A$$ or $$B$$ occurs) $$=P$$ (exactly one of the two events $$B$$ or $$C$$ occurs)$$=P$$ (exactly one of the events $$C$$ or $$A$$ occurs)$$=p$$ and $$P$$ (all the three events occur simultaneously) $$ = {p^2},$$ where $$0 < p < 1/2.$$ Then the probability of at least one of the three events $$A,B$$ and $$C$$ occurring is

Three of six vertices of a regular hexagon are chosen at random. The probability that the triangle with three vertices is equilateral, equals