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

If $$\overline E $$ and $$\overline F $$ are the complementary events of events $$E$$ and $$F$$ respectively and if $$0 < P\left( F \right) < 1,$$ then

If from each of the three boxes containing $$3$$ white and $$1$$ black, $$2$$ white and $$2$$ black, $$1$$ white and $$3$$ black balls, one ball is drawn at random, then the probability that $$2$$ white and $$1$$ black ball will be drawn is