question_answer

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

A) \[\frac{1}{3}\]

B) \[\frac{1}{6}\]

C) \[\frac{1}{2}\]

D) \[\frac{1}{4}\]

Correct Answer: A

Solution :

[a] The faulty machines can be identified in two tests only if both the tested machines are either all defective or all non-defective. See the following tree diagram. (Here D is for defective & ND is for non-defective) Reqd. Probability \[=\frac{2}{4}\times \frac{1}{3}+\frac{2}{4}\times \frac{1}{3}=\frac{1}{3}\] \[\therefore \]The probability that first machine is defective (or non-defective) is \[\frac{2}{4}\] and the probability that second machine is also defective (or non-defective) is \[\frac{1}{3}\]as 1 defective machine remains in total three machines.