Question
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.
$$\frac{1}{4}$$
Answer :
$$\frac{1}{3}$$
Solution :
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)
Required Probability $$ = \frac{2}{4} \times \frac{1}{3} + \frac{2}{4} \times \frac{1}{3} = \frac{1}{3}$$
$$\because $$ 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.
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)
Required Probability $$ = \frac{2}{4} \times \frac{1}{3} + \frac{2}{4} \times \frac{1}{3} = \frac{1}{3}$$
$$\because $$ 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.