Probability 3 Question 13
13. 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) $\frac{1}{4}$
(1998, 2M)
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Solution:
- Probability that only two tests are needed =Probability that the first machine tested is faulty $\times$ Probability that the second machine tested is faulty $=\frac{2}{4} \times \frac{1}{3}=\frac{1}{6}$
