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 will be required is-

1 / 3

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1 / 6

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Solution

The correct option is D $1 / 6$
The probability that only two tests are needed
$=$ (probability that the first machine tested is faulty) $\times$ (probability that the second machine tested is faulty given that the first machine tested is faulty)
$= \frac{2}{4} \times \frac{1}{3} = \frac{1}{6}$

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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 will be required is-

The correct option is D 1/6The probability that only two tests are needed =(probability that the first machine tested is faulty)×(probability that the second machine tested is faulty given that the first machine tested is faulty)=24×13=16

The correct option is D $1 / 6$
The probability that only two tests are needed
$=$ (probability that the first machine tested is faulty) $\times$ (probability that the second machine tested is faulty given that the first machine tested is faulty)
$= \frac{2}{4} \times \frac{1}{3} = \frac{1}{6}$