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(B) Let the four machines be $M_1, M_2, F_1, F_2$,where $F_1, F_2$ are faulty and $M_1, M_2$ are non-faulty.We need to identify both faulty machines.

Vedclass · Accepted Answer

$\frac{1}{6}$

Vedclass · Answer

(B) Let the four machines be $M_1, M_2, F_1, F_2$,where $F_1, F_2$ are faulty and $M_1, M_2$ are non-faulty.We need to identify both faulty machines. The total number of ways to pick two machines out of four in a specific order is $4 	imes 3 = 12$.For only two tests to be needed,the first two machines tested must be the two faulty ones ($F_1, F_2$ or $F_2, F_1$).The probability of picking a faulty machine in the first test is $\frac{2}{4}$.Given the first was faulty,the probability of picking the second faulty machine in the second test is $\frac{1}{3}$.Therefore,the probability that both faulty machines are identified in exactly two tests is $\frac{2}{4} 	imes \frac{1}{3} = \frac{2}{12} = \frac{1}{6}$.

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,until both the faulty machines are identified. The probability that only two tests are needed is:

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