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(B) Let the four machines be $M_1, M_2, F_1, F_2$,where $F$ denotes a faulty machine and $M$ denotes a working machine.We need to find the probability

Vedclass · Accepted Answer

$\frac{1}{6}$

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(B) Let the four machines be $M_1, M_2, F_1, F_2$,where $F$ denotes a faulty machine and $M$ denotes a working machine.We need to find the probability that both faulty machines are identified in exactly two tests.This means the first test must result in a faulty machine,and the second test must also result in a faulty machine.The total number of ways to choose $2$ machines out of $4$ for the first two tests is $^4P_2 = 4 	imes 3 = 12$.The number of ways to choose $2$ faulty machines out of $2$ for the first two tests is $^2P_2 = 2 	imes 1 = 2$.Therefore,the required probability is $\frac{2}{12} = \frac{1}{6}$.Alternatively,the probability that the first machine tested is faulty is $\frac{2}{4}$.Given that the first machine was faulty,the probability that the second machine tested is also faulty is $\frac{1}{3}$.Thus,the required probability is $\frac{2}{4} 	imes \frac{1}{3} = \frac{1}{6}$.

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. Then the probability that only two tests are needed is

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