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

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$\frac{1}{6}$

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(B) Let the four machines be $M_1, M_2, F_1, F_2$,where $F_1$ and $F_2$ are faulty and $M_1, M_2$ are non-faulty.We need to identify both faulty machines 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 pick $2$ machines out of $4$ in a specific order is $4 	imes 3 = 12$.The number of ways to pick the two faulty machines in the first two tests is $2 	imes 1 = 2$ (i.e.,$(F_1, F_2)$ or $(F_2, F_1)$).Therefore,the required probability is $\frac{2}{12} = \frac{1}{6}$.Alternatively,using combinations: The total number of ways to choose $2$ machines out of $4$ is $^4C_2 = 6$. Only one of these pairs consists of both faulty machines. Thus,the probability is $\frac{1}{6}$.

$Face$	$1$	$2$	$3$	$4$	$5$	$6$
$Probability$	$0.2$	$0.22$	$0.11$	$0.25$	$0.05$	$0.17$

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 faulty machines are identified. The probability that only two tests are needed is

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