There are four machines and it is known that exactly two of them are faulty. They are tested one by one, in…
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 need is
$\frac{1}{3}$
$\frac{1}{6}$
$\frac{1}{2}$
$\frac{1}{4}$
Solution
$\therefore$ Required probability $=P(\ln$ first two test either both are faulty or both are not faulty) $=P$ (First two are faulty)
$+P$ (First two are not faulty)
$\begin{aligned} & =\frac{2}{4} \times \frac{1}{3}+\frac{2}{4} \times \frac{1}{3} \\ & =\frac{4}{12}=\frac{1}{3}\end{aligned}$