A) \[\frac{1}{3}\]
B) \[\frac{1}{6}\]
C) \[\frac{1}{2}\]
D) \[\frac{1}{4}\]
Correct Answer: B
Solution :
This is a problem of without replacement. \[P=\frac{\text{one}\,\text{def}\,\text{.}\,\text{from}\,2\,\text{def}\text{.}}{\text{any}\,\text{one}\,\text{from}\,4}\times \frac{1\,\text{def}\,\text{.}\,\text{from}\,\text{remaining}\,\text{1}\,\text{def}\text{.}}{\text{any}\,\text{one}\,\text{from}\,\text{remaining}\,3}\] Hence required probability \[=\frac{2}{4}\times \frac{1}{3}=\frac{1}{6}\] Aliter : Number of ways in which two faulty machines may be detected (depending upon the test done to indentify the faulty machines) \[={}^{4}{{C}_{2}}=6\] Number of favourable cases = 1 [When faulty machines are identified in the first and the second test] Hence required probability \[=\frac{1}{6}.\]
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