A) \[\frac{1}{3}\]
B) \[\frac{1}{6}\]
C) \[\frac{1}{2}\]
D) \[\frac{1}{4}\]
Correct Answer: A
Solution :
[a] The faulty machines can be identified in two tests only if both the tested machines are either all defective or all non-defective. See the following tree diagram. (Here D is for defective & ND is for non-defective) Reqd. Probability \[=\frac{2}{4}\times \frac{1}{3}+\frac{2}{4}\times \frac{1}{3}=\frac{1}{3}\] \[\therefore \]The probability that first machine is defective (or non-defective) is \[\frac{2}{4}\] and the probability that second machine is also defective (or non-defective) is \[\frac{1}{3}\]as 1 defective machine remains in total three machines.You need to login to perform this action.
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