A) 1/27
B) 8/27
C) 19/27
D) 26/27
Correct Answer: C
Solution :
Probabiltiy of defective transistor \[=\frac{5}{15}=\frac{1}{3}\] and probability of non-defective transistor \[=1-\frac{1}{3}=\frac{2}{3}\] Probability that the inspector find a non-defective transistor \[=\,\frac{2}{3}\times \frac{2}{3}\times \frac{2}{3}=\frac{8}{27}\] Hence, probability that atleast one of d the inspectors find a defective transistors \[=1-\frac{8}{27}=\frac{19}{27}\]You need to login to perform this action.
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