Question:
A lot of 12 bulbs contains 4 defective bulbs. Three bulbs are drawn at random from the lot, one after the other. The probability that all three are non-defective is
A lot of 12 bulbs contains 4 defective bulbs. Three bulbs are drawn at random from the lot, one after the other. The probability that all three are non-defective is
Updated On: Jan 30, 2025
- \(\frac{14}{55}\)
- \(\frac{8}{12}\)
- 1/27\(\frac{1}{27}\)
- None of these
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The Correct Option is A
Solution and Explanation
Total possible cases 12C3
\(=\frac{12\times11\times10\times9!}{3!\times9!}=220\) ways
Total ways of selecting three non-defective bulbs
\(=\) 8C3\(=\frac{8\times7\times6\times5!}{3!\times5!}=56\)
Required probability \(=\frac{56}{220}=\frac{14}{55}\)
The correct option is (A)
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