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. \(\frac{14}{55}\)
• B. \(\frac{8}{12}\)
• C. 1/27\(\frac{1}{27}\)
• D. None of these

Answer 1

The Correct Option is A

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)