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

Answer: (A)

$\frac{14}{55}$

Answer 2

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)