A room has 3 lamps. From a collection of 10 light bulbs of which 6 are not good, a person selects 3 at random and puts them in a socket. The probability that he will have light is

The probability of picking up a defective lamp in the first attempt = $$\frac{6}{10}$$

The probability of picking up a defective lamp in the second attempt = $$\frac{5}{9}$$

The probability of picking up a defective lamp in the second attempt = $$\frac{4}{8}$$

The probability that he will have light = 1 - {$$\frac{6}{10} \times \frac{5}{9} \times \frac{4}{8}$$}

$$\Rightarrow$$ 1 - $$\frac{120}{720}$$ = 1 - $$\frac{1}{6}$$ = $$\frac{5}{6}$$

Hence, option A is the correct answer.