A consignment of 15 wrist watches contains 4 defectives. The wrist watches are selected at random, one by one and examined . The ones examined are not replaced back. What is the probability that ninth one examined is the last defective?

For the ninth one to be the last defective, among the first 8 watches, first 3 has to be defective.

So, total number of ways of selecting = $$^8C_3$$

And total number of possible cases = $$^{15}C_4$$

(Out of the 15 watches, examined in order, any 4 can be defective)

So, required probability = $$\dfrac{^8C_3}{^{15}C_4}=\dfrac{\dfrac{8!}{3!\times\ 5!}}{\dfrac{15!}{4!\times\ 11!}}=\dfrac{8}{195}$$

So, correct answer is option B.