A consignment of 15 record players contains 4 defective, the record players are selected at random, one-by-one and examined. Those examined are not replaced. What is the probability that the 9th one examined is last defective?

A consignment of 15 record players contains 4 defective, the record players are selected at random, one-by-one and examined. Those examined are not replaced. What is the probability that the  9th one examined is last defective.

Solution

 Let A be the event of getting exactly 3 defectives in examination of 8 record players.
 Let B the event that the 9th piece examined is a defective one. 

The record players are not replaced and since there are 4 defectives out of 15 CDs, 

we have 𝑃(A) =$\frac{\binom{4}{3}\binom{11}{5}}{\binom{15}{8}}$ 

Probability that the 9th examined CD is defective given that there were 3 defectives in the first 8 pieces examined = P($\frac{A}{B}$) =$\frac{1}{7}$(since there is only one defective is left among the remaining 15 - 8 = 7 record players)
The probability that the 9th one examined is the last defective = P(A$\cup$B) = P(A) P($\frac{B}{A}$) = $\frac{\binom{4}{3}\binom{11}{5}}{\binom{15}{8}}$ .$\frac{1}{7}$ = $\frac{8}{195}$