Defective Watches Inspection Probability

For the 9th watch to be the 4th defective, exactly 3 defectives must be chosen in the first 8 examinations, and the 9th watch must be defective. Ways to select 3 defectives and 5 non-defectives in the first 8 items = $\binom{4}{3} \times \binom{11}{5}$. Total ways to pick 8 items out of 15 = $\binom{15}{8}$. Probability of having 3 defectives in first 8 = $\frac{\binom{4}{3} \binom{11}{5}}{\binom{15}{8}}$. Then probability that 9th is the remaining 1 defective out of 7 remaining watches = $\frac{1}{7}$. Total probability = $\frac{\binom{4}{3} \binom{11}{5}}{\binom{15}{8}} \times \frac{1}{7} = \frac{8}{195}$.