Step 1: Understanding the Concept:
Total seating arrangements of 9 people in a row: \(9!\). We count the arrangements where the two end seats hold girls and no two girls are adjacent.
Step 2: Arrange the boys first:
The 6 boys can be arranged in \(6!\) ways. They create 7 gaps, including the 2 end gaps. Girls placed in separate gaps are never adjacent.
Step 3: Place the girls:
Two girls must sit in the two end gaps and the third girl in one of the 5 inner gaps (5 ways). The three girls can be assigned to these three gaps in \(3!\) ways.
Favourable arrangements = \(6!\times 5\times 3!\).
Step 4: Probability:
\[ P = \frac{6!\times 5\times 3!}{9!} = \frac{5\times 6}{9\times 8\times 7} = \frac{30}{504} = \frac{5}{84} \]
A girl at each end with the third girl in an inner gap automatically means no two girls are adjacent, because each is separated by at least one boy.
Final Answer:
The required probability is \(\frac{5}{84}\), option (D).
\[ \boxed{\frac{5}{84}} \]