There are 15 tennis balls in a box, of which 9 have not previously been used. 3 of these balls are randomly chosen, played with and then returned to the box. later, another 3 balls are randomly chosen from the box. Find the probability that none of these balls has ever been used.
I have decided to go simpler way and I have solved this problem another way: $$ (9/15)*(8/14)*(7/13)*(6/15)*(5/14)*(4/13)$$ and I got the answer 0.008114962
But book gives another answer - $0.083$.
Did I solve the problem right? If not, tell me please where I have made a mistake.
In the answer below, $n$ represents the number of unused balls chosen in the first phase:
$$\sum\limits_{n=0}^{3}\frac{\binom{9}{n}\cdot\binom{15-9}{3-n}}{\binom{15}{3}}\cdot\frac{\binom{9-n}{3}}{\binom{15}{3}}\approx0.089$$