We have 15 balls out of which 9 balls have not been previously used, so the remaining 6 balls are already used.
Let E1 be the first event when the 3 balls are randomly chosen and played with.
Now when choosing 3 balls, the combination can be
choosing all 3 unused, 2 unused and 1 used, 1 unused and 2 used & all 3 used.
So P(E1) = $\sum_{n=0}^{3} \binom{9}{n}*\binom{6}{3-n} / \binom{15}{3}$
Here n is the no of unused balls. For eg:- when n is 1, we can choose 1 unused ball from 9 unused in (9c1) ways and remaining 2 from 6 used balls in (6c2) ways .
Now let E2 be event of picking 3 unused balls after E1.
Then P(E2) = $\sum_{n=0}^{3} \binom{9-n}{3} / \binom{15}{3}$. Here n is the no of unused balls chosen after event E1. If suppose in E1, 1 unused and 2 used balls were chosen then P(E2) =((9-1)c3 /15c3) [9-1 because out of 9 unused balls, 1 ball is chosen]
Now total probability =P(E1)*P(E2)
=$\sum_{n=0}^{3} \binom{9}{n}*\binom{6}{3-n} /\binom{15}{3} * \binom{9-n}{3}/\binom{15}{3}$
=0.089