We consider the four possible cases of the initial draw.

Case 0: No used balls are drawn.

\[p_{0}=\frac{(\begin{array}{c}9\\ 3\end{array})}{(\begin{array}{c}15\\ 3\end{array})}\]

Case 1: 1 used ball is drawn.

\[p_{1}=\frac{(\begin{array}{c}9\\ 2\end{array}) \cdot 6}{(\begin{array}{c}15\\ 3\end{array})}\]

Case 2: 2 used balls are drawn.

\[p_{2}=\frac{9 \cdot (\begin{array}{c}6\\ 2\end{array})}{(\begin{array}{c}15\\ 3\end{array})}\]

Case 3: 3 used balls are drawn.

\[p_{3}=\frac{(\begin{array}{c}6\\ 3\end{array})}{(\begin{array}{c}15\\ 3\end{array})}\]

Step 2

We evaluate the probabilities and outcomes of each case.

Case 0: \(\displaystyle{p}_{{{0}}}={.1846}\), 6 new balls, 9 used left over.

Case 1: \(\displaystyle{p}_{{{1}}}={.4747}\), 7 new balls, 8 used.

Case 2: \(\displaystyle{p}_{{{2}}}={.2967}\), 8 new balls, 7 used.

Case 3: \(\displaystyle{p}_{{{3}}}={.044}\), 9 new balls, 6 used.

Step 3

We assess the probability of each case times the probability that no used balls turn up in the second draw, and add everything up.

\[p_{0} \frac{(\begin{array}{c}6\\ 3\end{array})}{(\begin{array}{c}15\\ 3\end{array})}+p_{1} \frac{(\begin{array}{c}7\\ 3\end{array})}{(\begin{array}{c}15\\ 3\end{array})}+p_{2} \frac{(\begin{array}{c}8\\ 3\end{array})}{(\begin{array}{c}15\\ 3\end{array})}+p_{3} \frac{(\begin{array}{c}9\\ 3\end{array})}{(\begin{array}{c}15\\ 3\end{array})}\]

Step 4

\(\displaystyle{.1846}\cdot{.044}+{.4747}\cdot{.0769}+{.2967}\cdot{.1231}+{.044}\cdot{.1846}={.0893}\)