1. **State the problem:** There are 16 tennis balls in total, 10 of which are new (not previously used). Two balls are chosen, played with, and put back. Then, two balls are chosen again. We want to find the probability that in the second selection, none of the balls have ever been used. 2. **Analyze the scenario:** - First pick: 2 balls chosen and played with, so these 2 balls are now "used". - The balls are returned to the box, so the total is still 16 balls. - Second pick: we want the probability that both balls chosen are from the original 10 unused balls (which excludes the 2 that were used in the first pick). 3. **Compute the probability step-by-step:** - First, find the probability distribution of how many unused balls were chosen in the first draw. - The number of unused balls chosen in the first pick can be 0, 1, or 2. 4. **Compute probability of choosing exactly $k$ unused balls in the first draw:** - Total ways to choose 2 balls: $\binom{16}{2} = 120$ - Ways to choose $k$ unused balls from 10: $\binom{10}{k}$ - Ways to choose $2-k$ used balls from 6 (since 16 total - 10 unused = 6 used): $\binom{6}{2-k}$ So, $$P(k) = \frac{\binom{10}{k} \cdot \binom{6}{2-k}}{120}$$ for $k = 0,1,2$. 5. **After first pick:** - Number of unused balls remaining (not chosen in first draw): $10 - k$ - Number of used balls remaining: $6 - (2 - k) + 2$ (The 2 balls from first pick become used and are replaced back), so total used balls after first pick stay at $6 + 2 - (2 - k) = 6 + k$ (since the 2 balls chosen are now used, incrementing the used count by 2, but 2-k of these come from previously used balls, so net used balls after first pick is $6 + k$). This means after the first pick: - Unused balls in box: $10 - k$ - Used balls in box: $6 + k$ - Total balls: $16$ 6. **Probability that the two balls chosen in the second pick are both unused, given $k$ unused balls were chosen first:** $$P(\text{2 unused second pick} | k) = \frac{\binom{10-k}{2}}{\binom{16}{2}}$$ 7. **Total probability (Law of Total Probability):** $$P = \sum_{k=0}^2 P(k) \times P(\text{2 unused second pick} | k) = \sum_{k=0}^2 \frac{\binom{10}{k} \binom{6}{2-k}}{120} \times \frac{\binom{10-k}{2}}{120}$$ 8. **Calculate terms:** - For $k=0$: $$P(0) = \frac{\binom{10}{0} \binom{6}{2}}{120} = \frac{1 \cdot 15}{120} = \frac{15}{120} = 0.125$$ $$P(\cdot|0) = \frac{\binom{10}{2}}{120} = \frac{45}{120} = 0.375$$ Contribution: $0.125 \times 0.375 = 0.046875$ - For $k=1$: $$P(1) = \frac{\binom{10}{1} \binom{6}{1}}{120} = \frac{10 \cdot 6}{120} = \frac{60}{120} = 0.5$$ $$P(\cdot|1) = \frac{\binom{9}{2}}{120} = \frac{36}{120} = 0.3$$ Contribution: $0.5 \times 0.3 = 0.15$ - For $k=2$: $$P(2) = \frac{\binom{10}{2} \binom{6}{0}}{120} = \frac{45 \cdot 1}{120} = 0.375$$ $$P(\cdot|2) = \frac{\binom{8}{2}}{120} = \frac{28}{120} \approx 0.2333$$ Contribution: $0.375 \times 0.2333 \approx 0.0875$ 9. **Sum contributions:** $$P = 0.046875 + 0.15 + 0.0875 = 0.284375$$ 10. **Final answer:** The probability that none of the balls in the second draw have ever been used is approximately $$\boxed{0.284}$$ or 28.4%.