1. **Problem Statement:** There are 16 tennis balls in total, 10 of which have never been used. Two balls are chosen and played with, then returned to the box. Later, two balls are chosen again. We need to calculate the probability that none of these later two balls have ever been used. 2. **Understanding the problem:** Since the first two balls are returned, the total number and composition of balls remains the same for the second draw. 3. **Calculating the probability:** We want the probability that both balls chosen in the second draw are from the 10 unused balls. Total balls = 16 Unused balls = 10 4. **Number of ways to choose 2 balls from 16:** $$\binom{16}{2} = \frac{16 \times 15}{2} = 120$$ 5. **Number of ways to choose 2 unused balls from the 10 unused balls:** $$\binom{10}{2} = \frac{10 \times 9}{2} = 45$$ 6. **Probability that both balls chosen are unused:** $$P = \frac{\binom{10}{2}}{\binom{16}{2}} = \frac{45}{120} = \frac{3}{8} = 0.375$$ **Final answer:** $$\boxed{\frac{3}{8} \text{ or } 0.375}$$