1. **Problem statement:** A bag contains only 10 cent and 20 cent coins. Josip draws two coins with replacement and records their values. The probability that the mean value of the two coins is 10 cents is given as $\frac{49}{121}$. We need to find the probability that the mean value of the two coins is 15 cents. 2. **Understanding the problem:** The mean value of two coins is 15 cents if one coin is 10 cents and the other is 20 cents. Since the coins are drawn with replacement, the draws are independent. 3. **Define variables:** Let $p$ be the probability of drawing a 10 cent coin, and $q$ be the probability of drawing a 20 cent coin. Since these are the only coins, $p + q = 1$. 4. **Given:** The probability that the mean is 10 cents means both coins are 10 cents. So, $$p^2 = \frac{49}{121}$$ 5. **Calculate $p$:** $$p = \sqrt{\frac{49}{121}} = \frac{7}{11}$$ 6. **Calculate $q$:** $$q = 1 - p = 1 - \frac{7}{11} = \frac{4}{11}$$ 7. **Calculate the probability that the mean is 15 cents:** This happens if one coin is 10 cents and the other is 20 cents. There are two ways: first coin 10 cents and second 20 cents, or first coin 20 cents and second 10 cents. So, $$P = 2 \times p \times q = 2 \times \frac{7}{11} \times \frac{4}{11} = \frac{56}{121}$$ 8. **Final answer:** $$\boxed{\frac{56}{121}}$$ This is the probability that the two coins have a mean value of 15 cents.