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To solve this problem, let's analyze it step-by-step.

### Problem Summary
- A bag contains only **10-cent** and **20-cent** coins.
- Josip randomly draws two coins one after the other (with replacement), records their values, and calculates the mean of the two values.
- We are given:
  - The probability that the mean value of the two coins is **10 cents** is \(\frac{49}{121}\).
- Our task:
  - Find the probability that the two coins have a mean value of **15 cents**.

### Solution

1. **Possible Outcomes and Mean Values**
   - Since Josip is drawing with replacement, each draw is independent. Let’s denote the two draws by the values they could have:
     - \(10 \text{ cents} \) and \(10 \text{ cents}\)
     - \(10 \text{ cents} \) and \(20 \text{ cents}\)
     - \(20 \text{ cents} \) and \(10 \text{ cents}\)
     - \(20 \text{ cents} \) and \(20 \text{ cents}\)

2. **Mean Values for Each Outcome**
   - For each possible outcome, we calculate the mean:
     - **(10, 10)**: Mean = \(\frac{10 + 10}{2} = 10\) cents
     - **(10, 20)**: Mean = \(\frac{10 + 20}{2} = 15\) cents
     - **(20, 10)**: Mean = \(\frac{20 + 10}{2} = 15\) cents
     - **(20, 20)**: Mean = \(\frac{20 + 20}{2} = 20\) cents

   Thus, the possible mean values are **10, 15,** and **20** cents.

3. **Probability Calculations**
   - Let:
     - \(p\) be the probability of drawing a **10-cent** coin,
     - \(1 - p\) be the probability of drawing a **20-cent** coin.
   
   - Given that the probability of the mean value being **10 cents** is \(\frac{49}{121}\), which happens only in the case **(10, 10)**:
     - Probability of drawing two 10-cent coins (mean = 10) is \(p \times p = p^2\).
     - So, we have:  
       \[
       p^2 = \frac{49}{121}
       \]
     - Solving for \(p\):
       \[
       p = \sqrt{\frac{49}{121}} = \frac{7}{11}
       \]
     - Therefore, \(p = \frac{7}{11}\) and \(1 - p = \frac{4}{11}\) (for drawing a 20-cent coin).

4. **Probability of a Mean Value of 15 Cents**
   - For the mean to be **15 cents**, we need either the outcome **(10, 20)** or **(20, 10)**:
     - Probability of **(10, 20)** = \(p \times (1 - p) = \frac{7}{11} \times \frac{4}{11} = \frac{28}{121}\)
     - Probability of **(20, 10)** = \((1 - p) \times p = \frac{4}{11} \times \frac{7}{11} = \frac{28}{121}\)
   
   - Adding these probabilities:
     \[
     \frac{28}{121} + \frac{28}{121} = \frac{56}{121}
     \]

### Final Answer
The probability that the two coins have a mean value of **15 cents** is:
\[
\frac{56}{121}
\]

Would you like a detailed explanation of any step, or do you have more questions?

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Here are 5 related questions:
1. What would be the probability if we added a third type of coin, say, a 5-cent coin?
2. How would the calculations change if Josip did not replace the coin after the first draw?
3. What if we wanted to find the probability of getting a mean value of 20 cents?
4. How would this problem be solved if the coin values were in percentages or fractions instead?
5. Could you find the probability of getting any specific mean if each coin had a different probability?

**Tip:** Always break down probability problems by identifying possible outcomes and calculating each scenario's probability separately.

A bag contains only 10-cent coins and 20-cent coins. Josip takes at random a coin from the bag, records its value and replaces it in the bag. He then takes at random a second coin from the bag, records its value and replaces it in the bag. Josip finds the mean value of the two coins. The probability that the two coins have a mean value of 10 cents is 49/121. Work out the probability that the two coins have a mean value of 15 cents.

Solve a probability problem involving a bag of 10-cent and 20-cent coins where Josip finds the mean value of two randomly drawn coins. Given that the probability of a mean of 10 cents is 49/121, this problem teaches the steps to calculate the probability for a mean of 15 cents, covering core concepts in probability and mean calculations.

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