To solve this problem, we need to calculate the probability that the two coins Josip draws from the bag have a mean value of 15 cents.

Step 1: Understanding the situation

The bag contains two types of coins:

• 10 cent coins
• 20 cent coins

Josip picks two coins (with replacement), and we need to find the probability that the mean value of the two coins is 15 cents. The mean value is calculated as:

$\text{Mean value} = \frac{\text{Value of first coin} + \text{Value of second coin}}{2}$

Given that the mean value should be 15 cents, we have:

$\frac{\text{Value of first coin} + \text{Value of second coin}}{2} = 15$

Multiplying both sides by 2:

$\text{Value of first coin} + \text{Value of second coin} = 30$

Step 2: Identifying possible coin combinations

The possible coin combinations that add up to 30 cents are:

• 10 cent + 20 cent = 30
• 20 cent + 10 cent = 30

These are the only combinations that satisfy the condition of the sum being 30 cents.

Step 3: Calculating the probability of each combination

Let's assume that the probability of drawing a 10 cent coin (denoted as $P(10)$) is $p$, and the probability of drawing a 20 cent coin (denoted as $P(20)$) is $1 - p$.

Scenario 1: First coin is 10 cents, second coin is 20 cents

The probability of this occurring is:

$P(\text{10, 20}) = P(10) \times P(20) = p \times (1 - p)$

Scenario 2: First coin is 20 cents, second coin is 10 cents

The probability of this occurring is:

$P(\text{20, 10}) = P(20) \times P(10) = (1 - p) \times p$

The total probability of these two scenarios is:

$P(\text{mean} = 15) = p \times (1 - p) + (1 - p) \times p = 2p(1 - p)$

Step 4: Using the given information

From the problem, we know the probability that the two coins have a mean value of 10 cents is $\frac{49}{121}$. We can use this information to find the value of $p$.

The mean value of 10 cents can occur in two ways:

• Both coins are 10 cents: $P(\text{10, 10}) = p \times p = p^2$
• Both coins are 20 cents: $P(\text{20, 20}) = (1 - p) \times (1 - p) = (1 - p)^2$

Thus, the total probability of the mean being 10 cents is:

$P(\text{mean} = 10) = p^2 + (1 - p)^2$

We are given that this probability is $\frac{49}{121}$, so:

$p^2 + (1 - p)^2 = \frac{49}{121}$

Expanding:

$p^2 + (1 - 2p + p^2) = \frac{49}{121}$

$2p^2 - 2p + 1 = \frac{49}{121}$

Multiplying both sides by 121 to eliminate the denominator:

$121(2p^2 - 2p + 1) = 49$

$242p^2 - 242p + 121 = 49$

$242p^2 - 242p + 72 = 0$

This is a quadratic equation. Solving this using the quadratic formula:

$p = \frac{-(-242) \pm \sqrt{(-242)^2 - 4(242)(72)}}{2(242)}$

$p = \frac{242 \pm \sqrt{58564 - 69696}}{484}$

$p = \frac{242 \pm \sqrt{-11132}}{484}$

The discriminant is negative, indicating no real solution, which suggests there was an error. I'll double-check the calculations and adjust.