To solve the problem, we need to calculate the probability that the sum of the numbers drawn from the balls marked 2, 4, and 6 equals 12 when three balls are drawn with replacement.
### Step-by-Step Solution:
1. **Identify the Total Outcomes**:
Each time a ball is drawn, there are 3 choices (2, 4, or 6). Since the ball is replaced each time, the total number of outcomes for drawing three balls is:
\[
\text{Total outcomes} = 3 \times 3 \times 3 = 27
\]
**Hint**: Remember that since the ball is replaced, the number of choices remains constant for each draw.
2. **Determine the Favorable Outcomes**:
We need to find the combinations of the three draws that sum up to 12. Let's denote the number of times we draw each ball:
- Let \( x_2 \) be the number of times we draw the ball marked 2.
- Let \( x_4 \) be the number of times we draw the ball marked 4.
- Let \( x_6 \) be the number of times we draw the ball marked 6.
The equation we need to satisfy is:
\[
2x_2 + 4x_4 + 6x_6 = 12
\]
with the constraint:
\[
x_2 + x_4 + x_6 = 3
\]
3. **Solve the System of Equations**:
We can express \( x_6 \) in terms of \( x_2 \) and \( x_4 \):
\[
x_6 = 3 - x_2 - x_4
\]
Substituting this into the sum equation gives:
\[
2x_2 + 4x_4 + 6(3 - x_2 - x_4) = 12
\]
Simplifying this:
\[
2x_2 + 4x_4 + 18 - 6x_2 - 6x_4 = 12
\]
\[
-4x_2 - 2x_4 + 18 = 12
\]
\[
-4x_2 - 2x_4 = -6
\]
Dividing by -2:
\[
2x_2 + x_4 = 3
\]
4. **Finding Non-negative Integer Solutions**:
Now we can find the non-negative integer solutions for \( x_2 \) and \( x_4 \):
- If \( x_2 = 0 \), then \( x_4 = 3 \) → (0, 3, 0)
- If \( x_2 = 1 \), then \( x_4 = 1 \) → (1, 1, 1)
- If \( x_2 = 2 \), then \( x_4 = 0 \) → (2, 0, 1)
The valid combinations that satisfy both equations are:
- (0, 3, 0)
- (1, 1, 1)
- (2, 0, 1)
Each of these combinations corresponds to a unique arrangement of draws.
5. **Count the Arrangements**:
- For (0, 3, 0): Only one arrangement (all 4s).
- For (1, 1, 1): The arrangements are \( \frac{3!}{1!1!1!} = 6 \).
- For (2, 0, 1): The arrangements are \( \frac{3!}{2!1!} = 3 \).
Therefore, the total number of favorable outcomes is:
\[
1 + 6 + 3 = 10
\]
6. **Calculate the Probability**:
The probability \( \lambda \) that the sum of the three numbers is 12 is given by:
\[
\lambda = \frac{\text{Favorable outcomes}}{\text{Total outcomes}} = \frac{10}{27}
\]
7. **Calculate \( 108\lambda \)**:
Now, we need to find \( 108\lambda \):
\[
108\lambda = 108 \times \frac{10}{27} = 4 \times 10 = 40
\]
### Final Answer:
\[
\boxed{40}
\]
