15 identical balls are placed in 3 different boxes, find the probability that each box contain at least 3 balls is p than 34p =

Explanation

To solve the problem of finding the probability that each of the 3 boxes contains at least 3 balls when distributing 15 identical balls, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Problem**:
We have 15 identical balls and 3 different boxes. We need to find the probability that each box contains at least 3 balls.

2. **Setting Up the Equation**:
Let \( x_1, x_2, x_3 \) be the number of balls in boxes 1, 2, and 3 respectively. We need to satisfy the equation:
\[ x_1 + x_2 + x_3 = 15 \]
with the condition that \( x_1, x_2, x_3 \geq 3 \).

3. **Adjusting for the Minimum Requirement**:
Since each box must contain at least 3 balls, we can allocate 3 balls to each box initially. This uses up \( 3 \times 3 = 9 \) balls. Therefore, we now have:
\[ x_1' + x_2' + x_3' = 6 \]
where \( x_i' = x_i - 3 \) (the number of balls in each box after the initial allocation).

4. **Finding the Total Number of Solutions**:
We need to find the number of non-negative integer solutions to the equation:
\[ x_1' + x_2' + x_3' = 6 \]
The formula for the number of solutions in non-negative integers is given by:
\[ \binom{n + r - 1}{r - 1} \]
where \( n \) is the total number of items to distribute (6 in this case) and \( r \) is the number of boxes (3). Thus, we have:
\[ \text{Total solutions} = \binom{6 + 3 - 1}{3 - 1} = \binom{8}{2} \]

5. **Calculating Total Solutions**:
\[ \binom{8}{2} = \frac{8 \times 7}{2 \times 1} = 28 \]

6. **Finding the Total Number of Unrestricted Solutions**:
Now, we find the total number of ways to distribute 15 balls into 3 boxes without any restrictions:
\[ x_1 + x_2 + x_3 = 15 \]
Using the same formula:
\[ \text{Total unrestricted solutions} = \binom{15 + 3 - 1}{3 - 1} = \binom{17}{2} \]

7. **Calculating Total Unrestricted Solutions**:
\[ \binom{17}{2} = \frac{17 \times 16}{2 \times 1} = 136 \]

8. **Finding the Probability**:
The probability \( P \) that each box contains at least 3 balls is given by the ratio of favorable outcomes to total outcomes:
\[ P = \frac{\text{Favorable outcomes}}{\text{Total outcomes}} = \frac{28}{136} \]

9. **Simplifying the Probability**:
\[ P = \frac{28}{136} = \frac{7}{34} \]

10. **Calculating \( 34P \)**:
We need to find \( 34P \):
\[ 34P = 34 \times \frac{7}{34} = 7 \]

### Final Answer:
The value of \( 34P \) is \( 7 \).