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 15 identical balls are distributed among them, we can follow these steps:

### Step 1: Determine the Total Ways to Distribute 15 Balls
We need to find the total number of ways to distribute 15 identical balls into 3 different boxes. This can be calculated using the "stars and bars" theorem, which states that the number of ways to distribute \( n \) identical items into \( r \) distinct groups is given by:

\[ \text{Total Ways} = \binom{n + r - 1}{r - 1} \]

In our case, \( n = 15 \) (the balls) and \( r = 3 \) (the boxes):

\[ \text{Total Ways} = \binom{15 + 3 - 1}{3 - 1} = \binom{17}{2} \]

### Step 2: Calculate the Total Ways
Now we calculate \( \binom{17}{2} \):

\[ \binom{17}{2} = \frac{17 \times 16}{2 \times 1} = 136 \]

### Step 3: Determine the Favorable Outcomes
Next, we need to find the number of ways to distribute the balls such that each box contains at least 3 balls. If we place 3 balls in each box first, we will have:

\[ 15 - 3 \times 3 = 6 \text{ balls remaining} \]

Now, we need to distribute these 6 remaining balls into the 3 boxes. Using the stars and bars theorem again:

\[ \text{Favorable Ways} = \binom{6 + 3 - 1}{3 - 1} = \binom{8}{2} \]

### Step 4: Calculate the Favorable Ways
Now we calculate \( \binom{8}{2} \):

\[ \binom{8}{2} = \frac{8 \times 7}{2 \times 1} = 28 \]

### Step 5: Calculate the Probability
Now we can find the probability \( p \) that each box contains at least 3 balls:

\[ p = \frac{\text{Favorable Ways}}{\text{Total Ways}} = \frac{28}{136} \]

### Step 6: Simplify the Probability
We can simplify \( \frac{28}{136} \):

\[ p = \frac{28 \div 28}{136 \div 28} = \frac{1}{4.857} \approx \frac{7}{34} \]

### Step 7: Calculate \( 34p \)
Finally, we need to find \( 34p \):

\[ 34p = 34 \times \frac{7}{34} = 7 \]

### Final Answer
Thus, the value of \( 34p \) is:

\[ \boxed{7} \]