Twelve balls are distributed among three boxes, find the probability that the first box will contains three balls.

Explanation

To solve the problem of finding the probability that the first box will contain exactly three balls when distributing twelve balls among three boxes, we can follow these steps:

### Step 1: Determine the Total Number of Ways to Distribute the Balls
Each of the 12 balls can go into any of the 3 boxes. Therefore, for each ball, there are 3 choices. The total number of ways to distribute the 12 balls is given by:
\[ \text{Total Ways} = 3^{12} \]

### Step 2: Determine the Favorable Outcomes
We want to find the number of ways in which exactly 3 balls are placed in the first box.

1. **Choose 3 Balls for the First Box**: We need to select 3 balls from the 12 to place in the first box. The number of ways to choose 3 balls from 12 is given by the combination formula:
\[ \text{Ways to choose 3 balls} = \binom{12}{3} \]

2. **Distribute the Remaining Balls**: After placing 3 balls in the first box, we have 9 balls left. These 9 balls can go into either the second box or the third box. Each of these 9 balls has 2 choices (either box 2 or box 3). Therefore, the number of ways to distribute these 9 balls is:
\[ \text{Ways to distribute 9 balls} = 2^9 \]

3. **Calculate the Favorable Outcomes**: The total number of favorable outcomes where the first box contains exactly 3 balls is:
\[ \text{Favorable Outcomes} = \binom{12}{3} \times 2^9 \]

### Step 3: Calculate the Probability
The probability that the first box contains exactly 3 balls is given by the ratio of the number of favorable outcomes to the total number of outcomes:
\[ \text{Probability} = \frac{\text{Favorable Outcomes}}{\text{Total Outcomes}} = \frac{\binom{12}{3} \times 2^9}{3^{12}} \]

### Final Answer
Thus, the probability that the first box contains exactly 3 balls is:
\[ \text{Probability} = \frac{\binom{12}{3} \times 2^9}{3^{12}} \]