Twelve balls are placed in three boxes. The probability that the first box contains three balls is

Options

A. `55/3((2)/(3))^(11)`
B. `55((2)/(3))^(10)`
C. `220((1)/(3))^(12)`
D. `22((1)/(3))^(11)`

Explanation

To solve the problem of finding the probability that the first box contains exactly three balls when twelve balls are placed in three identical boxes, we can follow these steps:

### Step 1: Understand the Problem
We have 12 indistinguishable balls and 3 identical boxes. We need to find the probability that the first box contains exactly 3 balls.

### Step 2: Determine the Selection of Balls
Since we want the first box to contain exactly 3 balls, we need to choose 3 balls out of the 12 to place in the first box. The number of ways to choose 3 balls from 12 is given by the combination formula:

\[ \text{Number of ways to choose 3 balls} = \binom{12}{3} \]

### Step 3: Calculate the Remaining Balls
After placing 3 balls in the first box, we have 9 balls left to distribute among the 3 boxes. The remaining 9 balls can go into any of the 3 boxes.

### Step 4: Calculate the Total Distributions
The total number of ways to distribute 9 indistinguishable balls into 3 indistinguishable boxes can be calculated using the "stars and bars" theorem. The formula for distributing \( n \) indistinguishable objects into \( r \) distinguishable boxes is:

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

In our case, \( n = 9 \) (remaining balls) and \( r = 3 \) (boxes):

\[ \text{Total distributions} = \binom{9 + 3 - 1}{3 - 1} = \binom{11}{2} \]

### Step 5: Calculate the Probability
Now, we can calculate the probability that the first box contains exactly 3 balls. The probability is given by the ratio of the favorable outcomes to the total outcomes.

The total number of ways to distribute all 12 balls into 3 boxes is given by:

\[ \text{Total distributions of 12 balls} = \binom{12 + 3 - 1}{3 - 1} = \binom{14}{2} \]

Thus, the probability \( P \) that the first box contains exactly 3 balls is:

\[ P = \frac{\text{Number of ways to choose 3 balls} \times \text{Total distributions of remaining balls}}{\text{Total distributions of 12 balls}} \]

Substituting the values we calculated:

\[ P = \frac{\binom{12}{3} \times \binom{11}{2}}{\binom{14}{2}} \]

### Step 6: Simplifying the Expression
Now we can compute the values:

1. \(\binom{12}{3} = \frac{12 \times 11 \times 10}{3 \times 2 \times 1} = 220\)
2. \(\binom{11}{2} = \frac{11 \times 10}{2 \times 1} = 55\)
3. \(\binom{14}{2} = \frac{14 \times 13}{2 \times 1} = 91\)

Now substituting these values into the probability expression:

\[ P = \frac{220 \times 55}{91} \]

### Step 7: Final Calculation
Calculating this gives us:

\[ P = \frac{12100}{91} = \frac{12100}{91} \approx 133.33 \]

### Conclusion
The final probability that the first box contains exactly 3 balls is:

\[ P = \frac{220 \times 55}{91} \text{ (exact value)} \]