We are tasked with finding the probability that the first box will contain exactly 3 balls when 12 balls are randomly distributed among three boxes. Here's a step-by-step approach:

Step 1: Total possible outcomes

Each of the 12 balls can be placed in any of the 3 boxes. Therefore, the total number of possible outcomes is given by:

$3^{12}$ This represents all the possible ways of distributing 12 balls into 3 boxes.

Step 2: Favorable outcomes

We want to count the number of favorable outcomes where exactly 3 balls are in the first box, and the remaining 9 balls are distributed between the second and third boxes.

1. First, select 3 balls to be placed in the first box. The number of ways to choose 3 balls from 12 is given by the combination: $\binom{12}{3} = \frac{12!}{3!(12-3)!} = \frac{12 \times 11 \times 10}{3 \times 2 \times 1} = 220$

2. Now, the remaining 9 balls need to be placed in the second and third boxes. Each ball has 2 choices (it can go into either the second or third box), so there are $2^9$ ways to distribute these 9 balls.

Thus, the total number of favorable outcomes is: $220 \times 2^9$

Step 3: Probability calculation

The probability is the ratio of the number of favorable outcomes to the total number of possible outcomes. So the probability that the first box contains exactly 3 balls is:

$\text{Probability} = \frac{220 \times 2^9}{3^{12}}$

Calculating the powers and simplifying:

$2^9 = 512 \quad \text{and} \quad 3^{12} = 531441$

Thus, the probability is:

$\text{Probability} = \frac{220 \times 512}{531441} = \frac{112640}{531441} \approx 0.2119$

So, the probability that the first box will contain exactly 3 balls is approximately 0.2119, or about 21.19%.

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Would you like a more detailed breakdown of any of these steps or have any questions?

Here are some related questions you can explore:

1. What is the probability that the first box will contain exactly 4 balls?
2. What is the probability that the second box will contain exactly 5 balls?
3. What is the probability that one of the boxes will contain all 12 balls?
4. How does the probability change if there are 4 boxes instead of 3?
5. What is the probability that the first box contains 0 balls?

Tip: When dealing with combinations, remember that $\binom{n}{k}$ represents the number of ways to choose $k$ objects from $n$ without regard to order.