After adding 4 red balls for each box, a box was chosen randomly and four balls
were taken out subsequently one after the other with replacements. Find the
probability of the balls taken out be 2 red balls, a white ball and a black ball.

Let's break down the problem step by step. We'll consider the setup and then calculate the probability for each scenario.

### Step 1: Setup After Adding Red Balls
- Originally, box \( A \) had \( n + 1 \) white balls and \( n + 1 \) black balls. After adding 4 red balls, box \( A \) now has \( n + 1 \) white balls, \( n + 1 \) black balls, and 4 red balls.
- The total number of balls in box \( A \) is \( 2n + 6 \).

- Similarly, box \( B \) originally had \( n + 2 \) white balls and \( n \) black balls. After adding 4 red balls, box \( B \) now has \( n + 2 \) white balls, \( n \) black balls, and 4 red balls.
- The total number of balls in box \( B \) is also \( 2n + 6 \).

### Step 2: Probabilities of Drawing Each Color from Box A
- Probability of drawing a red ball from box \( A \) is \( \frac{4}{2n+6} \).
- Probability of drawing a white ball from box \( A \) is \( \frac{n+1}{2n+6} \).
- Probability of drawing a black ball from box \( A \) is \( \frac{n+1}{2n+6} \).

### Step 3: Probabilities of Drawing Each Color from Box B
- Probability of drawing a red ball from box \( B \) is \( \frac{4}{2n+6} \).
- Probability of drawing a white ball from box \( B \) is \( \frac{n+2}{2n+6} \).
- Probability of drawing a black ball from box \( B \) is \( \frac{n}{2n+6} \).

### Step 4: Probability of Drawing 2 Red Balls, 1 White Ball, and 1 Black Ball
Since the balls are drawn with replacement, the order in which the balls are drawn matters. However, the problem states that we want the probability of drawing exactly 2 red balls, 1 white ball, and 1 black ball in any order.

The number of different orders in which this can occur is the number of permutations of the sequence "R, R, W, B", which is \( \frac{4!}{2!1!1!} = 12 \).

#### Probability for Box A:
The probability of drawing 2 red balls, 1 white ball, and 1 black ball from box \( A \) in a specific order (e.g., R, R, W, B) is:
\[
\left(\frac{4}{2n+6}\right) \times \left(\frac{4}{2n+6}\right) \times \left(\frac{n+1}{2n+6}\right) \times \left(\frac{n+1}{2n+6}\right)
\]
Since there are 12 different orders, the total probability for box \( A \) is:
\[
P(\text{2R, 1W, 1B from A}) = 12 \times \left(\frac{4^2 \times (n+1)^2}{(2n+6)^4}\right)
\]

#### Probability for Box B:
Similarly, the probability of drawing 2 red balls, 1 white ball, and 1 black ball from box \( B \) in a specific order (e.g., R, R, W, B) is:
\[
\left(\frac{4}{2n+6}\right) \times \left(\frac{4}{2n+6}\right) \times \left(\frac{n+2}{2n+6}\right) \times \left(\frac{n}{2n+6}\right)
\]
The total probability for box \( B \) is:
\[
P(\text{2R, 1W, 1B from B}) = 12 \times \left(\frac{4^2 \times (n+2) \times n}{(2n+6)^4}\right)
\]

### Step 5: Overall Probability (Considering Both Boxes)
The overall probability that the balls drawn are 2 red, 1 white, and 1 black is a weighted average of the probabilities for box \( A \) and box \( B \).

Since the box is chosen randomly, the probability of choosing either box \( A \) or \( B \) is \( \frac{1}{2} \).

Thus, the total probability \( P(\text{2R, 1W, 1B}) \) is:
\[
P(\text{2R, 1W, 1B}) = \frac{1}{2} \times P(\text{2R, 1W, 1B from A}) + \frac{1}{2} \times P(\text{2R, 1W, 1B from B})
\]
Substituting the values:
\[
P(\text{2R, 1W, 1B}) = \frac{1}{2} \times 12 \times \left(\frac{16 \times (n+1)^2}{(2n+6)^4}\right) + \frac{1}{2} \times 12 \times \left(\frac{16 \times (n+2) \times n}{(2n+6)^4}\right)
\]
Simplifying further:
\[
P(\text{2R, 1W, 1B}) = \frac{96}{(2n+6)^4} \times \frac{(n+1)^2 + (n+2)n}{2}
\]

Thus, the final probability of drawing 2 red balls, 1 white ball, and 1 black ball is:
\[
P(\text{2R, 1W, 1B}) = \frac{48 \left( n^2 + 3n + 2 \right)}{(2n+6)^4}
\]

Would you like more details or have any questions? Here are some related questions:

1. How would the probability change if there were no replacement?
2. What if we want to find the probability of drawing exactly 3 red balls and 1 white ball?
3. How does this probability change if only box \( A \) is considered?
4. What if \( n = 2 \)? What is the exact probability?
5. How would you calculate the expected number of red balls drawn in this scenario?

**Tip:** When dealing with problems involving multiple steps and different possible orders, it can be helpful to calculate the probability for a specific order and then multiply by the number of possible orders.

Calculate the probability of drawing 2 red balls, 1 white ball, and 1 black ball from a box after adding 4 red balls and considering replacements. This detailed solution covers the setup, probabilities from two boxes, and the overall probability calculation. Suitable for advanced college-level students and mathematicians.

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