Let `n_(1)and n_(2)` be the number of red and black balls, respectively, in box I. Let `n_(3) and n_(4)` be the numbers of red and black balls, respectively, in the box II.
A ball is drawn at random from box I and transferred to box II. If the probability of drawing a red ball from box I, after this transfer, is 1/3, then the correct options (s) with the possible values of `n_(1) and n_(2)` is (are)

Options

A. `n_(1)=4andn_(2)=6`
B. `n_(1)=2andn_(2)=3`
C. `n_(1)=10andn_(2)=20`
D. `n_(1)=3andn_(2)=6`

Explanation

To solve the problem, we need to analyze the situation step by step. We have two boxes, Box I and Box II, containing red and black balls. We denote the number of red balls in Box I as \( n_1 \) and the number of black balls in Box I as \( n_2 \). Similarly, we denote the number of red balls in Box II as \( n_3 \) and the number of black balls in Box II as \( n_4 \).

### Step-by-Step Solution:

1. **Understanding the Initial Setup**:
- In Box I, we have \( n_1 \) red balls and \( n_2 \) black balls.
- The total number of balls in Box I is \( n_1 + n_2 \).

2. **Calculating the Probability of Drawing a Red Ball**:
- The probability of drawing a red ball from Box I before any transfer is:
\[ P(\text{Red from Box I}) = \frac{n_1}{n_1 + n_2} \]

3. **Transferring a Ball**:
- A ball is drawn at random from Box I and transferred to Box II. This can either be a red ball or a black ball.
- If a red ball is transferred, the new counts in Box I will be \( n_1 - 1 \) red balls and \( n_2 \) black balls.
- If a black ball is transferred, the counts will remain \( n_1 \) red balls and \( n_2 - 1 \) black balls.

4. **Calculating the New Probability of Drawing a Red Ball from Box I**:
- After transferring a red ball:
\[ P(\text{Red from Box I}) = \frac{n_1 - 1}{(n_1 - 1) + n_2} = \frac{n_1 - 1}{n_1 + n_2 - 1} \]
- After transferring a black ball:
\[ P(\text{Red from Box I}) = \frac{n_1}{n_1 + (n_2 - 1)} = \frac{n_1}{n_1 + n_2 - 1} \]

5. **Using the Given Probability**:
- We are given that the probability of drawing a red ball from Box I after the transfer is \( \frac{1}{3} \). This means:
\[ \frac{n_1}{n_1 + n_2 - 1} = \frac{1}{3} \]

6. **Cross Multiplying to Solve for \( n_1 \) and \( n_2 \)**:
- Cross-multiplying gives us:
\[ 3n_1 = n_1 + n_2 - 1 \]
- Rearranging this equation:
\[ 3n_1 - n_1 - n_2 + 1 = 0 \implies 2n_1 - n_2 + 1 = 0 \]
- Therefore, we can express \( n_2 \) in terms of \( n_1 \):
\[ n_2 = 2n_1 + 1 \]

7. **Finding Possible Values**:
- Since \( n_1 \) and \( n_2 \) must be non-negative integers, we can choose values for \( n_1 \):
- If \( n_1 = 1 \), then \( n_2 = 2(1) + 1 = 3 \).
- If \( n_1 = 2 \), then \( n_2 = 2(2) + 1 = 5 \).
- If \( n_1 = 3 \), then \( n_2 = 2(3) + 1 = 7 \).
- And so on...

### Final Answer:
The possible values of \( n_1 \) and \( n_2 \) are:
- \( (1, 3) \)
- \( (2, 5) \)
- \( (3, 7) \)
- \( (k, 2k + 1) \) for any non-negative integer \( k \).