A bag contains 3 white and 4 black balls. One ball is drawn and kept aside. Then one white and one black balls are added to the bag. Now, one ball is drawn again. Find the probability that the second ball drawn is white.

Answer

To solve the problem step by step, we will use the concept of conditional probability and the law of total probability.

### Step-by-Step Solution:

1. **Identify the Initial Conditions:**
- The bag initially contains 3 white balls and 4 black balls.
- Total balls = 3 (white) + 4 (black) = 7 balls.

2. **Define Events:**
- Let \( E_1 \) be the event that the first ball drawn is white.
- Let \( E_2 \) be the event that the first ball drawn is black.

3. **Calculate Probabilities of Events:**
- Probability of drawing a white ball first:
\[ P(E_1) = \frac{3}{7} \]
- Probability of drawing a black ball first:
\[ P(E_2) = \frac{4}{7} \]

4. **Determine the Composition of the Bag After the First Draw:**
- If the first ball drawn is white (Event \( E_1 \)):
- Remaining balls: 2 white, 4 black.
- After adding 1 white and 1 black, the bag contains: 3 white and 5 black (total 8 balls).
- If the first ball drawn is black (Event \( E_2 \)):
- Remaining balls: 3 white, 3 black.
- After adding 1 white and 1 black, the bag contains: 4 white and 4 black (total 8 balls).

5. **Calculate Conditional Probabilities:**
- Probability of drawing a white ball second given that the first ball was white:
\[ P(W | E_1) = \frac{3}{8} \]
- Probability of drawing a white ball second given that the first ball was black:
\[ P(W | E_2) = \frac{4}{8} = \frac{1}{2} \]

6. **Apply the Law of Total Probability:**
- The total probability of drawing a white ball second is given by:
\[ P(W) = P(W | E_1) \cdot P(E_1) + P(W | E_2) \cdot P(E_2) \]
- Substituting the values:
\[ P(W) = \left(\frac{3}{8} \cdot \frac{3}{7}\right) + \left(\frac{1}{2} \cdot \frac{4}{7}\right) \]

7. **Calculate the Total Probability:**
- First term:
\[ \frac{3}{8} \cdot \frac{3}{7} = \frac{9}{56} \]
- Second term:
\[ \frac{1}{2} \cdot \frac{4}{7} = \frac{4}{14} = \frac{16}{56} \]
- Adding both terms:
\[ P(W) = \frac{9}{56} + \frac{16}{56} = \frac{25}{56} \]

### Final Answer:
The probability that the second ball drawn is white is \( \frac{25}{56} \).