An urn contains 4 white and 5 black balls, a second urn contains 5 white and 4 black balls. One ball is transferred from the first to second urn, then a ball is drawn from the second urn, what is the probability that it is white ?

Answer

To solve the problem, we need to find the probability of drawing a white ball from the second urn after transferring one ball from the first urn to the second urn. We will consider two cases based on the color of the ball transferred from the first urn.

### Step-by-Step Solution:

1. **Identify the contents of the urns:**
- First urn: 4 white balls and 5 black balls (total = 9 balls).
- Second urn: 5 white balls and 4 black balls (total = 9 balls).

2. **Case 1: A white ball is transferred from the first urn to the second urn.**
- Probability of transferring a white ball = Number of white balls in the first urn / Total number of balls in the first urn = \( \frac{4}{9} \).
- After transferring a white ball, the second urn will have:
- White balls = 5 + 1 = 6
- Black balls = 4
- Total balls in the second urn = 6 + 4 = 10.
- Probability of drawing a white ball from the second urn = Number of white balls in the second urn / Total number of balls in the second urn = \( \frac{6}{10} = \frac{3}{5} \).

3. **Case 2: A black ball is transferred from the first urn to the second urn.**
- Probability of transferring a black ball = Number of black balls in the first urn / Total number of balls in the first urn = \( \frac{5}{9} \).
- After transferring a black ball, the second urn will have:
- White balls = 5
- Black balls = 4 + 1 = 5
- Total balls in the second urn = 5 + 5 = 10.
- Probability of drawing a white ball from the second urn = Number of white balls in the second urn / Total number of balls in the second urn = \( \frac{5}{10} = \frac{1}{2} \).

4. **Combine the probabilities from both cases using the law of total probability:**
- Total probability of drawing a white ball from the second urn = (Probability of transferring a white ball × Probability of drawing a white ball from the second urn) + (Probability of transferring a black ball × Probability of drawing a white ball from the second urn).
- This can be expressed as:
\[ P(\text{White from 2nd urn}) = P(\text{White transferred}) \times P(\text{White drawn | White transferred}) + P(\text{Black transferred}) \times P(\text{White drawn | Black transferred}) \]
- Substituting the values:
\[ P(\text{White from 2nd urn}) = \left(\frac{4}{9} \times \frac{3}{5}\right) + \left(\frac{5}{9} \times \frac{1}{2}\right) \]
- Calculating each term:
- First term: \( \frac{4}{9} \times \frac{3}{5} = \frac{12}{45} \).
- Second term: \( \frac{5}{9} \times \frac{1}{2} = \frac{5}{18} = \frac{25}{90} \) (converting to a common denominator).
- Now, convert \( \frac{12}{45} \) to a common denominator of 90:
- \( \frac{12}{45} = \frac{24}{90} \).
- Now, add the two fractions:
\[ P(\text{White from 2nd urn}) = \frac{24}{90} + \frac{25}{90} = \frac{49}{90}. \]

### Final Answer:
The probability that the ball drawn from the second urn is white is \( \frac{49}{90} \).