A bag 'A' contains 6 white and 7 black balls while the other bag 'B' contains 4 white and 5 black balls. A ball is transferred from the bag A to the bag B. Then a ball is drawn from the bag B. Find the probability that the ball drawn is white.

Answer

To find the probability that the ball drawn from bag B is white after transferring a ball from bag A, we can break the problem down into two cases based on the color of the ball transferred from bag A.

### Step 1: Define the contents of the bags
- Bag A contains 6 white and 7 black balls.
- Bag B contains 4 white and 5 black balls.

### Step 2: Identify the two cases for transferring a ball from bag A to bag B
1. **Case 1**: A white ball is transferred from bag A to bag B.
2. **Case 2**: A black ball is transferred from bag A to bag B.

### Step 3: Calculate the probabilities for each case

**Case 1**: A white ball is transferred.
- Probability of transferring a white ball from bag A:
\[ P(\text{White from A}) = \frac{6}{13} \]
- After transferring a white ball, bag B will have:
- White balls: \(4 + 1 = 5\)
- Black balls: \(5\)

- Total balls in bag B after transfer: \(5 + 5 = 10\)
- Probability of drawing a white ball from bag B:
\[ P(\text{White from B | White transferred}) = \frac{5}{10} = \frac{1}{2} \]

**Case 2**: A black ball is transferred.
- Probability of transferring a black ball from bag A:
\[ P(\text{Black from A}) = \frac{7}{13} \]
- After transferring a black ball, bag B will have:
- White balls: \(4\)
- Black balls: \(5 + 1 = 6\)

- Total balls in bag B after transfer: \(4 + 6 = 10\)
- Probability of drawing a white ball from bag B:
\[ P(\text{White from B | Black transferred}) = \frac{4}{10} = \frac{2}{5} \]

### Step 4: Use the law of total probability to find the overall probability of drawing a white ball from bag B
\[ P(\text{White from B}) = P(\text{White from A}) \times P(\text{White from B | White transferred}) + P(\text{Black from A}) \times P(\text{White from B | Black transferred}) \]
Substituting the values:
\[ P(\text{White from B}) = \left(\frac{6}{13} \times \frac{1}{2}\right) + \left(\frac{7}{13} \times \frac{2}{5}\right) \]
Calculating each term:
- First term:
\[ \frac{6}{13} \times \frac{1}{2} = \frac{6}{26} = \frac{3}{13} \]
- Second term:
\[ \frac{7}{13} \times \frac{2}{5} = \frac{14}{65} \]

### Step 5: Find a common denominator and add the probabilities
The common denominator for \(13\) and \(65\) is \(65\):
- Convert \(\frac{3}{13}\) to have a denominator of \(65\):
\[ \frac{3}{13} = \frac{3 \times 5}{13 \times 5} = \frac{15}{65} \]
- Now add the two fractions:
\[ P(\text{White from B}) = \frac{15}{65} + \frac{14}{65} = \frac{29}{65} \]

### Final Answer
The probability that the ball drawn from bag B is white is:
\[ \frac{29}{65} \]