One bag a contains 4 red and 5 black balls. The other bag B contains 6 red and 3 black balls. A ball is taken rom bag A and transferred to bag B. Now a ball is from bag B. Find the probability that the ball drawn is red.

Explanation

To solve the problem step by step, we will calculate the probability of drawing a red ball from bag B after transferring a ball from bag A.

### Step 1: Identify the contents of the bags
- Bag A contains 4 red balls and 5 black balls.
- Bag B contains 6 red balls and 3 black balls.

### Step 2: Calculate the total number of balls in each bag
- Total balls in Bag A = 4 (red) + 5 (black) = 9 balls
- Total balls in Bag B = 6 (red) + 3 (black) = 9 balls

### Step 3: Define events
Let:
- Event E1: A red ball is transferred from Bag A to Bag B.
- Event E2: A black ball is transferred from Bag A to Bag B.

### Step 4: Calculate the probabilities of transferring a ball from Bag A
- Probability of transferring a red ball (E1):
\[ P(E1) = \frac{4}{9} \]
- Probability of transferring a black ball (E2):
\[ P(E2) = \frac{5}{9} \]

### Step 5: Calculate the probability of drawing a red ball from Bag B after transferring a ball
1. **If a red ball is transferred (Event E1)**:
- Bag B now has 7 red balls (6 original + 1 transferred) and 3 black balls.
- Total balls in Bag B = 7 (red) + 3 (black) = 10 balls.
- Probability of drawing a red ball from Bag B:
\[ P(\text{Red} | E1) = \frac{7}{10} \]

2. **If a black ball is transferred (Event E2)**:
- Bag B still has 6 red balls and now has 4 black balls (3 original + 1 transferred).
- Total balls in Bag B = 6 (red) + 4 (black) = 10 balls.
- Probability of drawing a red ball from Bag B:
\[ P(\text{Red} | E2) = \frac{6}{10} = \frac{3}{5} \]

### Step 6: Use the law of total probability to find the overall probability of drawing a red ball from Bag B
\[ P(\text{Red}) = P(E1) \cdot P(\text{Red} | E1) + P(E2) \cdot P(\text{Red} | E2) \]
Substituting the values:
\[ P(\text{Red}) = \left(\frac{4}{9} \cdot \frac{7}{10}\right) + \left(\frac{5}{9} \cdot \frac{3}{5}\right) \]

### Step 7: Calculate each term
1. First term:
\[ \frac{4}{9} \cdot \frac{7}{10} = \frac{28}{90} \]
2. Second term:
\[ \frac{5}{9} \cdot \frac{3}{5} = \frac{15}{45} = \frac{30}{90} \]

### Step 8: Combine the results
\[ P(\text{Red}) = \frac{28}{90} + \frac{30}{90} = \frac{58}{90} = \frac{29}{45} \]

### Final Answer
The probability that the ball drawn from Bag B is red is:
\[ \frac{29}{45} \]