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 from bag A and transferred to bag B. Now a ball is taken 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: Define the events
Let:
- Event \( E_1 \): A red ball is transferred from bag A to bag B.
- Event \( E_2 \): A black ball is transferred from bag A to bag B.

### Step 2: Calculate the probabilities of transferring each type of ball from bag A
Bag A contains 4 red balls and 5 black balls, making a total of 9 balls.

- Probability of transferring a red ball \( P(E_1) \):
\[ P(E_1) = \frac{4}{9} \]

- Probability of transferring a black ball \( P(E_2) \):
\[ P(E_2) = \frac{5}{9} \]

### Step 3: Analyze the situation after transferring a ball
1. **If a red ball is transferred (Event \( E_1 \))**:
- Bag B will then have \( 6 + 1 = 7 \) red balls and 3 black balls.
- Total balls in bag B = \( 7 + 3 = 10 \).

Probability of drawing a red ball from bag B given \( E_1 \):
\[ P(E | E_1) = \frac{7}{10} \]

2. **If a black ball is transferred (Event \( E_2 \))**:
- Bag B will then have 6 red balls and \( 3 + 1 = 4 \) black balls.
- Total balls in bag B = \( 6 + 4 = 10 \).

Probability of drawing a red ball from bag B given \( E_2 \):
\[ P(E | E_2) = \frac{6}{10} = \frac{3}{5} \]

### Step 4: Use the Total Probability Theorem
The total probability of drawing a red ball from bag B is given by:
\[ P(E) = P(E_1) \cdot P(E | E_1) + P(E_2) \cdot P(E | E_2) \]

Substituting the values we calculated:
\[ P(E) = \left(\frac{4}{9} \cdot \frac{7}{10}\right) + \left(\frac{5}{9} \cdot \frac{3}{5}\right) \]

Calculating each term:
1. For the first term:
\[ \frac{4}{9} \cdot \frac{7}{10} = \frac{28}{90} \]

2. For the second term:
\[ \frac{5}{9} \cdot \frac{3}{5} = \frac{15}{45} = \frac{30}{90} \]

### Step 5: Combine the results
Now, we combine the two probabilities:
\[ P(E) = \frac{28}{90} + \frac{30}{90} = \frac{58}{90} \]

### Step 6: Simplify the final probability
\[ P(E) = \frac{58}{90} = \frac{29}{45} \]

Thus, the probability that the ball drawn from bag B is red is:
\[ \boxed{\frac{29}{45}} \]