A bag contains 4 white and 5 red balls. Second bag contains 6 white and 3 red balls. A ball is transferred from first bag to second and then a ball is drawn from the second bag. Find the probability that this ball drawn is red.

Explanation

To solve the problem step by step, we will analyze the situation involving two bags of balls and calculate the probability of drawing a red ball from the second bag after transferring a ball from the first bag.

### Step 1: Understand the Composition of the Bags
- **Bag 1** contains 4 white balls and 5 red balls. Therefore, the total number of balls in Bag 1 is:
\[ 4 + 5 = 9 \text{ balls} \]
- **Bag 2** contains 6 white balls and 3 red balls. Therefore, the total number of balls in Bag 2 is:
\[ 6 + 3 = 9 \text{ balls} \]

### Step 2: Consider the Two Cases for Transferring a Ball
When transferring a ball from Bag 1 to Bag 2, there are two possible cases:
1. A white ball is transferred.
2. A red ball is transferred.

### Step 3: Calculate the Probability for Case 1 (Transferring a White Ball)
- The probability of transferring a white ball from Bag 1 to Bag 2 is:
\[ P(\text{White}) = \frac{4}{9} \]
- After transferring a white ball, Bag 2 will have:
- White balls: \(6 + 1 = 7\)
- Red balls: \(3\)
- Total balls in Bag 2: \(7 + 3 = 10\)

- The probability of drawing a red ball from Bag 2 now is:
\[ P(\text{Red | White transferred}) = \frac{3}{10} \]

- Therefore, the total probability for this case is:
\[ P(\text{Case 1}) = P(\text{White}) \times P(\text{Red | White transferred}) = \frac{4}{9} \times \frac{3}{10} = \frac{12}{90} = \frac{2}{15} \]

### Step 4: Calculate the Probability for Case 2 (Transferring a Red Ball)
- The probability of transferring a red ball from Bag 1 to Bag 2 is:
\[ P(\text{Red}) = \frac{5}{9} \]
- After transferring a red ball, Bag 2 will have:
- White balls: \(6\)
- Red balls: \(3 + 1 = 4\)
- Total balls in Bag 2: \(6 + 4 = 10\)

- The probability of drawing a red ball from Bag 2 now is:
\[ P(\text{Red | Red transferred}) = \frac{4}{10} = \frac{2}{5} \]

- Therefore, the total probability for this case is:
\[ P(\text{Case 2}) = P(\text{Red}) \times P(\text{Red | Red transferred}) = \frac{5}{9} \times \frac{2}{5} = \frac{10}{45} = \frac{2}{9} \]

### Step 5: Combine the Probabilities from Both Cases
Since the two cases are mutually exclusive, we can add the probabilities:
\[ P(\text{Red ball drawn}) = P(\text{Case 1}) + P(\text{Case 2}) = \frac{2}{15} + \frac{2}{9} \]

### Step 6: Find a Common Denominator and Add
To add \(\frac{2}{15}\) and \(\frac{2}{9}\), we need a common denominator. The least common multiple of 15 and 9 is 45:
- Convert \(\frac{2}{15}\) to have a denominator of 45:
\[ \frac{2}{15} = \frac{2 \times 3}{15 \times 3} = \frac{6}{45} \]
- Convert \(\frac{2}{9}\) to have a denominator of 45:
\[ \frac{2}{9} = \frac{2 \times 5}{9 \times 5} = \frac{10}{45} \]

Now we can add:
\[ P(\text{Red ball drawn}) = \frac{6}{45} + \frac{10}{45} = \frac{16}{45} \]

### Final Answer
The probability that the ball drawn from the second bag is red is:
\[ \frac{16}{45} \]