Bag 4 contains 3 red and 2 white balls, and Bag B contains 2 red and 5 white balls. A bag is selected at random, a ball is drawn and put into the other bag, then a ball is drawn from the second bag. Find the probability that both balls drawn are of the same colour.

Answer

To solve the problem step by step, we will analyze the events and calculate the probabilities accordingly.

### Step 1: Define the Events
Let:
- Bag A contains 3 red and 2 white balls.
- Bag B contains 2 red and 5 white balls.

We will consider the following events:
1. **Event E1**: Transfer a red ball from Bag A to Bag B.
2. **Event E2**: Transfer a white ball from Bag A to Bag B.
3. **Event E3**: Transfer a red ball from Bag B to Bag A.
4. **Event E4**: Transfer a white ball from Bag B to Bag A.

### Step 2: Calculate the Probabilities of Transferring Balls
1. **Probability of transferring a red ball from A to B (E1)**:
\[ P(E1) = \frac{3}{5} \quad \text{(3 red balls out of 5 total)} \]

2. **Probability of transferring a white ball from A to B (E2)**:
\[ P(E2) = \frac{2}{5} \quad \text{(2 white balls out of 5 total)} \]

3. **Probability of transferring a red ball from B to A (E3)**:
After transferring a red ball from A to B, Bag B will have 3 red and 5 white balls:
\[ P(E3) = \frac{2}{7} \quad \text{(2 red balls out of 7 total)} \]

4. **Probability of transferring a white ball from B to A (E4)**:
After transferring a white ball from A to B, Bag B will have 2 red and 6 white balls:
\[ P(E4) = \frac{5}{7} \quad \text{(5 white balls out of 7 total)} \]

### Step 3: Calculate the Probabilities of Drawing Balls
1. **Probability of drawing a red ball from B after transferring a red ball from A (Event A)**:
\[ P(A|E1) = \frac{3}{8} \quad \text{(3 red balls out of 8 total)} \]

2. **Probability of drawing a white ball from B after transferring a white ball from A (Event B)**:
\[ P(B|E2) = \frac{6}{8} = \frac{3}{4} \quad \text{(6 white balls out of 8 total)} \]

3. **Probability of drawing a red ball from A after transferring a red ball from B (Event C)**:
\[ P(C|E3) = \frac{4}{6} = \frac{2}{3} \quad \text{(4 red balls out of 6 total)} \]

4. **Probability of drawing a white ball from A after transferring a white ball from B (Event D)**:
\[ P(D|E4) = \frac{3}{6} = \frac{1}{2} \quad \text{(3 white balls out of 6 total)} \]

### Step 4: Calculate the Total Probability of Drawing Balls of the Same Color
Using the law of total probability, we can find the probability of drawing two balls of the same color.

1. **Probability of drawing same color when transferring from A to B**:
\[ P(\text{Same Color}|E1) = P(E1) \cdot P(A|E1) + P(E2) \cdot P(B|E2) \]
\[ = \left(\frac{3}{5} \cdot \frac{3}{8}\right) + \left(\frac{2}{5} \cdot \frac{6}{8}\right) \]
\[ = \frac{9}{40} + \frac{12}{40} = \frac{21}{40} \]

2. **Probability of drawing same color when transferring from B to A**:
\[ P(\text{Same Color}|E3) = P(E3) \cdot P(C|E3) + P(E4) \cdot P(D|E4) \]
\[ = \left(\frac{2}{7} \cdot \frac{2}{3}\right) + \left(\frac{5}{7} \cdot \frac{1}{2}\right) \]
\[ = \frac{4}{21} + \frac{5}{14} \]
To add these fractions, we need a common denominator (42):
\[ = \frac{8}{42} + \frac{15}{42} = \frac{23}{42} \]

### Step 5: Combine the Probabilities
Since either bag can be chosen with equal probability:
\[ P(\text{Same Color}) = \frac{1}{2} \left( \frac{21}{40} + \frac{23}{42} \right) \]
Finding a common denominator (840):
\[ = \frac{1}{2} \left( \frac{441}{840} + \frac{460}{840} \right) = \frac{1}{2} \cdot \frac{901}{840} = \frac{901}{1680} \]

### Final Answer
The probability that both balls drawn are of the same color is:
\[ \frac{901}{1680} \]