Given two bags A and B as follows : Bag A 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. The probability that both balls drawn are of the same colour is

Options

A. `187/1680`
B. `901/1680`
C. `439/1680`
D. None of these

Answer

To solve the problem step by step, we will analyze the situation with the two bags and calculate the probabilities accordingly.

### Step 1: Understand the contents of the bags
- Bag A contains:
- 3 Red balls
- 2 White balls
- Bag B contains:
- 2 Red balls
- 5 White balls

### Step 2: Calculate the probability of selecting each bag
Since a bag is selected at random, the probability of selecting either bag A or bag B is:
- P(A) = 1/2
- P(B) = 1/2

### Step 3: Calculate the probabilities for transferring a ball from Bag A to Bag B
1. **If we select Bag A and transfer a ball to Bag B:**
- Probability of transferring a Red ball from Bag A to Bag B = 3/5
- Probability of transferring a White ball from Bag A to Bag B = 2/5

After transferring a Red ball:
- Bag B will have 3 Red and 5 White balls (total 8 balls).
- Probability of drawing a Red ball from Bag B = 3/8
- Probability of drawing a White ball from Bag B = 5/8

After transferring a White ball:
- Bag B will have 2 Red and 6 White balls (total 8 balls).
- Probability of drawing a Red ball from Bag B = 2/8 = 1/4
- Probability of drawing a White ball from Bag B = 6/8 = 3/4

2. **Calculate the total probability of drawing the same color ball from Bag B after transferring from Bag A:**
- Probability of both balls being Red:
\[ P(\text{Red from A and Red from B}) = P(A) \cdot P(\text{Red from A}) \cdot P(\text{Red from B | Red from A}) = \frac{1}{2} \cdot \frac{3}{5} \cdot \frac{3}{8} = \frac{9}{80} \]

- Probability of both balls being White:
\[ P(\text{White from A and White from B}) = P(A) \cdot P(\text{White from A}) \cdot P(\text{White from B | White from A}) = \frac{1}{2} \cdot \frac{2}{5} \cdot \frac{3}{4} = \frac{3}{40} \]

- Total probability of drawing the same color after transferring from Bag A:
\[ P(\text{Same color from A}) = \frac{9}{80} + \frac{3}{40} = \frac{9}{80} + \frac{6}{80} = \frac{15}{80} = \frac{3}{16} \]

### Step 4: Calculate the probabilities for transferring a ball from Bag B to Bag A
1. **If we select Bag B and transfer a ball to Bag A:**
- Probability of transferring a Red ball from Bag B to Bag A = 2/7
- Probability of transferring a White ball from Bag B to Bag A = 5/7

After transferring a Red ball:
- Bag A will have 4 Red and 2 White balls (total 6 balls).
- Probability of drawing a Red ball from Bag A = 4/6 = 2/3
- Probability of drawing a White ball from Bag A = 2/6 = 1/3

After transferring a White ball:
- Bag A will have 3 Red and 3 White balls (total 6 balls).
- Probability of drawing a Red ball from Bag A = 3/6 = 1/2
- Probability of drawing a White ball from Bag A = 3/6 = 1/2

2. **Calculate the total probability of drawing the same color ball from Bag A after transferring from Bag B:**
- Probability of both balls being Red:
\[ P(\text{Red from B and Red from A}) = P(B) \cdot P(\text{Red from B}) \cdot P(\text{Red from A | Red from B}) = \frac{1}{2} \cdot \frac{2}{7} \cdot \frac{2}{3} = \frac{2}{21} \]

- Probability of both balls being White:
\[ P(\text{White from B and White from A}) = P(B) \cdot P(\text{White from B}) \cdot P(\text{White from A | White from B}) = \frac{1}{2} \cdot \frac{5}{7} \cdot \frac{1}{2} = \frac{5}{28} \]

- Total probability of drawing the same color after transferring from Bag B:
\[ P(\text{Same color from B}) = \frac{2}{21} + \frac{5}{28} \]
To add these fractions, we find a common denominator (84):
\[ P(\text{Same color from B}) = \frac{8}{84} + \frac{15}{84} = \frac{23}{84} \]

### Step 5: Combine the probabilities
Now, we combine the probabilities of both events:
\[ P(\text{Same color}) = P(\text{Same color from A}) + P(\text{Same color from B}) = \frac{3}{16} + \frac{23}{84} \]
Finding a common denominator (336):
\[ P(\text{Same color}) = \frac{63}{336} + \frac{92}{336} = \frac{155}{336} \]

### Final Answer
The probability that both balls drawn are of the same color is:
\[ \frac{155}{336} \]