( for explanation of law of total probability ) there are two bags . The first bag contains 4 white and 5 black balls and the second bag contains 5 white and 4 black balls , two balls are drawn at random from the first bag and are put into the second bag without noticing their colours , then two balls are drawn from the second bag . find the probability that the balls are white and black.

Answer

To solve the problem using the law of total probability, we will follow these steps:

### Step 1: Identify the scenarios
We have two bags:
- **Bag 1**: 4 white balls and 5 black balls.
- **Bag 2**: 5 white balls and 4 black balls.

When we draw 2 balls from Bag 1, there are three possible scenarios:
1. **Case A**: Both balls drawn are white (WW).
2. **Case B**: Both balls drawn are black (BB).
3. **Case C**: One ball is white and one ball is black (WB).

### Step 2: Calculate probabilities for each case
We need to calculate the probability of each case occurring when drawing 2 balls from Bag 1.

1. **Probability of Case A (WW)**:
\[ P(A) = \frac{\text{Number of ways to choose 2 white balls}}{\text{Total ways to choose 2 balls from Bag 1}} = \frac{4C2}{9C2} = \frac{6}{36} = \frac{1}{6} \]

2. **Probability of Case B (BB)**:
\[ P(B) = \frac{\text{Number of ways to choose 2 black balls}}{\text{Total ways to choose 2 balls from Bag 1}} = \frac{5C2}{9C2} = \frac{10}{36} = \frac{5}{18} \]

3. **Probability of Case C (WB)**:
\[ P(C) = \frac{\text{Number of ways to choose 1 white and 1 black ball}}{\text{Total ways to choose 2 balls from Bag 1}} = \frac{4C1 \cdot 5C1}{9C2} = \frac{20}{36} = \frac{5}{9} \]

### Step 3: Determine the composition of Bag 2 after transferring balls
Now, we need to find the probability of drawing one white and one black ball from Bag 2 after transferring the balls from Bag 1.

1. **If Case A occurs (WW)**:
- Bag 2 will have: 7 white and 4 black balls.
\[ P(D | A) = \frac{7C1 \cdot 4C1}{11C2} = \frac{28}{55} \]

2. **If Case B occurs (BB)**:
- Bag 2 will have: 5 white and 6 black balls.
\[ P(D | B) = \frac{5C1 \cdot 6C1}{11C2} = \frac{30}{55} \]

3. **If Case C occurs (WB)**:
- Bag 2 will have: 6 white and 5 black balls.
\[ P(D | C) = \frac{6C1 \cdot 5C1}{11C2} = \frac{30}{55} \]

### Step 4: Apply the law of total probability
Now we can find the total probability of drawing one white and one black ball from Bag 2:
\[ P(D) = P(A) \cdot P(D | A) + P(B) \cdot P(D | B) + P(C) \cdot P(D | C) \]

Substituting the values:
\[ P(D) = \left(\frac{1}{6} \cdot \frac{28}{55}\right) + \left(\frac{5}{18} \cdot \frac{30}{55}\right) + \left(\frac{5}{9} \cdot \frac{30}{55}\right) \]

Calculating each term:
1. \(\frac{1}{6} \cdot \frac{28}{55} = \frac{28}{330}\)
2. \(\frac{5}{18} \cdot \frac{30}{55} = \frac{150}{990} = \frac{50}{330}\)
3. \(\frac{5}{9} \cdot \frac{30}{55} = \frac{150}{495} = \frac{100}{330}\)

Adding them together:
\[ P(D) = \frac{28 + 50 + 100}{330} = \frac{178}{330} = \frac{89}{165} \]

### Final Answer
The probability that the balls drawn from Bag 2 are one white and one black is:
\[ \frac{89}{165} \]