Three urns respectively contain 3 white and 2 black, 2 white and 3 black, and 1 white and 4 black balls. One ball is drawn from each urn. Find the probability that the selection contains 1 black and 2 white balls.

Answer

To solve the problem of finding the probability that the selection contains 1 black and 2 white balls when one ball is drawn from each of the three urns, we can break down the solution into the following steps:

### Step 1: Identify the contents of each urn
- **Urn 1**: 3 white balls, 2 black balls
- **Urn 2**: 2 white balls, 3 black balls
- **Urn 3**: 1 white ball, 4 black balls

### Step 2: Determine the possible cases for drawing 1 black and 2 white balls
We can have the following combinations to achieve 1 black and 2 white balls:
1. **Case 1**: Black from Urn 1, White from Urn 2, White from Urn 3
2. **Case 2**: White from Urn 1, Black from Urn 2, White from Urn 3
3. **Case 3**: White from Urn 1, White from Urn 2, Black from Urn 3

### Step 3: Calculate the probability for each case
#### Case 1: Black from Urn 1, White from Urn 2, White from Urn 3
- Probability of drawing black from Urn 1:
\[ P(B_1) = \frac{2}{5} \]
- Probability of drawing white from Urn 2:
\[ P(W_2) = \frac{2}{5} \]
- Probability of drawing white from Urn 3:
\[ P(W_3) = \frac{1}{5} \]
- Combined probability for Case 1:
\[ P(\text{Case 1}) = P(B_1) \times P(W_2) \times P(W_3) = \frac{2}{5} \times \frac{2}{5} \times \frac{1}{5} = \frac{4}{125} \]

#### Case 2: White from Urn 1, Black from Urn 2, White from Urn 3
- Probability of drawing white from Urn 1:
\[ P(W_1) = \frac{3}{5} \]
- Probability of drawing black from Urn 2:
\[ P(B_2) = \frac{3}{5} \]
- Probability of drawing white from Urn 3:
\[ P(W_3) = \frac{1}{5} \]
- Combined probability for Case 2:
\[ P(\text{Case 2}) = P(W_1) \times P(B_2) \times P(W_3) = \frac{3}{5} \times \frac{3}{5} \times \frac{1}{5} = \frac{9}{125} \]

#### Case 3: White from Urn 1, White from Urn 2, Black from Urn 3
- Probability of drawing white from Urn 1:
\[ P(W_1) = \frac{3}{5} \]
- Probability of drawing white from Urn 2:
\[ P(W_2) = \frac{2}{5} \]
- Probability of drawing black from Urn 3:
\[ P(B_3) = \frac{4}{5} \]
- Combined probability for Case 3:
\[ P(\text{Case 3}) = P(W_1) \times P(W_2) \times P(B_3) = \frac{3}{5} \times \frac{2}{5} \times \frac{4}{5} = \frac{24}{125} \]

### Step 4: Sum the probabilities of all cases
To find the total probability of drawing 1 black and 2 white balls, we sum the probabilities of all three cases:
\[ P(\text{Total}) = P(\text{Case 1}) + P(\text{Case 2}) + P(\text{Case 3}) = \frac{4}{125} + \frac{9}{125} + \frac{24}{125} = \frac{37}{125} \]

### Final Answer
The probability that the selection contains 1 black and 2 white balls is:
\[ \frac{37}{125} \]