Three urns contain respectively 3 white and 1 black balls, 2 white and 2 black, 1 white and 3 black. One ball is selected at random from each urn. Find the chance that the three selected comprise 1 white and 2 black balls.

Answer

To solve the problem of finding the probability that the three selected balls comprise 1 white and 2 black balls from the three urns, we will follow these steps:

### Step 1: Identify the contents of each urn.
- **Urn 1**: 3 white balls (W) and 1 black ball (B)
- **Urn 2**: 2 white balls (W) and 2 black balls (B)
- **Urn 3**: 1 white ball (W) and 3 black balls (B)

### Step 2: Determine the total number of ways to select 1 white and 2 black balls.
To get 1 white and 2 black balls, we can have the following combinations:
1. White from Urn 1, Black from Urn 2, Black from Urn 3.
2. Black from Urn 1, White from Urn 2, Black from Urn 3.
3. Black from Urn 1, Black from Urn 2, White from Urn 3.

### Step 3: Calculate the probability for each combination.

**Combination 1**: (W from Urn 1, B from Urn 2, B from Urn 3)
- Probability of selecting W from Urn 1 = \( \frac{3}{4} \)
- Probability of selecting B from Urn 2 = \( \frac{2}{4} = \frac{1}{2} \)
- Probability of selecting B from Urn 3 = \( \frac{3}{4} \)

Thus, the probability for this combination:
\[ P_1 = \frac{3}{4} \times \frac{1}{2} \times \frac{3}{4} = \frac{9}{32} \]

**Combination 2**: (B from Urn 1, W from Urn 2, B from Urn 3)
- Probability of selecting B from Urn 1 = \( \frac{1}{4} \)
- Probability of selecting W from Urn 2 = \( \frac{2}{4} = \frac{1}{2} \)
- Probability of selecting B from Urn 3 = \( \frac{3}{4} \)

Thus, the probability for this combination:
\[ P_2 = \frac{1}{4} \times \frac{1}{2} \times \frac{3}{4} = \frac{3}{32} \]

**Combination 3**: (B from Urn 1, B from Urn 2, W from Urn 3)
- Probability of selecting B from Urn 1 = \( \frac{1}{4} \)
- Probability of selecting B from Urn 2 = \( \frac{2}{4} = \frac{1}{2} \)
- Probability of selecting W from Urn 3 = \( \frac{1}{4} \)

Thus, the probability for this combination:
\[ P_3 = \frac{1}{4} \times \frac{1}{2} \times \frac{1}{4} = \frac{1}{32} \]

### Step 4: Add the probabilities of all combinations.
Now, we add the probabilities of all three combinations:
\[ P = P_1 + P_2 + P_3 = \frac{9}{32} + \frac{3}{32} + \frac{1}{32} = \frac{13}{32} \]

### Final Answer
The probability that the three selected balls comprise 1 white and 2 black balls is \( \frac{13}{32} \).

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