Three groups of children contain respectively 3 girls and 1 boy, 2 girls and 2 boys and 1 girl and 3 boys. One child is selected at random from each group. The probability that the three selected consists of 1 girl and 2 boys is

Options

A. `13/32`
B. `12/32`
C. `15/32`
D. `11/32`

Answer

To find the probability that the three selected children consist of 1 girl and 2 boys from three groups of children, we can break down the problem step by step.

### Step 1: Identify the groups and their compositions
- Group A: 3 girls and 1 boy
- Group B: 2 girls and 2 boys
- Group C: 1 girl and 3 boys

### Step 2: Determine the possible combinations for selecting 1 girl and 2 boys
The combinations of selecting 1 girl and 2 boys from the three groups can be:
1. Girl from Group A, Boy from Group B, Boy from Group C
2. Boy from Group A, Girl from Group B, Boy from Group C
3. Boy from Group A, Boy from Group B, Girl from Group C

### Step 3: Calculate the probabilities for each combination

#### Combination 1: Girl from Group A, Boy from Group B, Boy from Group C
- Probability of selecting a girl from Group A: \( P(G_A) = \frac{3}{4} \)
- Probability of selecting a boy from Group B: \( P(B_B) = \frac{2}{4} = \frac{1}{2} \)
- Probability of selecting a boy from Group C: \( P(B_C) = \frac{3}{4} \)

The probability for this combination is:
\[ P(G_A, B_B, B_C) = P(G_A) \times P(B_B) \times P(B_C) = \frac{3}{4} \times \frac{1}{2} \times \frac{3}{4} = \frac{9}{32} \]

#### Combination 2: Boy from Group A, Girl from Group B, Boy from Group C
- Probability of selecting a boy from Group A: \( P(B_A) = \frac{1}{4} \)
- Probability of selecting a girl from Group B: \( P(G_B) = \frac{2}{4} = \frac{1}{2} \)
- Probability of selecting a boy from Group C: \( P(B_C) = \frac{3}{4} \)

The probability for this combination is:
\[ P(B_A, G_B, B_C) = P(B_A) \times P(G_B) \times P(B_C) = \frac{1}{4} \times \frac{1}{2} \times \frac{3}{4} = \frac{3}{32} \]

#### Combination 3: Boy from Group A, Boy from Group B, Girl from Group C
- Probability of selecting a boy from Group A: \( P(B_A) = \frac{1}{4} \)
- Probability of selecting a boy from Group B: \( P(B_B) = \frac{2}{4} = \frac{1}{2} \)
- Probability of selecting a girl from Group C: \( P(G_C) = \frac{1}{4} \)

The probability for this combination is:
\[ P(B_A, B_B, G_C) = P(B_A) \times P(B_B) \times P(G_C) = \frac{1}{4} \times \frac{1}{2} \times \frac{1}{4} = \frac{1}{32} \]

### Step 4: Add the probabilities of all combinations
Now, we sum the probabilities of all three combinations:
\[ P(\text{1 girl and 2 boys}) = P(G_A, B_B, B_C) + P(B_A, G_B, B_C) + P(B_A, B_B, G_C) \]
\[ = \frac{9}{32} + \frac{3}{32} + \frac{1}{32} = \frac{13}{32} \]

### Final Answer
The probability that the three selected children consist of 1 girl and 2 boys is \( \frac{13}{32} \).

---