There are two family each having two children. If there are at least two girls among the children, find the probability that all children are girls

Options

A. `(1)/(9)`
B. `(1)/(10)`
C. `(1)/(11)`
D. `(1)/(12)`

Answer

To solve the problem, we need to find the probability that all children are girls given that there are at least two girls among the children from two families, each having two children.

### Step-by-Step Solution:

1. **Identify the Total Number of Children**:
Each family has 2 children, and there are 2 families. Therefore, the total number of children is:
\[ 2 \text{ families} \times 2 \text{ children/family} = 4 \text{ children} \]

2. **Determine the Possible Combinations of Children**:
Each child can either be a boy (B) or a girl (G). The possible combinations for 4 children can be represented as:
- GGGG
- GGGB
- GGBG
- GBGG
- BGGG
- GBBG
- GBGB
- BBGG
- BGBG
- BGGG
- BBBB
- BBGB
- BGBB
- GBBB
- GBBG
- GBGB
- GGGG
- etc.

However, we only need to focus on the cases that have at least 2 girls.

3. **Count the Favorable Outcomes**:
We need to find cases where there are at least 2 girls. The possible distributions of girls (G) and boys (B) among 4 children can be:
- 4 girls (GGGG)
- 3 girls and 1 boy (GGGB, GGBG, GBGG, BGGG)
- 2 girls and 2 boys (GGBB, GBGB, BGBG, BBGG)

Let's count these cases:
- **All girls (4G)**: 1 way (GGGG)
- **3 girls and 1 boy**: There are 4 ways (choose 1 position for B from 4 positions).
- **2 girls and 2 boys**: There are \( \binom{4}{2} = 6 \) ways (choose 2 positions for G from 4 positions).

Therefore, the total number of cases with at least 2 girls is:
\[ 1 \text{ (4G)} + 4 \text{ (3G, 1B)} + 6 \text{ (2G, 2B)} = 11 \text{ cases} \]

4. **Identify the Favorable Case**:
The favorable case where all children are girls is just 1 case (GGGG).

5. **Calculate the Probability**:
The probability that all children are girls given that there are at least 2 girls is given by the ratio of favorable outcomes to the total outcomes:
\[ P(\text{All girls} | \text{At least 2 girls}) = \frac{\text{Number of favorable cases}}{\text{Total cases with at least 2 girls}} = \frac{1}{11} \]

### Final Answer:
Thus, the probability that all children are girls given that there are at least two girls is:
\[ \boxed{\frac{1}{11}} \]