Consider the following events for a family with children
A = {Children of both sexes}
B = {At-most one boy}
Then which of the following statements is true?

Options

A. A and B are independent events if a family has 3 children
B. A and B are not independent events if a family has 3 children
C. A and B are not independent events if a family has only 2 children
D. A and B are independent events if a family has only 2 children

Answer

To solve the problem, we need to analyze the two events given:

- Event A: Children of both sexes
- Event B: At most one boy

We will consider two cases: one where the family has 3 children and another where the family has 2 children.

### Case 1: Family with 3 Children

1. **Determine the Sample Space**:
The sample space for a family with 3 children can be represented as:
- BBB (all boys)
- BBG (two boys, one girl)
- BGB (two boys, one girl)
- GBB (two boys, one girl)
- BGG (one boy, two girls)
- GBG (one boy, two girls)
- GGB (one boy, two girls)
- GGG (all girls)

Thus, the sample space \( S \) is:
\( S = \{ BBB, BBG, BGB, GBB, BGG, GBG, GGB, GGG \} \)
Total outcomes = 8.

2. **Find Probability of Event A**:
Event A (children of both sexes) includes outcomes:
- BBG
- BGB
- GBB
- BGG
- GBG
- GGB

Number of favorable outcomes for A = 6.
\[ P(A) = \frac{6}{8} = \frac{3}{4} \]

3. **Find Probability of Event B**:
Event B (at most one boy) includes outcomes:
- BGG
- GBG
- GGB
- GGG

Number of favorable outcomes for B = 4.
\[ P(B) = \frac{4}{8} = \frac{1}{2} \]

4. **Find Probability of A Intersection B**:
The intersection of A and B (children of both sexes and at most one boy) includes:
- BGG
- GBG
- GGB

Number of favorable outcomes for \( A \cap B \) = 3.
\[ P(A \cap B) = \frac{3}{8} \]

5. **Check Independence**:
Two events A and B are independent if:
\[ P(A \cap B) = P(A) \times P(B) \]
Calculate \( P(A) \times P(B) \):
\[ P(A) \times P(B) = \frac{3}{4} \times \frac{1}{2} = \frac{3}{8} \]
Since \( P(A \cap B) = \frac{3}{8} \), the events A and B are independent when there are 3 children.

### Case 2: Family with 2 Children

1. **Determine the Sample Space**:
The sample space for a family with 2 children can be represented as:
- BB (both boys)
- BG (boy and girl)
- GB (girl and boy)
- GG (both girls)

Thus, the sample space \( S \) is:
\( S = \{ BB, BG, GB, GG \} \)
Total outcomes = 4.

2. **Find Probability of Event A**:
Event A (children of both sexes) includes:
- BG
- GB

Number of favorable outcomes for A = 2.
\[ P(A) = \frac{2}{4} = \frac{1}{2} \]

3. **Find Probability of Event B**:
Event B (at most one boy) includes:
- BG
- GB
- GG

Number of favorable outcomes for B = 3.
\[ P(B) = \frac{3}{4} \]

4. **Find Probability of A Intersection B**:
The intersection of A and B includes:
- BG
- GB

Number of favorable outcomes for \( A \cap B \) = 2.
\[ P(A \cap B) = \frac{2}{4} = \frac{1}{2} \]

5. **Check Independence**:
Calculate \( P(A) \times P(B) \):
\[ P(A) \times P(B) = \frac{1}{2} \times \frac{3}{4} = \frac{3}{8} \]
Since \( P(A \cap B) = \frac{1}{2} \) which is not equal to \( \frac{3}{8} \), the events A and B are not independent when there are 2 children.

### Conclusion
- For a family with 3 children, events A and B are independent.
- For a family with 2 children, events A and B are not independent.

### Final Answer
The true statements are:
- A and B are independent events if a family has 3 children.
- A and B are not independent events if a family has 2 children.