If at least one child in a family with 3 children is a boy, then the probability that 2 of the children are boys, is
A: 3/7
B: 4/7
C: 1/3
D: 3/8

To solve this problem, let's use the concept of conditional probability. The question states that there are 3 children in a family, and it is known that at least one of them is a boy. We are tasked to find the probability that 2 of the children are boys, given this information.

### Step 1: Define the sample space
Each child can either be a boy (B) or a girl (G). For a family with 3 children, there are \(2^3 = 8\) possible combinations:

1. BBB
2. BBG
3. BGB
4. BGG
5. GBB
6. GBG
7. GGB
8. GGG

### Step 2: Identify the events of interest
Let:
- \(A\) be the event that at least one child is a boy.
- \(B\) be the event that exactly 2 children are boys.

#### Event \(A\): At least one boy
The complement of \(A\) is the event that there are no boys, i.e., all children are girls (GGG). Therefore, the number of outcomes for \(A\) is all outcomes except GGG:

Number of favorable outcomes for \(A = 8 - 1 = 7.\)

So, the sample space reduces to:
BBB, BBG, BGB, BGG, GBB, GBG, GGB.

#### Event \(B\): Exactly 2 boys
For exactly 2 boys, the outcomes are:
BBG, BGB, GBB.

Number of favorable outcomes for \(B = 3.\)

### Step 3: Use conditional probability
The probability of \(B\) given \(A\) is calculated using the formula:

P(B | A) = P(B ∩ A) / P(A)

Here:
- \(P(A) = \frac{7}{8}\), since 7 outcomes out of 8 have at least one boy.
- \(P(B ∩ A) = \frac{3}{8}\), because all outcomes in \(B\) (BBG, BGB, GBB) are already part of \(A\).

Thus:
P(B | A) = (P(B ∩ A)) / P(A) = (3/8) / (7/8) = 3/7.

### Final Answer:
The probability that 2 of the children are boys, given that at least one is a boy, is **3/7**.

Correct option: A.