1. Model the sample space

Each family has $2$ children, so for one family the possible ordered outcomes are: $BB,\ BG,\ GB,\ GG$ Each is equally likely.

For two families, we can think of the total as $4$ children grouped into two pairs. Since each child is independently boy/girl with probability $\tfrac12$, there are $2^4=16$ equally likely outcomes.

2. Define the events

Let:

• $A$ = "all children are girls"
• $B$ = "at least two children are girls"

We need: $P(A\mid B)=\frac{P(A\cap B)}{P(B)}$

Since if all $4$ children are girls, then certainly at least two are girls, we have $A\subseteq B$ So, $P(A\mid B)=\frac{P(A)}{P(B)}$

3. Find $P(A)$

“All children are girls” means all $4$ children are girls: $GGGG$ This is just $1$ outcome out of $16$. So, $P(A)=\frac{1}{16}$

4. Find $P(B)$

“At least two girls” means number of girls among $4$ children is $2,3,$ or $4$.

Total favorable outcomes: $\binom{4}{2}+\binom{4}{3}+\binom{4}{4}=6+4+1=11$ Thus, $P(B)=\frac{11}{16}$

5. Compute the conditional probability

$P(A\mid B)=\frac{\frac{1}{16}}{\frac{11}{16}}=\frac{1}{11}$

6. Match with the options

$\frac{1}{11}$ corresponds to Option C.

7. Comparison with stored correct answer

Stored correct answer: C

Our derived answer is also C, so they agree.