Conditional probability elementary problem - Check if I am right? I need to help somebody prepare for some exam, and I need to find if the answers to the problem presented below are correct. I'm not very good at probabilities, and I want to check my reasoning

Someone visits a family with three children. We know that the probability that a child is boy or girl is $1/2$ and the sex of the children is independent. All children have different ages. We denote, for example, GBB the event that the first born is a girl, the second born is a boy and the third is a boy.
(A) We ring at the door and the child who answers is a girl. The set of possible cases is $\{ GGG,GGB,GBG,GBB,BGG,BGB,BBG\}$

My answer: this is correct, since one of the children must be a girl, which removes the possibility $BBB$.

(B) We ring, the child who opens is a girl. The probability that the other two children are boys is $3/7$.

My answer: this is correct, since if one of the children is a girl, we have three possibilities out of seven where the other two are boys.

(C) We find out that the girl who opened the door is the first born. The probability that the other two children are boys is $3/7$.

My answer: this is false, since if the girl is the first born, we have only one case out of four which is favorable. The probability is $1/4$.

(D) Another child appears behind his sister (who is first born). The probability that the remaining child is a boy is $1/2$.

My answer: False. We use conditional probabilities. Denote $X_1$ the random variable which denotes the sex of the first born, $Y$ the variable which denotes the sex of the child which appears behind his sister, $Z$ the variable which contains the sex of the third child. We have
$$P(Z = B | X_1 = G \cap Y = B) = \frac{P(X_1 = G \cap Y = B\cap Z = B)}{P(X_1 = G \cap Y=B)}=\frac{\frac{1}{4}}{\frac{3}{4}}=\frac{1}{3}$$
because we only have one case where the first born is a girl, and the other two are boys and we have three cases where the first born is a girl and there is another boy in the family.
Is what I did right?
 A: I would not agree with your analysis in (B).  
Suppose that there are $24$ families in the street with three of each type and, for each of the eight types of family, in one house the oldest opens the door, in another the middle child opens the door, and in the third the youngest opens the door. You choose a door at random.
Then in $12$ houses a girl opens the door and you have chosen one of them.  In $3$ of those she has two brothers.  So the conditional probability is $\dfrac{1}{4}$, just like your response to (C).  
You would get the same result if you said that there are $36$ girls and they are each equally to have opened the door. 
I assumed that girls and boys are equally likely to answer the door.  The $\frac37$ might be true in a society where boys are so lazy that they only open doors when they have no sisters. 
I would not agree with your answer to (D) either. There is a third child unseen and that individual may be male of female with equal probability assuming there is no geneder difference in going to the door. 
