Probability that girl who answers the door is the eldest girl?

There are several correct answers already; I just wanted to give what I think is a simpler argument.

The probability clearly wouldn't change if we instead had a boy answering the door, and wanted to know if he was the oldest boy. So we can equivalently ask what is the probability that the child who answered the door was the oldest child of their sex.

Now there are two possibilities. If all the children are the same sex (probability $1/4$), only one child is the oldest of their sex. Otherwise, two children will be the oldest of their sex. Thus the probability is $\frac14\times\frac13+\frac34\times\frac23=\frac7{12}$.

This easily generalises to $n$ children and gives an answer of $\frac{2^n-1}{n2^{n-1}}$.

There are $8*3=24$ equally likely situations, the $8$ gender orderings and for each one, one of {youngest, middle, eldest} child who answers the door. $12$ of those have a girl answering the door (count the $12$ G's in the list given in the answer above).

Among those $12$ equally likely G's, exactly $7$ are the eldest girl (one from each of the $7$ non-BBB orderings). So the desired probability is $7/12$.

The condition "a random child turns out to be a girl" is NOT equivalent to a YES answer to the question "is there at least one girl?"