Suppose that from an arbitrary family of $3$ children we choose one of them at random. If the chosen one is a girl, what is the probability that she has a sister and a brother?
(Note: suppose that the probability of being a boy or girl is the same and that the events are independent. i.e the sex of one child doesn't affect the probability of the sex of the other ones.)
What I have tried:
So this is probably wrong, but my approach was saying this is the same as asking what's the probability of two of them being girls and the other one a boy. What is really bugging me is that BGG, GGB, and GBG appear to be the same, so I don't really know what the sample space is...
So my next approach was saying, this can be rephrased as saying: $$P(2\text{ girls, }1\text{ boy)}=P(2\text{ girls})-P(3\text{ girls})$$ So $P(2\text{ girls})$ must be equal to $P(2\text{ boys})$ since any family has at least 2 boys or 2 girls and neither can happen in the same family, meaning $P(2\text{ girls}) + P(2\text{ boys})=1$, then $P(2\text{ girls})=\frac{1}{2}$. $P(3\text{ girls})$ must be $\frac{1}{8}$ and so $P(2\text{ girls, }1\text{ boy})= \frac{1}{2} - \frac{1}{8} = \frac{3}{8}$
Is this correct or am I misunderstanding the problem? Or could it be $P(1\text{ boy, }2\text{ girls}|1\text{ girl})$ ? Thanks in advance.
Oh, and the problem says you can use Bayes to solve this but I dont see how to be honest...