Continuous Random Variable question, Probability and Statistics a little help please
A couple decide they really want a daughter. So, they decide to start having children and continue until they have their first daughter. Assuming having either a boy or girl is equally likely, answer the following:
(a) In the end, will the couple be more likely to have more boys or more girls? Explain why.
(b) Give a formula for the probability that they end up with exactly k boys. 
So for (a), I feel like the intuitive answer would be that the couple is more likely to have a boy. This is because once you have a girl, your done. So the first child is 1/2 likely to be a boy or girl. But, then you have to think about the cases in which multiple boys are born for a girl. {G, BG, BBG, BBBG, etc..} So if I add up all these probabilites, does the chance of having a boy outweight that of having a girl?  
For b, I feel like the geometric distribution is the distribution that I need. p(k) = ((1-p)^(k-1))p
This finds the kth success. So if we consider a girl k, then we can find the number of boys? I'm not exactly sure how to think about this, especially without knowing p
 A: For (a), you are thinking along the right path, but you need to be a bit more formal and consider what the question is really asking.  Out of the set of possible outcomes that you wrote, $$\{G, BG, BBG, BBBG, \ldots \},$$ which such outcomes correspond to:


*

*Having more girls than boys?  

*Having more boys than girls?

*Having an equal number of boys and girls?


For each of the above three cases, what is the sum of all the probabilities of these outcomes?  Now, can you answer the question in part (a)?  Does your answer validate or refute your intuition?  In retrospect, what is the key insight that tells you what the answer ought to be?  (Hint:  the couple is always guaranteed to have exactly one girl, but half the time, the couple does not have a boy!)
For part (b), this should be a simple exercise once you know that the only way that the couple has exactly $k$ boys is if they obtain the outcome $B\ldots BG$, where there are exactly $k$ $B$s and one $G$.
A: (a) Expectation on the girl is of course $1$. And for the boys we have something like $P\{B=k\}=\frac{1}{2^{k+1}}$, where $k\in\{0,1,...\}$, $B$- distribution that counts boys. $$\mathbb{E}[B]=\sum_{k=0}^\infty \frac{k}{2^{k+1}}=1.$$
It's exactly the $1/2$ that they have alone girl, and $1/2$ that girl+any positive natural number of boys. 
And the (b) answer has already been given.
