A model for the number of kids in a family and conditional probability 
$\textbf{The Problem:}$ Let $0<r<1.$ We model the number of children in a randomly chosen family with the following distribution
  $$P(\text{exactly $n$ kids})=\frac{1}{2}r^n\quad\text{if }n\geq1,$$
  and $P(\text{no kids})=1-\sum^\infty_{n=1}\frac{1}{2}r^n$. We assume that the gender of each children is male or female with probability $\frac{1}{2}$ independently of the genders of the other siblings.
What is the probability that a family has no boys given that there are $k$ kids?

$\textbf{My Thoughts:}$ Using the definition of conditional probability we have that
\begin{equation}\begin{split}
P(\text{no boys}\,|\,\text{exactly $k$ kids})
&=\frac{P(\text{no boys}\cap\text{exactly $k$ kids})}{P(\text{exaclty $k$ kids})}\\
&=\frac{\frac{1}{2}r^k\cdot(\frac{1}{2})^k}{\frac{1}{2}r^k}\\
&=\frac{1}{2^k}.
\end{split}\end{equation}
My only concern is with the probability of the event $\{\text{no boys}\cap\text{exactly $k$ kids}\}$. My reasoning for my answer was that since none of the $k$ kids can be a boy, by the assumed independence, we can multiply the $1/2$'s.

Do you agree with my attempt at a solution of this problem? If not, please point me in the right direction with a hint, only a hint.
Thank you for your time and appreciate any feedback.
 A: You get the right answer. To make the reasoning more rigorous we can start with considering what is given indeed. The assumption 

gender of each children is male or female with probability $\frac12$ independently of the genders of the other siblings

can be rewritten as assumption on conditional distribution: given any fixed number $k$ kids in a family, we have $k$ Bernoulli trials with success probability $\frac12$ to determine the number of boys or girls. So the distribution of a number of boys in a family given that there are exactly $k$ kids is binomial: 
$$
\mathbb P(m\text{ boys in a family}\mid \text{exactly } k \text{ kids}) = \binom{k}{m}\frac{1}{2^k}, \; m=0,1,\ldots,k.
$$
After rewritten condition in this way, you can factor the probability of intersection as
\begin{equation}\begin{split}
P(\text{no boys}\,|\,\text{exactly $k$ kids})
&=\frac{P(\text{no boys}\cap\text{exactly $k$ kids})}{P(\text{exactly $k$ kids})}\\
&=\frac{P(\text{no boys}\mid \text{exactly $k$ kids})\cdot\mathbb P(\text{exactly $k$ kids})}{P(\text{exactly $k$ kids})}\\ 
&=P(\text{no boys}\mid \text{exactly $k$ kids})\\
&=\frac{1}{2^k}.
\end{split}\end{equation}
