Given that a family have $n$ childern for $n\in \mathbb{N}$ is in the probability $$C\cdot\left({\frac{9}{17}}\right)^n$$ for $C>0$. the probability for getting male or female is equal.
What is the probablity that there is $7$ childrens in the family given that there is no females
My attempt
First of all I found $C$ in the following way: $$P(\mathbb{N})=1=\large\Sigma_{n=0}^{\infty}C\cdot\left({\frac{9}{17}}\right)^n$$ $$C=\frac{8}{17}$$
Let $A$ and $B$ be an events :
$A$ - there is 7 childrens in the familiy
$B$ - there is no females in the familiy
I want to calculate $$P(A|B)$$ for this I want use the formula $$P(A|B)=\frac{P(A\cap B)}{P(B)}$$ now calculate $P(B)$ is easy it is $\frac{1}{2}$
Now how should I calculate $P(A\cap B)$? is it true to argue that $A$ and $B$ are independent and than use the formula $$P(A\cap B)= P(A) \cdot P(B)?$$ not sure how should I do this step
$P(B)$ is not $\frac{1}{2}$, that would be the probability of one of the children being male or female, but from what you said I think $B$ is the event where none of the children are female (so all of them are male). You need to consider how many children are (and you cannot say 7 because $B$ has nothing to do with $A$ in $P(B)$).
If you call $C_i$ for $i=0,1,2,\ldots$ the event that there are $i$ children, then $P(B)=\sum\limits_{i=0}^\infty P(B\mid C_i)P(C_i)=\sum\limits_{i=0}^\infty\frac{1}{2}^i\frac{8}{17}\frac{9}{17}^i=\frac{8}{17}\sum\limits_{i=0}^\infty(\frac{1}{2}\frac{9}{17})^i=\ldots$
Note that I substituted $P(B\mid C_i)=\frac{1}{2}^i$, since that's the probability of having no females given that there are $i$ children. From that summation you get $P(B)$.
Now, for the rest, you just need one more step:
$P(A|B)=\dfrac{P(A\cap B)}{P(B)}=\dfrac{P(B\mid A)P(A)}{P(B)}$
Note $A=C_7$.