hard conditional probability question It is known that a woman has $7$ children and $2$ of them are girls , the rest is unknown.
a-) What is the probability that the intermediate child is boy (intemediate boy means it is in amid when they ordered as to their ages)
b-)What is the probability that woman has  three boys
I know that it is conditional probability question but i stuck in it.I do not even how to calculate "a woman has $7$ children and $2$ of them are girls , the rest is unknown."
Can you help me to handle this question.
 A: $$P(\textrm{Boy in the middle})=\sum_{n=2}^6P(\textrm{Boy in the middle}|\textrm{#Girls}=n)P(\textrm{#Girls}=n)$$
Assume two girls and five boys. Total arrangements: $\frac{7!}{2!5!}$
Arrangements in which the middle one is a boy: $\frac{6!}{2!4!}$
For the additional girls, the probability is binomial with success probability $p=0.5$ and trials $N=5$. So in the end
$$P=\sum_{n=2}^6\left(\frac{7!}{n!(7-n)!}\right)^{-1}\left(\frac{6!}{n!(7-n-1)!}\right)\binom{5}{n-2}0.5^{n-2}0.5^{5-(n-2)}$$
A: Let $X_k$, for $k\in\{1,\ldots,5\}$, a list of independent Bernoulli r.v. with probability $p=1/2$ each one, such that $X_k=0$ if the $k$-th kid is a boy, and $X_k=1$ if it is a girl.
Then for part a) we want to calculate $\Pr [X_3=0|\sum_{k=1}^5X_k\geqslant 2]$, this gives
$$
\Pr \left[X_3=0\bigg |\sum_{k=1}^5X_k\geqslant 2\right]=\frac1{\Pr \left[\sum_{k=1}^5X_k\geqslant 2\right]}\cdot \Pr \left[\{X_3=0\}\cap \left\{\sum_{k=1}^5X_k\geqslant 2\right\}\right]\\
=\frac1{\Pr \left[\sum_{k=1}^5X_k\geqslant 2\right]}\cdot \Pr \left[\sum_{k=1}^4X_k\geqslant 2\right]
$$
And for b) you want to calculate $\Pr \left[\sum_{k=1}^5 X_k=2|\sum_{k=1}^5 X_k\geqslant 2\right]$.
I leave to you the remaining calculations.
A: Followup to my comments, which followed the question.  Suppose that two random children were selected, and both children were seen to be girls.  This is as opposed to examining all of the genders at once, and noticing that there are less than 6 boys.
Then, to attack the problem of the chances that the total number of girls is $x$, where $x \in \{2,3,4,5,6,7\}$, I would proceed as follows:
Let $E_x$ denote the event that there are exactly $x$ girls. 
Let $F$ denote the event that two randomly inspected children were seen to be girls.
Then, I would calculate $p(E_x|F)$ as follows:
Let $\displaystyle T_k = \left[\frac{\binom{7}{k}}{2^7} \times \frac{\binom{k}{2}}{\binom{7}{2}}\right].$
Then, $\displaystyle p(E_x|F) = \frac{p(E_x F)}{p(F)} = \frac{T_x}{\sum_{k=2}^7 T_k}.$
