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.

$$P(\textrm{Boy in the middle})=\sum_{n=2}^6P(\textrm{Boy in the middle}|\textrm{#Girls}=n)P(\textrm{#Girls}=n)$$
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
$$P=\sum_{n=2}^6\left(\frac{7!}{n!(7-n)!}\right)^{-1}\left(\frac{6!}{n!(7-n-1)!}\right)\binom{5}{n-2}0.5^{n-2}0.5^{5-(n-2)}$$
