Children in school throwing darts at target I'm stuck on the last bullet of a question:

Children play a game of throwing darts at target. The probability of practiced child to hit the target is $0.8$. The probability for not practiced child to hit the target is $0.4$. We choose randomly $100$ children from a school where $70\%$ are practiced. Each child throws $10$ darts at the target. What is the probability to get more than $700$ hits?

As I mentioned it's the last bullet of a question (previous bullets not related to this one). I started with marking $A$ is the event of a practiced child hitting the target and $B$ to be the event of non practiced child to hit the target. So we get $P(A)=0.8$ and $P(B)=0.4$. Of course $P(A^c)=0.2$ and $P(B^c)=0.6$. But How do I continue from here? I believe it's some distribution which is used here but I can't find it because of the "practiced/not practiced" and "hit/ didn't hit".
 A: I would define the events in a different way. Let $H$ be the event of a child hitting the target. And $E$ the event of a practiced/experienced child.
Then $P(H|E)=0.8,  P(H|\overline E)=0.4$  and $P(E)=0.7$
Next we can calculate $P(E\cap H)$ and $P(\overline E\cap H)$ with the Bayes theorem.
$P(H|E)=\frac{P(E\cap H)}{P(E)} \Rightarrow 0.8=\frac{P(E\cap H)}{0.7}\Rightarrow P(E\cap H)=0.56$
$P(H|\overline E)=\frac{P(\overline E\cap H)}{P(\overline E)} \Rightarrow 0.4=\frac{P(\overline E\cap H)}{0.3}\Rightarrow P(\overline E\cap H)=0.12$
Therefore $P(H)=P(E\cap H)+P(\overline E\cap H)=0.56+0.12=0.68$ and $P(\overline H)=0.32$
We can handle the total 1000 throws as 1000 independent bernoulli distributed variables ($X_i$), where $\mu_i=0.68$ and $\sigma_i^2=0.68\cdot 0.32$. Therefore the sum of the  $1000$ random variables is distributed as $S_{1000}\sim \textrm{Bin}\left(n, p \right)=\textrm{Bin}\left(1000, 0.68 \right)$. Finally apply the central limit theorem to calculate (approximately) $P(S_{1000}>700)=1-P(S_{1000}\leq 700)$.
