Central limit theorem - Coin toss We toss $n=200$ Euro coins on the table. 
I want to calculate, using the central limit theorem, the probability that at least $110$ coins have tutned on the same side. 
$$$$ 
Do we have to consider one side, without loss of generality, to determine the random variables $X_i$ as follows: \begin{equation*}X_i=\begin{cases} 
1 , & \text{coin } i \text{ shows head}\\ 
0 , & \text{coin } i \text{ does not show head}
\end{cases}\end{equation*} 
Or do we not have to consider one specific side of the coin? But how would the random variables be then defined? 
 A: You can define $X_i$ as you suggest though not all Euros have a head (is it the map side they all have? or the other side which sometimes has a head?) Let's define $X_i$ as an indicator of the map side, so you either want at least $110$ or no more than $90$ map sides showing
You then want $\mathbb P \left(\sum X_i \ge 110\right)+\mathbb P \left(\sum X_i \le 90\right)$
Assuming independence, you can then use the binomial distribution for an exact calculation, or a Gaussian approximation with a cuttoffs at $90.5$ and $109.5$ (or using symmetry, double no more than $90.5$).  The probability would be almost $18\%$ 
For example in R:
> pbinom(90, size=200, prob=0.5) + 1 - pbinom(109, size=200, prob=0.5)
[1] 0.178964

> 2 * pnorm(90.5, mean=200*0.5, sd=sqrt(200*0.5*0.5)) 
[1] 0.1791092

A: By using the CLT:
Let $\sum_{i=1}^{200} X_i$ be the number of "heads" in $200$ coin flips. Hence, 
$\sum X_i \sim Bin (n,p)$ or 
$$
\sum X_i\overset{approx.}{\sim}\mathcal{N}  (np, np(1-p)),
$$
thus \begin{align}
\mathbb{P}( \sum X_i \ge 110) &= 1- \mathbb{P}( \sum X_i < 110)\\
                              &\approx 1- \Phi\left( \frac{109.5 - np }{\sqrt{npq}} \right). 
\end{align}
For tails it will be the same calculations. Thus assuming a fair coin and independent tosses, the probability of the event if interest approx equals
$$
2 \left( \approx 1- \Phi\left( \frac{109.5 - 100 }{\sqrt{50}} \right) \right)
$$
