We flip 2 coins:Our random value is the number of heads
Possible values: $$\begin{array}{|c|c|c|c|}\hline &TT& HT ~~~ TH & HH\\\hline X& 0& 1& 2\\ \hline p(X=x)& 0.25& 0.5& 0.25\\\hline\end{array}$$
To get the expected value we find this sum : $$\begin{align}\left.\sum_{x=0}^2 \middle(x\cdot p(X=x)\right)~&=~ 0\cdot 0.25+1\cdot 0.5+2\cdot 0.25\\[1ex]&=0.5+0.5\\[1ex]&=1\end{align}$$
Is that correct? I just started studying probabilities so go easy on me.
Another way to look at the problem is to observe that flipping 2 fair once coins or flipping twice a single fair coin is the same. So the number of H (or T) is distributed with a Binomial law
$$X\sim Bin(n;p)=Bin(2;\frac{1}{2})$$
with mean $\mathbb{E}[X]=n\cdot p=1$