<s>[INST] Throwing a coin - expected value of tails There are $10$ coins: $8$ of them are fair (equal probability for heads and tails) and for $2$ probability of heads is two times higher than for tails (so $\frac{2}{3}$ for heads). We draw one coin and throw it three times. Let $X$ denote a number of tails - I need to find a $\Bbb EX$. So:
$\Bbb P(X = k) = \Bbb P(X = k| \text{fair coin} ) \cdot \Bbb P(\text{fair coin}) + \Bbb P(X = k| \text{unfair coin} )  \cdot \Bbb P(\text{unfair coin})$
With
$\Bbb P(X = k| \text{fair coin} )={3 \choose k} (\frac{1}{2})^k(\frac{1}{2})^{3-k}$
$\Bbb P(\text{fair coin}) = \frac{8}{10} $
$\Bbb P(X = k| \text{unfair coin} )={3 \choose k} (\frac{1}{3})^k(\frac{2}{3})^{3-k} $ 
$\Bbb P(\text{unfair coin})= \frac{2}{10} $
So
$\Bbb P(X = k) = {3 \choose k} (\frac{1}{2})^k(\frac{1}{2})^{3-k} \cdot \frac{4}{5} + {3 \choose k} (\frac{1}{3})^k(\frac{2}{3})^{3-k}\cdot \frac{1 }{5} $
And then obviously $\Bbb EX = \sum\limits_{i=0}^3 i\cdot \Bbb P(X=i)$
Am I correct?
 [/INST] You know that the expectation of a binomial is $\mu=np$ thus you have
$$E(X)=\frac{8}{10}\times 3 \times \frac{1}{2}+\frac{2}{10}\times 3 \times \frac{1}{3}=\frac{14}{10}$$

this easy method works as you pmf is a mixture of two pmf's
$$p_X(x)=0.8p_{X_1}(x)+0.2p_{X_2}(x)$$
</s>