# conditional probability on multiple random variables

I am trying to solve the following question and I am not sure if I am thinking of the correct solution.
Given X that could take any of the following values with the corresponding probabilities:

$$X= \begin{cases} 0, & p =0.3 \\ 1, & p=0.4 \\ 2, & p=0.3 \end{cases}$$

Let $$Y = 5 - X^{2} + \varepsilon$$ Where $$X ~and ~ \varepsilon$$ are independent and

$$\varepsilon= \begin{cases} -1, & p =0.5 \\ 1, & p=0.5 \end{cases}$$

find

$$g(x) = E[Y|X=x]\\$$ and the least squares linear predictor of Y.

Since the variables are independent can I say that

$$Y= \begin{cases} 4-X^{2}, & p =0.5 \\ 6-X^{2}, & p=0.5 \end{cases}$$

and would

$$h(X) = \mu_Y + b\mu_X$$

The third and fourth line in @nafizh's answer are incorrect. In the third line the value of $X = x$ is given and therfore no expectation is needed. In the fourth line the values of $\epsilon$ do not need to be squared. \begin{eqnarray*} g(x) = E[Y|X=x] &=& E[5 - X^2 + \epsilon|X =x]\\ &=& E[5] - E[X^2|X=x] + E[\epsilon|X=x]) \\ &=& 5 - x^2 + E[\epsilon] \\ &=& 5 - x^2 + [-1(0.5) + 1(0.5)] \\ &=& 5 - x^2 \end{eqnarray*}