Consider two different random variables, say $X_1$ and $X_2$.
$P[X_1=1]=0.01$ and $X_1=0$ otherwise.
$P[X_2=2]=0.05$ and $X_2=0$ otherwise.
I'm trying to find the third central moment of both random variables, $E[(X_1-E[X_1])^3]$ and $E[(X_2-E[X_2])^3]$
I determined:
$$E[X_1]=1\cdot0.01+0\cdot0.99=0.01$$ $$E[X_2]=2\cdot0.05+0\cdot0.95=0.1$$
But I don't know how to proceed, so any help is appreciated.
Hints: So for the first variable you have $P[X=0] = 0.99$ and $P[X=1]=0.01$ (writing $X$ instead of $X_1$ for simplicity). $X$ only takes the values $0$ and $1$, evidently.
You should easily be able to compute $E[X^k]$ for any $k$. Recall that's just $$E[X^k] = \sum_{}w^k\cdot P[X=w]$$ where $w$ ranges over all possible values of $X$ (there are only two such values in this problem).
Once you have that, just expand $$(X-E[X])^3 = X^3 - 3X^2\cdot E[X]+3X\cdot E[X]^2-E[X]^3$$ Rearranging slightly to group coefficients at the beginning of each term, we have $$(X-E[X])^3 = X^3 - \underbrace{3E[X]}_{\textrm{const}}\cdot X^2+\underbrace{3E[X]^2}_{\textrm{const}}\cdot X-\underbrace{E[X]^3}_{\textrm{const}} $$ The indicated coefficients are constants you figured out above. Now apply $E[\cdot]$, keeping in mind that in the expression above, the parts involving $E[X^k]$ for various $k$ are constants, and $E[\cdot]$ is linear. For example, $$E[aX^2 + bX+c] = aE[X^2] + bE[X] + c$$ (this last is just an example of how you apply linearity to a sum of terms -- apply it to your specific expression above).