Moment Generating Function I am little unsure if I am doing the following question right:

Suppose Y is a discrete random variable such that 
$P(Y=0)=\frac{1}{3}$;  $P(Y=1)=\frac{1}{6}$; $P(Y=2)=\frac{1}{6}$;
  $P(Y=3)=\frac{1}{3}$;
a) Write down the moment generating function of $Y$.
b) Find the mean and variance of $Y$.
c) Suppose that $X_1$ and $X_2$ are independent random variables with
  the same distribution as $Y$. Compute $P(X_1 + X_2 = 3)$.

My solutions so far:
a) Defintion of MGF $$M(t) = E[e^{tX}] = \sum_{x}e^{tx}p(x)$$
So
$$M_Y(s) = \frac{1}{3} + \frac{1}{6}e^{t} + \frac{1}{6}e^{2t} + \frac{1}{3}e^{3t}$$
b) Using the following defintions:
$$ M'(0) = E[X]$$
$$M''(0) = E[X^2]$$
So
$M'(t) = e^{3t} + \frac{1}{3}e^{2t} + \frac{1}{6}e^t \implies M'(0) = \frac{3}{2}$
$M''(t) = 3e^{3t} + \frac{2}{3}e^{2t} + \frac{1}{6}e^t \implies M''(0) = \frac{23}{6}$
$Var(X) = \frac{23}{6} - (\frac{3}{2})^2 = \frac{19}{12}$
I am not sure how to do part c though.
 A: a) looks good.
I'm not checking your answers for b), but the methodology looks fine.
For c), suppose $X_1$, $X_2$ are independent, are distributed as $Y$ is, and $Z = X_1 + X_2$.
$Y$ takes on only the following values: $\{0, 1, 2, 3\}$.
If $Z = 3$, what are the possible values that $X_1$ and $X_2$ can take?
Well, $X_1, X_2$ are both distributed as $Y$ is, so they take on the same values as $Y$ (note here I don't mean that they are equal to $Y$).
Remember, we're only looking at cases where $Z = 3$.
If $X_1 = 0$, $X_2$ must be $3$. I will represent this with the pair $(0, 3)$.
$(3, 0)$ is another way to get $Z = 3$.
We can also find $(1, 2), (2, 1)$ as possibilities.
So:
$$P((X_1, X_2) = (0, 3)) = P((X_2, X_1) = (3, 0) = P(X_1 = 0 \cap X_2 = 3) = P(X_1 = 0)P(X_2 = 3)$$
using independence and you can do something similar for $(1, 2)$ and $(2, 1)$.
A: Let $Z=X_1+X_2$. Since they are independent random variables with the same distribution as $Y$
\begin{align}
M_Z(t) &= E[e^{t(X_1+X_2)}] 
\\
&= E[e^{tX_1}] \:E[e^{tX_2}] 
\\
&=(\frac{1}{3} + \frac{1}{6}e^{t} + \frac{1}{6}e^{2t} + \frac{1}{3}e^{3t})^2
\\
&=\frac1{36}(4+4e^t+5e^{2t}+10e^{3t}+5e^{4t}+4e^{5t}+4e^{6t})
\end{align}
Thus $P(Z=3)$ is the coefficient of $e^{3t}$, i.e.
$$
P(Z=3)=\frac{5}{18}
$$
