Q: Moment Generating Functions for beginners Problem: Let  $X$  be a random variable with MGF  $M_X(2)=\frac{1}{3}e^{-t}+\frac{1}{6}+\frac{1}{2}e^{2t}$. What is  $P(X≤1)$?
I tried taking $P(X=0)$ and $P(X=1)$ and plugging those numbers into the equation for $t$ but that didn't work.
I tried:
$$M_X(2)=\frac{1}{3}e^{-(0)}+\frac{1}{6}+\frac{1}{2}e^{2(0)}=0.3333+0.16666+0.5=0.999999 (wrong)$$
$$M_X(2)=\frac{1}{3}e^{-(1)}+\frac{1}{6}+\frac{1}{2}e^{2(1)}=0.1226+0.16666+3.6945=4.1944 (wrong)$$.
I might be confusing what $t$ is with what $X$ is.  Any suggestions would be helpful.

A: If... the MGF is the following
$$M_X(t)=\frac{1}{3}e^{-t}+\frac{1}{6}+\frac{1}{2}e^{2t}$$
This is the MGF of hte following discrete rv
$$ X =
\begin{cases}
\frac{2}{6},  & \text{if $x=-1$ } \\
\frac{1}{6},  & \text{if $x=0$ } \\
\frac{3}{6},  & \text{if $x=2$ } 
\end{cases}$$
Thus
$$\mathbb{P}[X\leq 1]=\frac{3}{6}$$
