$X$ is a random variable with MGF $$ M_X(t) = \frac{1}{6}e^{-2t} + \frac{1}{4}e^{2t} + \frac{1}{3}e^{-t} + \frac{1}{4}e^{t} $$ I'm asked to find $P(|X| \leq 1)$.
I've been struggling with this one. I know that $E[X] = \frac{d}{d s}M_X(s)|_{s=0} = \frac{1}{12}$. And I thoguth maybe this is related to Markov inequality, because it gives $P(X \geq 1) \leq E[X]$. But this is only a bound.
Looking at the common MGFs for known distributions, I wasn't able to deduce what X's distribution should be. Any help would be appreciated.
Guide:
Note that $$M_X(t)=E[e^{Xt}]=\sum P(X=x)e^{xt}$$
Hence I can read of from the first term of MGF that $P(X=-2)=\frac16$. Try to read off the other terms and you should be able to answer the question.