Assume that the random variable $X$ has moment generating function,
$M_X(t) = \frac{1}{6}e^{-2t} + \frac{1}{5}e^{-t} + \frac{3}{10}e^{t} + \frac{1}{3}e^{2t}$
Find $Pr(|X| \leq 1)$.
I've tried using the moment generating function technique by going over known distributions to see if I am some how successful in finding which distribution this MGF belongs to but haven't succeeded.
$X$ takes the values $-2, -1,1,2$ with probabilities $\frac 1 6, \frac 1 5, \frac 3 {10}$ and $\frac 1 3$ respectively. Hence $P(|X| \leq 1)=P(X=1)+P(X=-1)=\frac 1 5+ \frac 3 {10}=\frac 1 2$.