\(f(x)=\begin{cases}\dfrac{125}{31}&\text{for }x\in\{1,2,3\}\\\\0&\text{otherwise}\end{cases}\)
The mean is
\(\mathbb E(X)=\displaystyle\frac{125}{31}\sum_{x=1}^3x\left(\frac15\right)^x=\dfrac{38}{31}\)
You then have
\(\mathbb E(X^2)=\displaystyle\frac{125}{31}\sum_{x=1}^3x^2\left(\frac15\right)^x=\dfrac{54}{31}\approx1.74\)
so the variance is
\(\mathbb V(X)=\dfrac{54}{31}-\left(\dfrac{38}{31}\right)^2=\dfrac{230}{961}\approx0.239\)
