If probability distribution is given by,

x	0	1	2	3
p(x)	$\frac{8a-1}{30}$	$\frac{4a-1}{30}$	$\frac{2a+1}{30}$	b

$$\begin{gathered} \text{If it is given that}{\mathrm{ \sigma }}^2+{\mu }^2=2\text{ where }\mathrm{\sigma }\text{is standard deviation and}\mathrm{ \mu }\text{is mean of distribution then }\frac{a}{b}\text{ is} \\ \end{gathered}$$