A random varaible X takes values 0,1,2,3 with probabilities $$\frac{2a+1}{30},\frac{8a-1}{30},\frac{4a+1}{30}$$, b respectively, where $$a,b \epsilon R$$. let $$\mu$$ and $$\sigma$$ respectively be the mean and standard deviation of X such that $$\sigma^{2}+\mu^{2}=2$$. Then $$\frac{a}{b}$$ is equal to :

X takes values 0, 1, 2, 3 with probabilities $$\frac{2a+1}{30}, \frac{8a-1}{30}, \frac{4a+1}{30}, b$$.

We enforce that the sum of these probabilities equals one: $$\frac{2a+1+8a-1+4a+1}{30} + b = 1$$, which simplifies to $$\frac{14a+1}{30} + b = 1$$ and hence $$b = 1 - \frac{14a+1}{30} = \frac{29-14a}{30}$$ ... (i)

We compute the mean $$\mu = E(X)$$ as $$\mu = 0 \cdot \frac{2a+1}{30} + 1 \cdot \frac{8a-1}{30} + 2 \cdot \frac{4a+1}{30} + 3b = \frac{8a-1+8a+2}{30} + 3b = \frac{16a+1}{30} + 3 \cdot \frac{29-14a}{30} = \frac{88-26a}{30}$$

The second moment is $$E(X^2) = 0 + 1 \cdot \frac{8a-1}{30} + 4 \cdot \frac{4a+1}{30} + 9b = \frac{8a-1+16a+4}{30} + 9 \cdot \frac{29-14a}{30} = \frac{264-102a}{30}$$

Using $$\sigma^2 = E(X^2) - \mu^2$$ and the condition $$\sigma^2 + \mu^2 = 2$$ gives $$E(X^2) = 2$$, so $$\frac{264-102a}{30} = 2 \Rightarrow 264 - 102a = 60 \Rightarrow 102a = 204 \Rightarrow a = 2$$

Substituting into (i) gives $$b = \frac{29-28}{30} = \frac{1}{30}$$

Therefore, $$\frac{a}{b} = \frac{2}{1/30} = 60$$

The answer is Option 2: 60.