If the mean of the following probability distribution of a random variable $$X$$:

is $$\frac{46}{9}$$, then the variance of the distribution is

Sum of Probabilities = 1.

$$a + 2a + (a+b) + 2b + 3b = 1 \implies 4a + 6b = 1$$.

Use Mean.

$$E(X) = 0(a) + 2(2a) + 4(a+b) + 6(2b) + 8(3b) = 4a + 4a + 4b + 12b + 24b = 8a + 40b$$.

$$8a + 40b = \frac{46}{9} \implies 4a + 20b = \frac{23}{9}$$.

Solve for $$a, b$$.

Subtract $$(4a+6b=1)$$ from $$(4a+20b=23/9)$$:

$$14b = \frac{23}{9} - 1 = \frac{14}{9} \implies \mathbf{b = \frac{1}{9}}$$.

$$4a + 6(1/9) = 1 \implies 4a = 1 - 2/3 = 1/3 \implies \mathbf{a = \frac{1}{12}}$$.

$$Var(X) = E(X^2) - [E(X)]^2$$.

$$E(X^2) = 0^2(a) + 2^2(2a) + 4^2(a+b) + 6^2(2b) + 8^2(3b) = 8a + 16a + 16b + 72b + 192b = 24a + 280b$$.

$$E(X^2) = 24(1/12) + 280(1/9) = 2 + \frac{280}{9} = \frac{298}{9}$$.

$$Var(X) = \frac{298}{9} - (\frac{46}{9})^2 = \frac{2682 - 2116}{81} = \mathbf{\frac{566}{81}}$$ (Option B)