If the mean of the following probability distribution of a random variable X;  

X 0 2 4 6 8
P(X) a 2a a + b 2b 3b

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

1
\(\frac{581}{81}\)
2
\(\frac{566}{81}\)
3
\(\frac{173}{27}\)
4
\(\frac{151}{27}\)
5
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