For a Binomial distribution with mean 6 and standard deviation 2 ,...

Let $n$ be the number of trials and $p$ be the probability of success.

Mean $= np = 6$

Standard deviation $= \sqrt{npq} = \sqrt{2}$

Variance $= npq = 2$

Dividing the variance by the mean:

$\dfrac{npq}{np} = \dfrac{2}{6}$

$q = \dfrac{1}{3}$

Since $p + q = 1$, we get $p = \dfrac{2}{3}$.

Substituting $p$ into the mean equation:

$n\left(\dfrac{2}{3}\right) = 6 \Rightarrow n = 9$

The probability of $X = 0$ is given by:

$P(X=0) = ^{n}C_{0} p^0 q^n$

$P(X=0) = ^{9}C_{0} \left(\dfrac{2}{3}\right)^0 \left(\dfrac{1}{3}\right)^9 = \left(\dfrac{1}{3}\right)^9$

Answer: \left(\dfrac{1}{3}\right)^9