Given,

E(X) = np = 2

Var(x) = npq = $\frac{4}{3}$

To find :- Probability distribution of X = ?

Since np and npq are given. So, let's divide them.

$\implies q = \frac{npq}{np} = \frac{(\frac{4}{3})}{2} = \frac{4}{6} = \frac{2}{3}$

$\therefore\ p = 1- \frac{2}{3} = \frac{1}{3}$

Now, let's find the value of n.

$\implies np = 2\ (Given)$

$\implies n \cdot \frac{1}{3} = 2$

$\implies i.e.\ n = 6$

Now,

The distribution is p(X) = $^nC_x\ p^x\ q^{n-x} $

$p(X) = (^6C_x) (\frac{1}{3})^x (\frac{2}{3})^{6-x}.$

$\therefore\ Probability\ distribution\ of\ X\ = (^6C_x) (\frac{1}{3})^x (\frac{2}{3})^{6-x}\ Ans.$