Let the probability mass function of random variable $x$ be
$P\left( x \right)=\left\{ \begin{matrix}
   {}^{4}{{C}_{x}}{{\left( \dfrac{5}{9} \right)}^{x}}{{\left( \dfrac{4}{9} \right)}^{4-x}} \\
   0,\text{ otherwise} \\
\end{matrix},\text{ }x=0,1,2,3,4 \right.$
Find the $E\left( x \right)$ and $Var\left( x \right)$ .

Hint: For solving this we will compare the given probability mass function with the formula for the probability mass function for the binomial distribution. After that, we will apply the direct formula for mean and variance to get the correct answer.

Complete step-by-step answer:
Given:
We have the probability mass function $P\left( x \right)=\left\{ \begin{matrix}
   {}^{4}{{C}_{x}}{{\left( \dfrac{5}{9} \right)}^{x}}{{\left( \dfrac{4}{9} \right)}^{4-x}} \\
   0,\text{ otherwise} \\
\end{matrix},\text{ }x=0,1,2,3,4 \right.$ for a random variable $x$ and we have to find the value of $E\left( x \right)$ and $Var\left( x \right)$ .
Now, before we proceed we should know the probability mass function for a random variable which follows binomial distribution is given by $P\left( x=r \right)={}^{n}{{C}_{r}}{{\left( p \right)}^{r}}{{\left( q \right)}^{n-r}}$ where $r\in \left[ 0,n \right]$ and $p,q>0$ such that $p+q=1$ . And mean $\left( E\left( x \right) \right)$ and variance $\left( Var\left( x \right) \right)$ of this probability mass function is written below:
$\begin{align}
  & E\left( x \right)=\sum\limits_{x=0}^{n}{x\cdot P\left( x \right)=}np....................\left( 1 \right) \\
 & Var\left( x \right)=\sum\limits_{x=0}^{n}{{{x}^{2}}\cdot P\left( x \right)=}npq...............\left( 2 \right) \\
\end{align}$
Now, if we analyse the given probability mass function then we can say that it is also following binomial distribution and we compare the given expression with the standard expression of the binomial distribution. Then,
$\begin{align}
  & {}^{4}{{C}_{x}}{{\left( \dfrac{5}{9} \right)}^{x}}{{\left( \dfrac{4}{9} \right)}^{4-x}}={}^{n}{{C}_{r}}{{\left( p \right)}^{r}}{{\left( q \right)}^{n-r}} \\
 & \Rightarrow n=4;p=\dfrac{5}{9};q=\dfrac{4}{9} \\
\end{align}$
Now, from the formula in the equation (1) and equation (2) to find the value of mean $\left( E\left( x \right) \right)$ and variance $\left( Var\left( x \right) \right)$ . Then,
$\begin{align}
  & E\left( x \right)=np=4\times \dfrac{5}{9}=\dfrac{20}{9} \\
 & Var\left( x \right)=npq=4\times \dfrac{5}{9}\times \dfrac{4}{9}=\dfrac{80}{81} \\
\end{align}$
Thus, the value of mean will be $E\left( x \right)=\dfrac{20}{9}$ and the value of variance will be $Var\left( x \right)=\dfrac{80}{81}$ .

Note: Here, the student should first try to understand what is asked in the problem. After that, we should try to analyse the given probability mass function then apply the suitable formula of mean and variance then, put suitable values of each variable and avoid calculation mistakes while solving the problem to get the correct answer. Moreover, in some questions, we should apply the general formula for the mean and variance to find the answer.