Question

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

Answer 1

Hint: The given p.m.f will be of the form $P\left( x \right){{=}^{n}}{{C}_{x}}{{p}^{x}}{{q}^{\left( n-x \right)}}$ which is the formula for binomial distribution of random variable x, where n is the number of experiments, $x=0,1,2,3,...$ , p is the probability of success in a single experiment and q is the probability of failure in a single experiment, $q=1-p$ . We have to compare the given p.m.f with this formula and find the value of n, p and q. $E\left( x \right)$ can be found using the formula of mean of binomial distribution which is $E\left( x \right)=np$ and variance can be obtained from the formula $\text{Var}\left( x \right)=npq$ .

Complete step by step answer:
We are given with probability mass function of random variable x as $P\left( x \right)=\left\{ \begin{matrix}
   \left( \dfrac{4}{x} \right){{\left( \dfrac{5}{9} \right)}^{x}}{{\left( \dfrac{4}{9} \right)}^{4-x}},\begin{matrix}
   {} & x=0,1,2,3,4 \\
\end{matrix} \\
   0\text{ }\begin{matrix}
   {} & {} & , & \text{Otherwise} \\
\end{matrix} \\
\end{matrix} \right.$
We can see that the p.m.f. is of the form $P\left( x \right){{=}^{n}}{{C}_{x}}{{p}^{x}}{{q}^{\left( n-x \right)}}$ which is the formula for binomial distribution of random variable x. In this formula, n is the number of experiments, $x=0,1,2,3,...$ , p is the probability of success in a single experiment and q is the probability of failure in a single experiment, $q=1-p$ . Therefore, we can compare the given p.m.f with the standard formula. We will get $n=4,p=\dfrac{5}{9},q=\dfrac{4}{9}$.
For a binomial distribution, we know that mean is given by the formula
$E\left( x \right)=np$
Let us substitute the values in the above formula.
$\begin{align}
  & \Rightarrow E\left( x \right)=4\times \dfrac{5}{9} \\
 & \Rightarrow E\left( x \right)=\dfrac{20}{9} \\
 & \Rightarrow E\left( x \right)=2.22 \\
\end{align}$
Now, let us find the variance of x. We know that for a binomial distribution, variance is given by
$\text{Var}\left( x \right)=npq$
Let us substitute the values in the above formula.
$\begin{align}
  & \Rightarrow \text{Var}\left( x \right)=4\times \dfrac{5}{9}\times \dfrac{4}{9} \\
 & \Rightarrow \text{Var}\left( x \right)=\dfrac{80}{81} \\
 & \Rightarrow \text{Var}\left( x \right)=0.9876 \\
\end{align}$
Hence, $E\left( x \right)=2.22$ and $\text{Var}\left( x \right)=0.9876$ .

Note: Students must note that they can only compare the given p.m.f with the binomial formula by checking whether $q=1-p$ or not since all other values are similar to the formula. If the given p.m.f does not resemble the binomial formula, we have to use the formula $E\left( x \right)=\sum\limits_{x=0}^{i}{{{x}_{i}}P\left( {{x}_{i}} \right)}$ and $\text{Var}\left( x \right)=\sum\limits_{x=0}^{i}{{{x}_{i}}^{2}P\left( {{x}_{i}} \right)}$ for mean and variance respectively.