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Question: Let the p.m.f. ( probability mass function ) of random variable x be \(P\left( x \right)=\left\\{ \b...

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(x)E\left( x \right) and Var(x)\text{Var}\left( x \right) .

Explanation

Solution

The given p.m.f will be of the form P(x)=nCxpxq(n−x)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,...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−pq=1-p . We have to compare the given p.m.f with this formula and find the value of n, p and q. E(x)E\left( x \right) can be found using the formula of mean of binomial distribution which is E(x)=npE\left( x \right)=np and variance can be obtained from the formula Var(x)=npq\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(x)=nCxpxq(n−x)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,...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−pq=1-p . Therefore, we can compare the given p.m.f with the standard formula. We will get n=4,p=59,q=49n=4,p=\dfrac{5}{9},q=\dfrac{4}{9}.
For a binomial distribution, we know that mean is given by the formula
E(x)=npE\left( x \right)=np
Let us substitute the values in the above formula.
⇒E(x)=4×59 ⇒E(x)=209 ⇒E(x)=2.22 \begin{aligned} & \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{aligned}
Now, let us find the variance of x. We know that for a binomial distribution, variance is given by
Var(x)=npq\text{Var}\left( x \right)=npq
Let us substitute the values in the above formula.
⇒Var(x)=4×59×49 ⇒Var(x)=8081 ⇒Var(x)=0.9876 \begin{aligned} & \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{aligned}
Hence, E(x)=2.22E\left( x \right)=2.22 and Var(x)=0.9876\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−pq=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(x)=∑x=0ixiP(xi)E\left( x \right)=\sum\limits_{x=0}^{i}{{{x}_{i}}P\left( {{x}_{i}} \right)} and Var(x)=∑x=0ixi2P(xi)\text{Var}\left( x \right)=\sum\limits_{x=0}^{i}{{{x}_{i}}^{2}P\left( {{x}_{i}} \right)} for mean and variance respectively.