If a random variable $X$ has the probability distribution given by $P(X=0)=3 C^3$, $P(X=2)=5 C-10 C^2$ and $P(X=4)=4 C-1$, then the variance of that distribution is

• A

  $\frac{68}{9}$
• B

  $\frac{22}{9}$
• C

  $\frac{612}{81}$
• Correct answer
  D

  $\frac{128}{81}$

Solution

Given, $$ \begin{aligned} & P(X=0)=3 C^3 \\ & P(X=2)=5 C-10 C^2 \\ & \text { and } \quad P(X=4)=4 C-1 \\ & \text { We know that, } \\ & \Sigma P(X)=1 \\ & \Rightarrow \quad 3 C^3+\left(5 C-10 C^2\right)+(4 C-1)=1 \\ & \Rightarrow \quad 3 C^3-10 C^2+9 C-2=0 \\ & \end{aligned} $$ $$ \begin{array}{ll} \Rightarrow & (C-1)\left(3 C^2-7 C+2\right)=0 \\ \Rightarrow & (C-1)(3 C-1)(C-2)=0 \\ \Rightarrow & C=1, \frac{1}{3^{\prime}}, 2 \\ \therefore & C=\frac{1}{3} \end{array} $$ Now,
Hence, variance $=\Sigma X_P^2-\left(\Sigma X_P\right)^2$ $$ \begin{aligned} & =\left(0^2 \times \frac{1}{9}+4 \times \frac{5}{9}+16 \times \frac{1}{3}\right)-\left(\frac{10}{9}+\frac{4}{3}\right)^2 \\ & =\left(\frac{20}{9}+\frac{16}{3}\right)-\left(\frac{66}{27}\right)^2=\frac{60+144}{27}-\frac{484}{81} \\ & =\frac{204}{27}-\frac{484}{81}=\frac{612-484}{81}=\frac{128}{81} \end{aligned} $$

Hence, the correct answer is option: 'D'.