Question

The probability distribution of a discrete random variable $X$ is given in the following table:

$\begin{array}{cccc}X=x& 0& 1& 2\\ P\left(x\right)& 4C^3& 4C-13C^2& 7C-1\end{array}$

If $C>0$, then $C=$____

• A. $2$
• B. $1$
• C. $\frac{1}{4}$
• D. $1$ and $-\frac{1}{4}$

Answer 1

The correct option is C $\frac{1}{4}$

For a probability distribution,
$\sum P\left(x_i\right)=1$
$\Rightarrow 4C^3+4C-13C^2+7C-1=1$
$\Rightarrow 4C^3-13C^2+11C-2=0$
$\Rightarrow (C-1)(C-2)(C-\frac{1}{4})=0$
For $C=1,2,P\left(x_i\right)$ has values greater than $1$
Hence, $C=\frac{1}{4}$