Let a random variable X take values $0,1,2,3$ with $\mathrm{P}(\mathrm{X}=0)=\mathrm{P}(\mathrm{X}=1)=\mathrm{p}, \mathrm{P}(\mathrm{X}=2)=\mathrm{P}(\mathrm{X}=3)$ and $\mathrm{E}\left(\mathrm{X}^2\right)=2 \mathrm{E}(\mathrm{X})$. Then the value of $8 \mathrm{p}-1$ is :

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Aicharya · Accepted Answer

$\begin{aligned} & 2 p+2 q=\frac{1}{2} \ & p+q \ & E\left(x^2ight)=\sum_{i=0}^3 x_i^2 p\left(x_iight)=0 \cdot p+1 \cdot p+4 \cdot q+9 q \ & =p+13 q \ & E(x)=\sum_{i=0}^3 x_i^2 p\left(x_iight)=0 \cdot p+1 \cdot p+2 q+3 q=p+5 q \ & p+13 q=2(p+5 q) \ & p=3 q \ & 	ext { So, } q=\frac{1}{8} \& p=\frac{3}{8} \quad \ & 	ext { So, } 8 p-1=2 \quad 	ext { Option (2) }\end{aligned}$ ,

Let a random variable X take values $0,1,2,3$ with…

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