If a discrete random variable X has probability distribution as follows
\(\begin{array}{|c|c|c|c|c|}
\hline \mathrm{X}=x & 0 & 1 & 2 & 3\mathrm{P}[\mathrm{X}=x] & \mathrm{k} & 3 \mathrm{k} & 3 \mathrm{k} & \mathrm{k} \\ \hline\end{array}\)
Then \(\operatorname{var}(\mathrm{X})=\)

Here \(\mathrm{k}+3 \mathrm{k}+3 \mathrm{k}+\mathrm{k}=8 \mathrm{k}=1 \Rightarrow \mathrm{k}=\frac{1}{8}\)
\(\sum \mathrm{P}_{\left(\mathrm{x}_{1}\right)} \mathrm{x}_{\mathrm{i}} =0 \times \frac{1}{8}+1 \times \frac{3}{8}+2 \times \frac{3}{8}+3 \times \frac{1}{8}=\) \(\frac{12}{8}=\frac{3}{2} \text { and } \)
\( \sum \mathrm{P}_{\left(\mathrm{x}_{i}\right)} \mathrm{x}_{\mathrm{i}}^{2} =\left(0 \times \frac{1}{8}\right)+\left(1 \times \frac{3}{8}\right)+\left(4 \times \frac{3}{8}\right)~+\) \(\left(9 \times \frac{1}{8}\right)=\frac{24}{8}=3 \)
\( \text { Variance } =\mathrm{V}(\mathrm{X}) \)
\( =\sum \mathrm{P}_{\left(\mathrm{x}_{i}\right)} \mathrm{x}_{\mathrm{i}}^{2}-\left[\sum \mathrm{P}_{\left(\mathrm{x}_{i}\right)} \mathrm{x}_{\mathrm{i}}\right]^{2} \)
\( =3-\left(\frac{3}{2}\right)^{2}=3-\frac{9}{4}=\frac{3}{4}\)