Three fair coins numbered 1 and 0 are tossed simultaneously. Then variance \(\operatorname{Var}(\mathrm{X})\) of the probability distribution of random variable \(\mathrm{X}\), where \(X\) is the sum of numbers on the uppermost faces, is

Three fair coins numbered 1,0 are tossed.
\(\therefore \) Sample space \(=\{111,110,101,011,100,010\), \(001,000\}\)
\(\therefore \mathrm{n}(\mathrm{S})=8\)
\(\mathrm{X}\) represents the sum of numbers on upper most face
\(\therefore P(X=0)=\frac{1}{8}, \)
\( P(X=1)=\frac{3}{8}, \)
\( P(X=2)=\frac{3}{8}, \)
\( P(X=3)=\frac{1}{8}\)
\(\therefore \) Probability distribution of \(\mathrm{X}\) is
\(\begin{array}{|c|c|c|c|c|}
\hline \mathrm{X} & 0 & 1 & 2 & 3 \\
\hline \mathrm{P}(\mathrm{X}) & \frac{1}{8} & \frac{3}{8} & \frac{3}{8} & \frac{1}{8} \\
\hline
\end{array}\)
\(\mathrm{E}(\mathrm{X}) =\sum_{x=0}^3 x_{\mathrm{i}} \mathrm{P}\left(x_{\mathrm{i}}\right) \)
\( \mathrm{E}(\mathrm{X}) =0 \times \frac{1}{8}+1 \times \frac{3}{8}+2 \times \frac{3}{8}+3 \times \frac{1}{8} \)
\( =\frac{12}{8}=\frac{3}{2} \)
\( \mathrm{E}\left(\mathrm{X}^2\right) =\sum_{x=0}^3 x_1^2 \mathrm{P}\left(x_{\mathrm{i}}\right)=3 \)
\( \text { Variance of } \mathrm{X} =\mathrm{E}\left(\mathrm{X}^2\right)-[\mathrm{E}(\mathrm{X})]^2 \)
\( =3-\left(\frac{3}{2}\right)^2 \)
\( =3-\frac{9}{4} \)
\( =\frac{3}{4}=0.75\)