Let, random variable $\mathrm{X}$ is the number of tails in three tosses of the coin.
$$
\begin{aligned}
&\therefore \quad \mathrm{X}=0,1,2,3 \\
&\Rightarrow \quad \mathrm{P}(\mathrm{X}=\mathrm{x})={ }^{\mathrm{n}} C_{\mathrm{x}}(\mathrm{p})^{\mathrm{x}} \mathrm{q}^{\mathrm{n}-\mathrm{x}} \\
&\text { where } \\
&\quad \mathrm{n}=3, \mathrm{p}=1 / 2, \mathrm{q}=1 / 2 \text { and } \mathrm{x}=0,1,2,3 .
\end{aligned}
$$
\begin{array}{|l|l|l|l|l|}
\hline \mathbf{X} & \mathbf{0} & \mathbf{1} & \mathbf{2} & \mathbf{3} \\
\hline \mathrm{P}(\mathrm{X}) & \frac{1}{8} & \frac{3}{8} & \frac{3}{8} & \frac{1}{8} \\
\mathrm{XP}(\mathrm{X}) & 0 & \frac{3}{8} & \frac{3}{4} & \frac{3}{8} \\
\mathrm{X}^2 \mathrm{P}(\mathrm{X}) & 0 & \frac{3}{8} & \frac{3}{2} & \frac{9}{8} \\
\hline
\end{array}
We know that, $\operatorname{Var}(X)=E\left(X^2\right)-[E(X)]^2$
Also $\mathrm{E}\left(\mathrm{X}^2\right)$
$$
\begin{aligned}
&=\sum_{\mathrm{i}=1}^{\mathrm{n}} \mathrm{x}_{\mathrm{i}}^2 \mathrm{P}\left(\mathrm{X}_{\mathrm{i}}\right)=0+\frac{3}{8}+\frac{3}{2}+\frac{9}{8}=\frac{24}{8}=3 \\
&\text { and }[\mathrm{E}(\mathrm{X})]^2 \\
&=\left[\sum_{\mathrm{i}=0}^{\mathrm{n}} \mathrm{x}_{\mathrm{i}}^2 \mathrm{P}\left(\mathrm{x}_{\mathrm{i}}\right)\right]^2=\left[0+\frac{3}{8}+\frac{3}{4}+\frac{3}{8}\right]^2=\left[\frac{12}{8}\right]^2=\frac{9}{4} \\
&\left.\therefore \operatorname{Var}(\mathrm{X})=3-\frac{9}{4}=\frac{3}{4} \quad \text { [using Eq. (i) }\right]
\end{aligned}
$$
and S.D of $X=\sqrt{\operatorname{Var}(X)}=\sqrt{\frac{3}{4}}=\frac{\sqrt{3}}{2}$