Three fair coins numbered 1 and 0 are tossed simultaneously. Then variance $\operatorname{Var}(\mathrm{X})$…
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
0.7
0.75
0.65
0.62
Solution
Three fair coins numbered 1,0 are tossed.
$\therefore \quad$ Sample space $=\{111,110,101,011,100,010$, $001,000\}$
$\therefore \quad \mathrm{n}(\mathrm{S})=8$
$\mathrm{X}$ represents the sum of numbers on upper most face
$\begin{aligned}
\therefore \quad & P(X=0)=\frac{1}{8}, \\
& P(X=1)=\frac{3}{8}, \\
& P(X=2)=\frac{3}{8}, \\
& P(X=3)=\frac{1}{8}
\end{aligned}$
$\therefore \quad$ 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}
$\begin{aligned}
\mathrm{E}(\mathrm{X}) & =\sum_{x=0}^3 x_{\mathrm{i}} \mathrm{P}\left(x_{\mathrm{i}}\right) \\
\mathrm{E}(\mathrm{X}) & =0 \times \frac{1}{8}+1 \times \frac{3}{8}+2 \times \frac{3}{8}+3 \times \frac{1}{8} \\
& =\frac{12}{8}=\frac{3}{2} \\
\mathrm{E}\left(\mathrm{X}^2\right) & =\sum_{x=0}^3 x_1^2 \mathrm{P}\left(x_{\mathrm{i}}\right)=3 \\
\text { Variance of } \mathrm{X} & =\mathrm{E}\left(\mathrm{X}^2\right)-[\mathrm{E}(\mathrm{X})]^2 \\
& =3-\left(\frac{3}{2}\right)^2 \\
& =3-\frac{9}{4} \\
& =\frac{3}{4}=0.75
\end{aligned}$