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Three fair coins with faces numbered 1 and 0 are tossed simultaneously. Then variance $(\mathrm{X})$ of the probability distribution of random variable $\mathrm{X}$, where $\mathrm{X}$ is the sum of numbers on the upper most faces, is
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The correct answer is:
$0.75$
Possible value of $\mathrm{X}$ are $0,1,2,3$ Here,
$\mathrm{S}=\{000,001,010,1,00,111,110,101,011\}$
$\mathrm{n}(\mathrm{S})=8$
$$
\begin{aligned}
& \therefore \mathrm{E}(\mathrm{X})=\sum x_{\mathrm{i}} \mathrm{p}_{\mathrm{i}}=0+\frac{3}{8}+\frac{6}{8}+\frac{3}{8}=\frac{12}{8} \\
& \mathrm{E}\left(\mathrm{X}^2\right)= \\
& \therefore \quad \sum x_{\mathrm{i}}^2 \mathrm{p}_{\mathrm{i}}^2=0+\frac{3}{8}+\frac{12}{8}+\frac{9}{8}=\frac{24}{8} \\
& \quad=3-\left(\frac{3}{2}\right)^2 \\
& =3-\frac{9}{4} \\
& =\frac{3}{4} \\
& =0.75
\end{aligned}
$$
$\mathrm{S}=\{000,001,010,1,00,111,110,101,011\}$
$\mathrm{n}(\mathrm{S})=8$
$$
\begin{aligned}
& \therefore \mathrm{E}(\mathrm{X})=\sum x_{\mathrm{i}} \mathrm{p}_{\mathrm{i}}=0+\frac{3}{8}+\frac{6}{8}+\frac{3}{8}=\frac{12}{8} \\
& \mathrm{E}\left(\mathrm{X}^2\right)= \\
& \therefore \quad \sum x_{\mathrm{i}}^2 \mathrm{p}_{\mathrm{i}}^2=0+\frac{3}{8}+\frac{12}{8}+\frac{9}{8}=\frac{24}{8} \\
& \quad=3-\left(\frac{3}{2}\right)^2 \\
& =3-\frac{9}{4} \\
& =\frac{3}{4} \\
& =0.75
\end{aligned}
$$
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