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A biased die is such that $\mathrm{P}(4)=\frac{1}{10}$ and other scores being equally likely. The die is tossed twice. If $X$ is the 'number of fours seen', then find the variance of the random variable $\mathrm{X}$.
Solution:
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Verified Answer
Let $X=$ Number of fours seen On tossing two die, $\mathrm{X}=0,1,2$.
So, $\mathrm{P}(\mathrm{X}=0)=\frac{9}{10} \cdot \frac{9}{10}=\frac{81}{100}$
$$
\begin{aligned}
&\mathrm{P}(\mathrm{X}=1)=\frac{9}{10} \cdot \frac{1}{10}+\frac{1}{10} \cdot \frac{9}{10}=\frac{18}{100} \\
&\mathrm{P}(\mathrm{X}=2)=\frac{1}{10} \cdot \frac{1}{10}=\frac{1}{100}
\end{aligned}
$$
Thus,
\begin{array}{|l|c|c|c|}
\hline \mathbf{X} & 0 & 1 & 2 \\
\hline \mathbf{P}(\mathbf{X}) & \frac{81}{100} & \frac{18}{100} & \frac{1}{100} \\
\mathbf{X P}(\mathbf{X}) & 0 & 18 / 100 & 2 / 100 \\
\mathbf{X}^2 \mathbf{P}(\mathbf{X}) & 0 & 18 / 100 & 4 / 100 \\
\hline
\end{array}
$\therefore \quad \operatorname{Var}(\mathrm{X})=\Sigma \mathrm{X}^2 \mathrm{P}(\mathrm{X})-[\Sigma \mathrm{XP}(\mathrm{X})]^2$
$=\left[0+\frac{18}{100}+\frac{4}{100}\right]-\left[0+\frac{18}{100}+\frac{2}{100}\right]^2$
$=\frac{11}{50}-\frac{1}{25}=\frac{9}{50}=\frac{18}{100}=0.18$
So, $\mathrm{P}(\mathrm{X}=0)=\frac{9}{10} \cdot \frac{9}{10}=\frac{81}{100}$
$$
\begin{aligned}
&\mathrm{P}(\mathrm{X}=1)=\frac{9}{10} \cdot \frac{1}{10}+\frac{1}{10} \cdot \frac{9}{10}=\frac{18}{100} \\
&\mathrm{P}(\mathrm{X}=2)=\frac{1}{10} \cdot \frac{1}{10}=\frac{1}{100}
\end{aligned}
$$
Thus,
\begin{array}{|l|c|c|c|}
\hline \mathbf{X} & 0 & 1 & 2 \\
\hline \mathbf{P}(\mathbf{X}) & \frac{81}{100} & \frac{18}{100} & \frac{1}{100} \\
\mathbf{X P}(\mathbf{X}) & 0 & 18 / 100 & 2 / 100 \\
\mathbf{X}^2 \mathbf{P}(\mathbf{X}) & 0 & 18 / 100 & 4 / 100 \\
\hline
\end{array}
$\therefore \quad \operatorname{Var}(\mathrm{X})=\Sigma \mathrm{X}^2 \mathrm{P}(\mathrm{X})-[\Sigma \mathrm{XP}(\mathrm{X})]^2$
$=\left[0+\frac{18}{100}+\frac{4}{100}\right]-\left[0+\frac{18}{100}+\frac{2}{100}\right]^2$
$=\frac{11}{50}-\frac{1}{25}=\frac{9}{50}=\frac{18}{100}=0.18$
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