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  • A biased die is such that P (4) = 1/10 and other scores being equally likely. The die is tossed twice. If X is the ‘number of fours seen’, find the variance of the random variable X.

A biased die is such that P (4) = 1/10 and other scores being equally likely. The die is tossed twice. If X is the ‘number of fours seen’, find the variance of the random variable X.

Answers (1)

Solution

Given-

 X= number of four seen

On tossing to die, X=0,1,2

\mathrm{P}_{4}=\frac{1}{10}\ \ , \mathrm{P}_{\mathrm{not} 4}=\frac{9}{10}
Therefore,

\mathrm{P}(\mathrm{X}=0)=\mathrm{P}_{\mathrm{not} 4} \cdot \mathrm{P}_{\mathrm{not} 4}=\frac{9}{10} \times \frac{9}{10}=\frac{81}{100}
\\ \mathrm{P}(\mathrm{X}=1)=\mathrm{P}_{\mathrm{not} 4} \cdot \mathrm{p}_{4} +P_4\cdot \mathrm{P}_{\mathrm{not} 4}\\=\frac{9}{10} \times \frac{1}{10}+\frac{1}{10} \times \frac{9}{10}\\=\frac{18}{100} \\\\\mathrm{P}(\mathrm{X}=2)=\mathrm{P}_{4} \cdot \mathrm{P}_{4} =\frac{1}{10} \times \frac{1}{10}=\frac{1}{100}
Thus, the table is derived

 \begin{aligned} &\begin{array}{|l|l|l|l|} \hline \mathrm{X} & 0 & 1 & 2 \\ \hline \mathrm{P}(\mathrm{X}) & \frac{81}{100} & \frac{18}{100} & \frac{1}{100} \\ \hline \mathrm{XP}(\mathrm{X}) & 0 & \frac{18}{100} & \frac{2}{100} \\ \hline \mathrm{X}^{2} \mathrm{p}(\mathrm{x}) & 0 & \frac{18}{100} & \frac{4}{100} \\ \hline \end{array}\\ &\therefore \operatorname{Var}(\mathrm{X})=\mathrm{E}(\mathrm{X})^{2}-\left[\mathrm{E}(\mathrm{X})^{2}\right]=\mathrm{X}^{2} \mathrm{P}(\mathrm{x})-[\mathrm{XP}(\mathrm{X})]^{2}\\ &\Rightarrow\left[0+\frac{18}{100}+\frac{4}{100}\right]-\left[0+\frac{18}{100}+\frac{2}{100}\right]^{2}\\ &=\frac{22}{100}-\left(\frac{20}{100}\right)^{2}=\frac{11}{50}-\frac{1}{25}\\ &=\frac{11-2}{50}=\frac{9}{50}=\frac{18}{100}=0.18 \end{aligned}

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