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  • Find the probability distribution of the maximum of the two scores obtained when a die is thrown twice. Determine also the mean of the distribution.

Find the probability distribution of the maximum of the two scores obtained when a die is thrown twice. Determine also the mean of the distribution.

Answers (1)

Let X be the random variable score obtained when a die is thrown twice.

∴ X= 1,2,3,4,5,6

The sample space is
\\S= {(1,1), (1,2)... (2,1) (2,2).. (3,1) (3,2) (3,3),..... (6,6)} \\\therefore P(X=1)=\frac{1}{6} \times \frac{1}{6}=\frac{1}{36}$ \\$P(X=2)=\frac{1}{6} \times \frac{1}{6}+\frac{1}{6} \times \frac{1}{6}+\frac{1}{6} \times \frac{1}{6}=\frac{3}{36}$ \\$\mathrm{P}(\mathrm{X}=3)=\frac{5}{36}
In the same way,
\\P(X=4)=\frac{7}{36} \\ P(X=5)=\frac{9}{36}$ \\$P(X=6)=\frac{11}{36}
Therefore, the required distribution is,

\begin{aligned} &\begin{array}{|l|l|l|l|l|l|l|} \hline \mathrm{X} & 1 & 2 & 3 & 4 & 5 & 6 \\ \hline \mathrm{P}(\mathrm{X}) & \frac{1}{36} & \frac{3}{36} & \frac{5}{36} & \frac{7}{36} & \frac{9}{36} & \frac{11}{36} \\ \hline \end{array}\\ &\text { Also, it is known that, }\\ &\operatorname{Mean}\left\{\mathrm{E}(\mathrm{X})=\Sigma \mathrm{XP}(\mathrm{X})=\frac{1}{36}+\frac{6}{36}+\frac{15}{36}+\frac{28}{36}+\frac{45}{36}+\frac{66}{36}\right.\\ &=\frac{161}{36} \end{aligned}

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