The expected value of the sum of the two numbers obtained on the uppermost faces, when two fair dice are rolled, is

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Aicharya · Accepted Answer

In a single throw of a pair of dice, the sum of the numbers on them can be $2,3,4,5,6,7,8$, $9,10,11,12$. So X can take values $2,3,4, \ldots$, 12. The probability distribution of $X$ is

$\begin{aligned} \therefore \quad & \mathrm{E}(\mathrm{X}) \\ = & \sum x_i \cdot \mathrm{P}\left(x_{\mathrm{i}}\right) \\ = & \frac{1}{36} \times 2+\frac{2}{36} \times 3+\frac{3}{36} \times 4+\frac{4}{36} \times 5 \\ & +\frac{5}{36} \times 6+\frac{6}{36} \times 7+\frac{5}{36} \times 8+\frac{4}{36} \times 9 \\ & +\frac{3}{36} \times 10+\frac{2}{36} \times 11+\frac{1}{36} \times 12\end{aligned}$ $\begin{aligned} & \Rightarrow \mathrm{E}(\mathrm{X})=\frac{1}{36}(2+6+12+20+30+42+40 \\ & \Rightarrow \mathrm{E}(\mathrm{X})=\frac{252}{36}=7\end{aligned}$

The expected value of the sum of the two numbers obtained on the uppermost faces, when two fair dice are…

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