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

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

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