Let \(X_1\) be the outcome of the first die and \(X_2\) be the outcome of the second die. We want to find the expected value of the sum, \(E(X_1 + X_2)\).
By the linearity of expectation, \(E(X_1 + X_2) = E(X_1) + E(X_2)\).
First, find the expected value of a single fair die roll. The possible outcomes are \{1, 2, 3, 4, 5, 6\}, each with a probability of \(1/6\).
The expected value \(E(X_1)\) is:
\[ E(X_1) = \sum x \cdot P(x) = 1(\frac{1}{6}) + 2(\frac{1}{6}) + 3(\frac{1}{6}) + 4(\frac{1}{6}) + 5(\frac{1}{6}) + 6(\frac{1}{6}) \]
\[ E(X_1) = \frac{1+2+3+4+5+6}{6} = \frac{21}{6} = 3.5 \]
Since the second die is also fair, its expected value is the same: \(E(X_2) = 3.5\).
Now, the expected value of the sum is:
\[ E(X_1 + X_2) = E(X_1) + E(X_2) = 3.5 + 3.5 = 7 \]