Let \(X\) be the sum of the outcomes obtained when \(2\) fair dice are thrown.
One constructs the possibility diagram that represents the outcomes of this experiment.
| Sum | \(1\) | \(2\) | \(3\) | \(4\) | \(5\) | \(6\) |
|---|---|---|---|---|---|---|
| \(1\) | \(2\) | \(3\) | \(4\) | \(5\) | \(6\) | \(7\) |
| \(2\) | \(3\) | \(4\) | \(5\) | \(6\) | \(7\) | \(8\) |
| \(3\) | \(4\) | \(5\) | \(6\) | \(7\) | \(8\) | \(9\) |
| \(4\) | \(5\) | \(6\) | \(7\) | \(8\) | \(9\) | \(10\) |
| \(5\) | \(6\) | \(7\) | \(8\) | \(9\) | \(10\) | \(11\) |
| \(6\) | \(7\) | \(8\) | \(9\) | \(10\) | \(11\) | \(12\) |
By counting the frequency of each occurrence of the sum, we make the following tabulation that shows the probability distribution of \(X\):
| \(x\) | \(2\) | \(3\) | \(4\) | \(5\) | \(6\) | \(7\) | \(8\) | \(9\) | \(10\) | \(11\) | \(12\) |
|---|---|---|---|---|---|---|---|---|---|---|---|
| \(P(X = x)\) | \(\frac{1}{36}\) | \(\frac{2}{36}\) | \(\frac{3}{36}\) | \(\frac{4}{36}\) | \(\frac{5}{36}\) | \(\frac{6}{36}\) | \(\frac{5}{36}\) | \(\frac{4}{36}\) | \(\frac{3}{36}\) | \(\frac{2}{36}\) | \(\frac{1}{36}\) |
It is straightforward to observe that the set of values this random variable can take is finite, and each of the probability associated to the values is in the interval \([0, 1]\). Additionally,
\[\sum_{\text{all}\hspace{0.5em} x} P(X = x) = \frac{1}{36} + \frac{2}{36} + \frac{3}{36} + \frac{4}{36} + \frac{5}{36} + \frac{6}{36} + \frac{5}{36} + \frac{4}{36} + \frac{3}{36} + \frac{2}{36} + \frac{1}{36} = 1.\]
Given a probability density function of random variable \(X\) whose set of values is finite, we can use a table such as in Example \(2\) to display the probability distribution of \(X\). Another more visual representation is to use a histogram.

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