Discrete Random Variables (DRV) - Constructing and Using Discrete Probability Distributions

Discrete Random Variables (DRV) - Constructing and Using Discrete Probability Distributions

IB Year 5 | Grade 11

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.

Possibility diagram for the sum of two dice (Die 1 vs Die 2)
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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