91Ó°ÊÓ

A discrete random variable is a variable that can take on a countable number of distinct values. Unlike continuous random variables, which can take any value within a range, discrete random variables have specific separate values. In the given exercise, the random variable \(X\) is discrete because it only takes on the values 1, 2, and 3 from a definite set. Each of these values has a specific probability associated with it, given by the probability mass function (pmf). This pmf is defined by \(p(x) = \frac{1}{3}\) for \(x = 1, 2, 3\).

• A discrete random variable lists all possible outcomes. In this case, these outcomes are finite.
• The pmf explains how the probability is distributed across these discrete outcomes.

Understanding discrete random variables helps us describe processes and situations where results are specific. For instance, rolling a die results in a discrete random variable because the outcomes are limited to integers between 1 and 6.

In probability, transforming variables allows us to model different scenarios by changing how we compute outcomes from given data. Here, the transformation helps us find the probability mass function (pmf) of a new random variable \(Y\) derived from the original \(X\). The transformation equation given is \(Y = 2X + 1\). This means we multiply each value of \(X\) by 2, then add 1 to find the corresponding \(Y\) values.

• When \(X = 1\), \(Y = 2(1) + 1 = 3\).
• When \(X = 2\), \(Y = 2(2) + 1 = 5\).
• When \(X = 3\), \(Y = 2(3) + 1 = 7\).

The transformation preserves the probability distribution of \(X\) because all converted \(Y\) values have the same probability as their original \(X\) counterparts. This new pmf for \(Y\) still reflects uniform probability across its possible values 3, 5, and 7, each having \(p(y) = \frac{1}{3}\).

The sum of probabilities in any probability distribution must always equal 1. This fundamental principle ensures we account for all possible outcomes. After transforming \(X\) into \(Y\), we check that the sum of the probabilities in the pmf for \(Y\) is 1. The exercise solution confirmed this by showing:

• \(p(3) = \frac{1}{3}\)
• \(p(5) = \frac{1}{3}\)
• \(p(7) = \frac{1}{3}\)

Adding these probabilities gives \(\frac{1}{3} + \frac{1}{3} + \frac{1}{3} = 1\).
This verification step is critical to confirming that the transformation and the probability assignments were done correctly. Ensuring probabilities across all outcomes add up to 1 guarantees that our model fully represents all potential happenings.