91Ó°ÊÓ

Discrete probability refers to the probability of outcomes that are countable and separate. Specifically, it deals with events that can be expressed as a list of distinct values, like rolling a die, where the outcomes are distinct integers from 1 to 6. In this context, we focus on a particular example—a random variable \(X\) that can take on the values 1, 2, or 3. Each of these outcomes is considered "discrete," because it's a separate, distinct result.

These possibilities are represented by a probability mass function, or PMF, which assigns a probability to each discrete outcome. Each outcome in a discrete probability distribution has a non-negative probability associated with it. Moreover, the sum of all these probabilities must equal one. This ensures the function accurately represents all potential outcomes of the random variable. In our exercise, the PMF was given as \(f(x) = \left(\frac{8}{7}\right)\left(\frac{1}{2}\right)^x\), for \(x = 1, 2, 3\). Calculating these gives us valid probabilities that sum to 1, confirming it's a valid PMF.

PMF verification is a crucial step to ensure that a probability mass function correctly represents a probability scenario. The PMF must satisfy two conditions:

• All probabilities must be non-negative: This ensures that there's no negative chance for any event, as probability values range from 0 to 1.
• The sum of probabilities for all possible outcomes must equal 1: This ensures completeness, meaning all outcomes thought of by the random variable are included and their total equals certainty.

In our given example, we calculated the probabilities for each possible value of \(X\).
For \(x = 1\), \(f(1) = \frac{4}{7}\), for \(x = 2\), \(f(2) = \frac{2}{7}\), and for \(x = 3\), \(f(3) = \frac{1}{7}\). Summing these, \(f(1) + f(2) + f(3) = 1\), confirms the function's completeness and validity.

This verification ensures that all probabilities are realistic and account for every possibility.

Probability calculation involves determining the likelihood of specific events based on the probability mass function. In the given problem, we are asked to calculate four probabilities using the PMF.

For event \(P(X \leq 1)\), we look at the value when \(X = 1\), which is simply \(f(1) = \frac{4}{7}\).

• For \(P(X > 1)\), we sum the probabilities for \(x = 2\) and \(x = 3\), giving us \(f(2) + f(3) = \frac{3}{7}\).
• For \(P(2 < X < 6)\), we only consider \(x = 3\) since it falls within the range, resulting in \(f(3) = \frac{1}{7}\).
• The probability \(P(X \leq 1\) or \(X > 1)\) includes all outcomes, therefore it equals 1.

The calculation of these probabilities is straightforward due to the structure of a PMF, and each sum or individual probability calculation reflects the likelihood specific to the context set within the PMF definition.