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Calculating probability events involves understanding the likelihood of various outcomes in a given situation. In this context, when we're dealing with disks that have different resistance levels, the goal is to determine how many disks fit specific criteria. For example, knowing the total number of disks and how they are distributed across different resistance levels helps us calculate frequencies related to specific events.

In probability, events are often represented by letters such as \( A \) and \( B \). In our scenario:

• Event \( A \) is defined as disks having high shock resistance.
• Event \( B \) refers to disks having high scratch resistance.

The total number of outcomes is based on the 100 disks from the table. From there, we can directly count the number of disks that meet certain conditions, such as both events happening simultaneously or neither event happening. This sets the stage for more complex calculations, such as finding intersections and unions of events.

Understanding the intersection \( A \cap B \) and union \( A \cup B \) of events is crucial in probability. The intersection \( A \cap B \) relates to scenarios where both events occur together. This is viewed by looking at the overlap. In our exercise, the intersection consists of disks with both high shock and high scratch resistance, and we find that 70 disks meet this criteria.

On the other hand, the union \( A \cup B \) relates to instances where either one or both events occur. The calculation of the union includes all disks that have at least one high resistance, which considers:

• The number of disks with high shock resistance \( |A| = 79 \).
• The number of disks with high scratch resistance \( |B| = 86 \).
• Minus those counted twice in \( A \cap B \) which is 70.

This brings us to an overarching count of disks meeting at least one of the resistance criteria: \( |A \cup B| = 95 \). By visualizing intersections and unions, we understand the shared and distinct characteristics of the datasets.

A discrete probability distribution involves determining probabilities for outcomes in a countable sample space. In the case of the disks with given conditions on resistance, each disk represents a discrete outcome. These outcomes form the basis for our understanding of probability distributions.

The contingency table provided offers a clear layout of how many disks fall under each category of resistance. This kind of tabular data aids in constructing discrete distributions by allowing us to assign probabilities based on frequencies observed. For instance:

• The probability of picking a disk with high shock resistance is the number of such disks (79) divided by the total disks (100), making it 0.79.
• Similarly, the probability for high scratch resistance becomes 0.86, given 86 disks fall under this category.

Each specific outcome—like high shock and low scratch—can also be determined using the table, offering a detailed view into the likelihood of each individual event or combination of events. This clearly demonstrates how probabilities are derived from empirical data, forming the cornerstone of decision-making and prediction in probability studies.