91Ó°ÊÓ

In the realm of probability, two or more events are considered independent if the occurrence, or non-occurrence, of any of these events does not affect the likelihood of the other events taking place. This is a fundamental concept when dealing with probability calculations and it simplifies computations immensely.

For instance, if we consider events \( A \), \( B \), and \( C \) as independent, it implies:

• The probability of event \( A \) occurring does not alter the probability of \( B \) or \( C \) occurring, and vice versa.
• The combined probability of all three events happening at the same time can be calculated by simply multiplying their individual probabilities.

Thus, to find \( P(A \cap B \cap C) \), the product of their individual probabilities is calculated as follows:\[ P(A \cap B \cap C) = P(A) \times P(B) \times P(C) \]. For this case: \[ P(A \cap B \cap C) = 0.88 \times 0.65 \times 0.44 \approx 0.25024 \].
This calculation shows that when you have independent events, handling them in probability equations can become much simpler.

A complementary event occurs when the event in question does not take place. For any event \( A \), its complement, represented as \( A^c \), denotes every outcome not covered by \( A \). Understanding complements helps in calculating probabilities for events not happening.

The relation between an event and its complement is straightforward:

• The sum of the probabilities of an event and its complement always equals 1, \( P(A) + P(A^c) = 1 \).
• If you know the probability of an event happening, you can easily find out the probability of it not happening by subtracting from 1, e.g., \( P(A^c) = 1 - P(A) \).

In the provided exercise, we calculated the complements for events \( A \), \( B \), and \( C \):

• \( P(A^c) = 0.12 \)
• \( P(B^c) = 0.35 \)
• \( P(C^c) = 0.56 \)

Knowing how to manage complementary events is crucial for solving probability problems where you are interested in the likelihood of an event not occurring.

Intersection of events refers to situations where two or more events occur simultaneously. In probability terms, it examines the overlap between sets of outcomes.

The intersection of events \( A \) and \( B \) is represented by \( A \cap B \) and includes only the outcomes shared by both events. When applying this to independent events, the concept appears straightforward due to multiplication properties.

Consider the task to find the probability of all three complementary events \( A^c \cap B^c \cap C \) happening. It involves multiplying their individual probabilities, considering they are independent. This applies the same principle as calculating an ordinary intersection: \[ P(A^c \cap B^c \cap C) = P(A^c) \times P(B^c) \times P(C) \].

This problem demonstrates the intersection of not only standard but also complementary events, thereby expanding the application of this concept. Here, it helps to calculate the likelihood of none of the three main events \( A \), \( B \) occurring and \( C \) occurring:\[ P(A^c \cap B^c \cap C) = 0.12 \times 0.35 \times 0.44 \approx 0.01848 \]. Understanding and using intersections effectively is key in mastering probability calculations.