91Ó°ÊÓ

When exploring probability, complementary events play a crucial role. Let's consider an event, say "B." Its complementary event, denoted as \(\bar{B}\), is the event that "B" does not occur. The probability of a complementary event is found by subtracting the probability of the original event from one. In mathematical terms, \(P(\bar{B}) = 1 - P(B)\). This equation simply expresses that either the event occurs, or it does not—a foundational concept in probability.
This definition becomes especially useful when analyzing independent events. If two events, like "A" and "B" in this exercise, are independent, then calculating the probability of "A" and \(\bar{B}\) together requires understanding these complementary principles. They help us extend our analysis from original pairs of events to the complements of those events effortlessly.

In probability, certain rules help establish relationships between events and their probabilities. One key probability rule is for independent events, where the occurrence of one event does not affect the probability of another event. Mathematically, for two independent events "A" and "B", the relationship is expressed as \( P(A \cap B) = P(A) \cdot P(B) \).
Applying this knowledge, the probability for the joint occurrence of "A" and \(\bar{B}\)—the complement of "B"—should adjust to reflect these independence concepts. This means we reformulate the probability to fit the complementary event: \(P(A \cap \bar{B}) = P(A) \cdot (1 - P(B))\). This calculation maintains the independence established between "A" and "B" even when one of the events is replaced by its complement, showcasing the flexibility and consistency of probability rules.

Developing a mathematical proof involves logical reasoning, expressed through equations and derived results. In this case, we explore the proof by applying known probability rules and complementary event principles.
First, we validate the rule for "A" and \(\bar{B}\). Using known independence of "A" and "B", we find \( P(A \cap \bar{B}) = P(A) (1 - P(B))\), matching \(P(A) \cdot P(\bar{B})\). Checking this confirms "A" and \(\bar{B}\) are independent.
Next, explore "\bar{A}" and "\bar{B}". Starting with \(P(\bar{A}\cap\bar{B}) = 1 - (P(A) + P(B) - P(A \cap B))\), applying \(P(A \cap B)=P(A)P(B)\), simplifies it to \(1 - P(A) - P(B) + P(A)P(B)\). Notice this equals \((1-P(A))(1-P(B))\), proving independence for "\bar{A}" and "\bar{B}". This step-by-step logical progression highlights the elegance and rigor of mathematical proof.