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Chapter 37: Problem 4

Zeige: Sind \(A\) und \(B\) unabh盲ngig, dann sind auch \(A\) und \(B^{C}\) unabh?ngig, ebenso \(B\) und \(A^{\text {C }}, A^{C}\) und \(B^{C}\)

Short Answer

Expert verified
Question: Show that if events A and B are independent, then A and B^C, B and A^C, and A^C and B^C are also independent. Answer: We have proved that if A and B are independent, then the following pairs of events are also independent: A and B^C, B and A^C, and A^C and B^C.

Step by step solution

01

Recall the definition of independence

To prove that two events are independent, we need to show that the probability of their joint event is equal to the product of their individual probabilities. Mathematically, this can be written as: P(A鈭〣) = P(A)P(B) Since we are given that A and B are independent, we know this equation is true.
02

Prove that A and B^C are independent

To prove that A and B^C are independent, we need to show: P(A鈭〣^C) = P(A)P(B^C) Using the fact that B^C = 1 - B: P(A鈭〣^C) = P(A) - P(A鈭〣) Using the given information that A and B are independent, we have: P(A鈭〣) = P(A)P(B) So, P(A) - P(A鈭〣) = P(A) - P(A)P(B) = P(A)(1 - P(B)) Using the same fact that B^C = 1 - B: P(B^C) = 1 - P(B) So, P(A)P(B^C) = P(A)(1 - P(B)) Since we have proved that P(A鈭〣^C) = P(A)P(B^C), A and B^C are independent.
03

Prove that B and A^C are independent

Similarly, we need to show: P(B鈭〢^C) = P(B)P(A^C) Using the fact that A^C = 1 - A: P(B鈭〢^C) = P(B) - P(B鈭〢) Using the given information that A and B are independent, we have: P(B鈭〢) = P(B)P(A) So, P(B) - P(B鈭〢) = P(B) - P(B)P(A) = P(B)(1 - P(A)) Using the same fact that A^C = 1 - A: P(A^C) = 1 - P(A) So, P(B)P(A^C) = P(B)(1 - P(A)) Since we have proved that P(B鈭〢^C) = P(B)P(A^C), B and A^C are independent.
04

Prove that A^C and B^C are independent

To prove that A^C and B^C are independent, we need to show: P(A^C鈭〣^C) = P(A^C)P(B^C) We can use De Morgan's law, which states that (A^C鈭〣^C) = (A鈭狟)^C. Therefore, we have: P((A鈭狟)^C) = P(1 - (A鈭狟)) Using the complementary rule, P(A鈭狟) = P(A) + P(B) - P(A鈭〣), and given that A and B are independent: P(1 - (A鈭狟)) = 1 - (P(A) + P(B) - P(A)P(B)) Now using the fact that A^C = 1 - A and B^C = 1 - B: P(A^C)P(B^C) = (1 - P(A))(1 - P(B)) Since we have proved that P(A^C鈭〣^C) = P(A^C)P(B^C), A^C and B^C are independent. We have now shown that if A and B are independent, then A and B^C, B and A^C, and A^C and B^C are also independent.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding Independence in Probability Theory
In probability theory, understanding the concept of independence is crucial for analyzing the likelihood of various events occurring. Independence between two events, say event A and event B, means that the occurrence of one does not affect the occurrence of the other. This relationship is denoted mathematically as
\[ P(A \cap B) = P(A)P(B) \]
where
\( P(A \cap B) \) represents the probability of both A and B happening together, while \( P(A) \) and \( P(B) \) are the probabilities of A and B occurring independently.

To improve comprehension, imagine rolling two dice. The result of one die has no impact on the result of the other; therefore, the events represented by each die roll are independent. This foundational concept is vital when it comes to analyzing more complex probability problems, such as the one presented in the exercise involving complements of events.
De Morgan's Laws in Probability
De Morgan's laws play a significant role in probability theory, especially when dealing with complements. These laws are essential for understanding the relationships between sets and their complements. The laws are typically presented in two parts:
  • \( (A \cup B)^C = A^C \cap B^C \)
  • \( (A \cap B)^C = A^C \cup B^C \)

Applied to probability, De Morgan's laws help us to find the probability of the complement of combined events, by simplifying expressions involving unions (OR) and intersections (AND) of sets and their complements. For example, if events A and B are independent, De Morgan's laws can assist in determining the probability that neither A nor B occurs, a concept critical to understanding the latter steps of the exercise presented.
Applying the Complementary Rule in Probability
The complementary rule is a fundamental concept in probability that states the probability of the complement of an event is equal to one minus the probability of the event itself. In mathematical terms, for any event A:
\[ P(A^C) = 1 - P(A) \]
Using this rule is particularly advantageous when dealing with complement events, as seen in our exercise.

For example, suppose we wanted to calculate the likelihood of not rolling a six on a single die. The probability of rolling a six is \( \frac{1}{6} \), so using the complementary rule, we find the probability of not rolling a six is \( 1 - \frac{1}{6} = \frac{5}{6} \). This rule has direct implications in the exercise, where understanding the independence of events and their complements is beautifully illustrated by applying the complementary rule to break down the complexities of probabilities involving events and their complements.

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