Problem 36 - Calculus for Biology and Medicine

Question

(a) Use a Venn diagram to show that $$ (A \cup B)^{c}=A^{c} \cap B^{c} $$ (b) Use your result in (a) to show that if \(A\) and \(B\) are independent, then \(A^{c}\) and \(B^{c}\) are independent. (c) Use your result in (b) to show that if \(A\) and \(B\) are independent, then $$ P(A \cup B)=1-P\left(A^{c}\right) P\left(B^{c}\right) $$

Step-by-Step Solution

Answer

(A) Use diagram to prove the set equality. (B) Complements are independent. (C) Formula for union using complement independence.

Step 1: Venn Diagram Explanation

To demonstrate \((A \cup B)^{c}=A^{c} \cap B^{c}\) using a Venn diagram, draw two overlapping circles labeled \(A\) and \(B\) inside a rectangle representing the universal set. The region outside both circles represents \((A \cup B)^c\). According to the formula, it should equal the intersection of \(A^c\) (everything outside \(A\)) and \(B^c\) (everything outside \(B\)). These two complement areas overlap exactly in the region that is not part of either \(A\) or \(B\), effectively proving the statement.

Step 2: Independence of Complements

Given two independent sets \(A\) and \(B\), we need to show \(A^c\) and \(B^c\) are independent. For independence, the probability condition is \(P(A \cap B) = P(A)P(B)\). Since \(P(A^c) = 1 - P(A)\) and \(P(B^c) = 1 - P(B)\), calculate \(P(A^c \cap B^c)\): \(P(A^c \cap B^c) = P((A \cup B)^c) = 1 - P(A \cup B)\). From Step 1, logically this equals \(P(A^c)P(B^c)\), proving \(A^c\) and \(B^c\) are independent.

Step 3: Probability of Union

To show \(P(A \cup B) = 1 - P(A^c)P(B^c)\) using the result from Step 2, use the complement rule: \(P(A \cup B) = 1 - P((A \cup B)^c)\). From Step 1, \(P((A \cup B)^c) = P(A^c \cap B^c)\). Since \(A^c\) and \(B^c\) are independent, \(P(A^c \cap B^c) = P(A^c)P(B^c)\). Substitute this back to get \(P(A \cup B) = 1 - P(A^c)P(B^c)\), completing the proof.

Key Concepts

Set Theory

Set theory is a fundamental branch of mathematical logic that provides the basis for many areas of mathematics. A Venn diagram is a useful tool in set theory to visually represent relationships between sets. When working with sets, we use operations such as union, intersection, and complement to combine or subtract elements.

- **Union (\(A \cup B\))**: This is the set containing all elements that are in set \(A\), or set \(B\), or in both.- **Intersection (\(A \cap B\))**: This consists of elements found in both set \(A\) and set \(B\).- **Complement (\(A^c\))**: This includes everything that is not in the set \(A\).

Venn diagrams help illustrate concepts like the complement of a union such as \((A \cup B)^{c}\), which equals the intersection of the complements \(A^{c} \cap B^{c}\). This means any area outside both sets \(A\) and \(B\). By visualizing with two overlapping circles, one can see how these operations work together.

Independent Events

Independent events are central to probability theory. Two events are independent if the occurrence of one doesn’t affect the probability of the other. For events \(A\) and \(B\), they are said to be independent if
\[ P(A \cap B) = P(A) \cdot P(B) \]

When discussing complements, \(A^c\) and \(B^c\) are complements of \(A\) and \(B\) respectively. Provided \(A\) and \(B\) are independent, it is crucial to understand that their complements remain independent.
This means:
- The probability that neither event happens, \(P(A^c \cap B^c)\), equals \(P(A^c) \cdot P(B^c)\).
- This result is derived from the identity \( A^c \cap B^c = (A \cup B)^c \). Thus, independence conditions are retained in the complements.

Probability Theory

Probability theory is the study of uncertainty and the likelihood of different outcomes. It uses mathematical principles to evaluate how likely events are to occur.

Given \(A\) and \(B\) are independent, we can understand their probabilities using the complement rule.

• The probability of a union, or either event \(A\) or \(B\) happening, is given by:
  \[ P(A \cup B) = 1 - P(A^c)P(B^c) \]
• This relates to the fact that the probability of at least one occurring equals the complement of neither occurring.

The talent of probability theory lies in finding these relationships between events through formulations and logical rules, making it easier to predict the behavior of complex combinations of events.