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 $$ P(A \cup B)=1-P\left(A^{c}\right) P\left(B^{c}\right) $$ \(12.3 .4\)

Step-by-Step Solution

Answer

(a) Use a Venn diagram to verify \((A \cup B)^c = A^c \cap B^c\). (b) If \(A\) and \(B\) are independent, then \(A^c\) and \(B^c\) are also independent. (c) Thus, \(P(A \cup B) = 1 - P(A^c)P(B^c)\).

Step 1: Understanding Venn Diagram for Set Complementation

To solve part (a), let's first understand the Venn diagram representation. We need to show that \((A \cup B)^{c} = A^{c} \cap B^{c}\). In a Venn diagram, the union \(A \cup B\) includes all elements that are in either set \(A\) or set \(B\). The complement \((A \cup B)^{c}\) will then include all elements that are not in the union \(A \cup B\). This should equal to the intersection of elements that are not in \(A\) or \(B\), which is represented by \(A^{c} \cap B^{c}\).

Step 2: Proving Venn Diagram Complements Property

The complement of the union \((A \cup B)^{c}\) means taking everything outside of the area covered by the union of \(A\) and \(B\). Since no element outside \(A\) or \(B\) would reside in the union, the only elements outside \(A \cup B\) will necessarily reside in both \(A^{c}\) and \(B^{c}\), proving that \((A \cup B)^{c} = A^{c} \cap B^{c}\).

Step 3: Using Complement Independence

Given \(A\) and \(B\) are independent, we must now show \(A^{c}\) and \(B^{c}\) are independent. By definition, two events \(A\) and \(B\) are independent if \(P(A \cap B) = P(A)P(B)\). From Step 1, we observe \((A \cup B)^{c} = A^{c} \cap B^{c}\), hence:\[P(A^{c} \cap B^{c}) = P((A \cup B)^{c}) = 1 - P(A \cup B)\]Using De Morgan's Laws and the independence of \(A\) and \(B\), \(A^{c}\) and \(B^{c}\) are independent since the complement rule aligns with their separability.

Step 4: Calculating Probability of a Union Based on Complements

Expanding on Step 3, if \(A\) and \(B\) are independent, we showed that \(A^{c}\) and \(B^{c}\) are independent, leading to:\[P(A \cup B) = 1 - P(A^{c} \cap B^{c})\]Since \(P(A^{c} \cap B^{c}) = P(A^{c})P(B^{c})\), then:\[P(A \cup B) = 1 - P(A^{c})P(B^{c})\]This is achieved by complementing to find the probability of the union using given expressions for independent events.

Key Concepts

Venn Diagram

A Venn Diagram is a visual tool used to show relationships between different sets. It's extremely useful for understanding basic operations like union, intersection, and complementation. In a Venn diagram, sets are usually depicted as overlapping circles in an enclosed rectangle, representing the universal set. The union of sets, denoted by \(A \cup B\), includes all elements in either set \(A\) or set \(B\). Conversely, a complement of a set consists of all elements not in the set.

To depict the complement of a union \((A \cup B)^{c}\), we shade all areas in the Venn diagram that are not part of \(A \cup B\). This effectively leaves all elements in both \(A^{c}\) and \(B^{c}\)—the elements that are outside of both \(A\) and \(B\). Thus, a Venn diagram vividly illustrates the equality \((A \cup B)^{c} = A^{c} \cap B^{c}\). This insight is valuable for understanding how complementation in set theory visually manifests in diagrams.

Set Complementation

Set Complementation involves finding all elements that are not part of a specific set. Formally, for any set \(A\), its complement \(A^{c}\) includes all elements not in \(A\). This concept plays a crucial role in probability and mathematics. Knowing the complement of a set can help determine probabilities of events not happening.

In the context of probabilistic events, if \(A\) and \(B\) are two events, their complements \(A^{c}\) and \(B^{c}\) cover the scenarios where \(A\) or \(B\) do not happen, respectively. Utilizing set complementation with the intersection of complements—\((A \cup B)^{c} = A^{c} \cap B^{c}\)—illustrates that everything outside the union of two sets is exactly the intersection of everything outside each individual set. This understanding simplifies solving complex probability problems, especially those involving independent events.

De Morgan's Laws

De Morgan's Laws are fundamental rules of logic and set theory, providing a relationship between unions, intersections, and complementation. These laws aid in transforming expressions involving complex set operations, offering an alternative way to view them.

The laws state that:

• The complement of the union of two sets equals the intersection of their complements: \( (A \cup B)^{c} = A^{c} \cap B^{c} \)
• The complement of the intersection of two sets equals the union of their complements: \( (A \cap B)^{c} = A^{c} \cup B^{c} \)

These principles extend well to probability, explaining behaviors and relationships between probabilistic events. For independent events \(A\) and \(B\), originally defined with \(P(A \cap B) = P(A)P(B)\), we can apply De Morgan's Laws to see how independent complements function. If we know \(A\) and \(B\) are independent, we find \(A^{c}\) and \(B^{c}\) are also independent because \( P(A^{c} \cap B^{c}) = P(A^{c})P(B^{c}) \). These logical transformations form the basis for proving various probability formulas, enabling clearer visualization and simplification of complex problems.