Problem 9
Question
Let \(A\) and \(B\) be subsets of a universal set \(U\) and suppose \(n(U)=200, n(A)=100, n(B)=80\), and \(n(A \cap B)=40\). Compute: a. \(n(A \cup B)\) b. \(n\left(A^{c}\right)\) c. \(n\left(A \cap B^{c}\right)\)
Step-by-Step Solution
Verified Answer
Based on the given information and following the step-by-step solution provided, we can conclude that:
a. \(n(A \cup B) = 140\)
b. \(n(A^c) = 100\)
c. \(n(A \cap B^c) = 60\)
1Step 1: Identify Known Information
We are given the cardinalities of the sets U, A, B, and A ∩ B:
- n(U) = 200
- n(A) = 100
- n(B) = 80
- n(A ∩ B) = 40
2Step 2: Compute n(A ∪ B)
We can use the formula for the cardinality of the union of two sets:
n(A ∪ B) = n(A) + n(B) - n(A ∩ B)
Now, we can plug in the known values and compute n(A ∪ B):
n(A ∪ B) = 100 + 80 - 40 = 140
3Step 3: Compute n(A^c)
To find the cardinality of the complement of set A with respect to universal set U, we can use the following formula:
n(A^c) = n(U) - n(A)
Now, plug in the values of n(U) and n(A) to compute n(A^c):
n(A^c) = 200 - 100 = 100
4Step 4: Compute n(A ∩ B^c)
To find the cardinality of the intersection of set A and the complement of set B with respect to universal set U, we can first recognize that this is the same as finding the number of elements in A which are not in B. We can use the following formula:
n(A ∩ B^c) = n(A) - n(A ∩ B)
Now, plug in the values of n(A) and n(A ∩ B) to compute n(A ∩ B^c):
n(A ∩ B^c) = 100 - 40 = 60
Now we have the answer to all three requested values:
a. n(A ∪ B) = 140
b. n(A^c) = 100
c. n(A ∩ B^c) = 60
Key Concepts
CardinalitySet OperationsUniversal Set
Cardinality
In set theory, cardinality refers to the number of elements in a set. It's like the headcount of all the unique items within it. For example, if we have a set \( A \) with the elements \{1, 2, 3\}, the cardinality of \( A \), denoted as \( n(A) \), is 3. By knowing the cardinalities of different sets, we gain valuable insight into their size and relationships.
Consider this: If two sets \( A \) and \( B \) are part of a universal set \( U \), and you know how many elements \( A \) and \( B \) contain, you can figure out how many elements they share or how many elements are outside these sets.
Consider this: If two sets \( A \) and \( B \) are part of a universal set \( U \), and you know how many elements \( A \) and \( B \) contain, you can figure out how many elements they share or how many elements are outside these sets.
- Cardinality is essential for understanding how sets interact with each other and with the universal set.
- For complex problems, cardinality gives us straightforward numbers we can manipulate using mathematical formulas.
Set Operations
Set operations are like tools in a toolbox, each performing a specific function to help us understand and manipulate sets. Let's explore the most common set operations featured in our exercise.
The **Union** operation, represented as \( A \cup B \), combines all elements from both sets \( A \) and \( B \). To find the cardinality of the union, use the formula \( n(A \cup B) = n(A) + n(B) - n(A \cap B) \), which ensures you do not double-count elements found in both sets.
The **Complement** operation involves elements not in the set in question, relative to the universal set. For set \( A \), the complement is \( A^c \), and you can find its cardinality with \( n(A^c) = n(U) - n(A) \).
Lastly, the **Intersection** operation \( A \cap B^c \) captures elements in \( A \) but not in \( B \). Its cardinality is found using \( n(A \cap B^c) = n(A) - n(A \cap B) \). Each of these operations helps in visualizing and solving problems involving multiple sets.
The **Union** operation, represented as \( A \cup B \), combines all elements from both sets \( A \) and \( B \). To find the cardinality of the union, use the formula \( n(A \cup B) = n(A) + n(B) - n(A \cap B) \), which ensures you do not double-count elements found in both sets.
The **Complement** operation involves elements not in the set in question, relative to the universal set. For set \( A \), the complement is \( A^c \), and you can find its cardinality with \( n(A^c) = n(U) - n(A) \).
Lastly, the **Intersection** operation \( A \cap B^c \) captures elements in \( A \) but not in \( B \). Its cardinality is found using \( n(A \cap B^c) = n(A) - n(A \cap B) \). Each of these operations helps in visualizing and solving problems involving multiple sets.
Universal Set
The universal set, often denoted by \( U \), is like the master list that contains all the elements under consideration. Imagine it as a giant set that holds every possible element you might deal with in a particular discussion or problem.
The concept of a universal set is fundamental because it's the reference point for discussing complements of other sets. When you talk about the complement of a set \( A \), you're referring to elements that are found in the universal set but not in \( A \).
The concept of a universal set is fundamental because it's the reference point for discussing complements of other sets. When you talk about the complement of a set \( A \), you're referring to elements that are found in the universal set but not in \( A \).
- The universal set allows us to frame our problem accurately and to ensure all elements are accounted for.
- It acts as the backdrop that helps to measure and analyze other sets.
Other exercises in this chapter
Problem 9
Evaluate the given expression. $$ P(n, 1) $$
View solution Problem 9
A new state employee is offered a choice of ten basic health plans, three dental plans, and two vision care plans. How many different health-care plans are ther
View solution Problem 9
State whether the statements are true or false. a. \(\\{a, b, c\\}=\\{c, a, b\\}\) b. \(A \in A\)
View solution Problem 10
An experiment consists of selecting a card at random from a 52-card deck. Refer to this experiment and find the probability of the event. A red face card is dra
View solution