Problem 15 - Calculus for Biology and Medicine

Question

In Problems 13-15, assume that $$ \Omega=\\{1,2,3,4\\} $$ and \(P(1)=0.1 .\) Furthermore, assume that \(A=\\{2,3\\}\) and \(B=\) \\{3. 4\\}. \(P(A)=0.7\), and \(P(B)=0.5\). Find \(P\left((A \cap B)^{c}\right.\)

Step-by-Step Solution

Answer

The probability \(P((A \cap B)^c)\) is 0.8.

Step 1: Define the Intersection of Sets A and B

To find the complement of the intersection of sets \(A\) and \(B\), first determine the intersection \(A \cap B\). Here, \(A = \{2,3\}\) and \(B = \{3,4\}\). The intersection, \(A \cap B\), includes all elements common to both sets. Thus, \(A \cap B = \{3\}\).

Step 2: Calculate the Probability of the Intersection

Given that \(A \cap B = \{3\}\), find \(P(A \cap B)\). Since only element 3 is in the intersection and it is part of the overall set \(\Omega\), we need to determine its probability. However, we are not explicitly given this probability, and it is not required for finding \(P((A \cap B)^c)\) due to the formula we'll use next.

Step 3: Use the Complement Rule

To find \(P((A \cap B)^c)\), apply the complement rule. The complement rule states that \(P((A \cap B)^c) = 1 - P(A \cap B)\). Since we know that \(A\) and \(B\) probabilities intersect to 0.3, use the formula: \(P(A \cap B) = P(A) + P(B) - P(A \cup B)\). Thus, \(P(A \cap B) = 0.7 + 0.5 - P(A \cup B)\).

Step 4: Calculate Probabilities

Completeness suggests \(P(A \cup B) = P(A) + P(B) - P(A \cap B)\). Hence \(P(A \cap B) = 0.7 + 0.5 - 1\), given that \(P(A \cup B) = 1\). \(P(A \cap B) = 0.2\).

Step 5: Find the Probability of Complement of the Intersection

Finally, calculate \(P((A \cap B)^c)\) using the complement calculation: \(1 - 0.2 = 0.8\). Therefore, the probability that an element is not in both \(A\) and \(B\) is 0.8.

Key Concepts

Set Theory

Set theory is the foundation of probability and involves understanding and manipulating sets, which are collections of distinct objects or elements. In the context of probability, these sets often represent events within a sample space, such as the one defined in this problem by the universal set \( \Omega=\{1,2,3,4\} \).
Here, two sets are defined:

• Set \(A = \{2, 3\}\) includes the elements 2 and 3.
  
• Set \(B = \{3, 4\}\) includes the elements 3 and 4.

Understanding set theory involves operations like unions, intersections, and complements, which allow mathematical operations between these sets. When you grasp these concepts well, it becomes easier to compute probabilities involving the sets in question, such as in this exercise.

Complement Rule

The complement rule is a fundamental concept in probability. It states that the probability of an event not occurring is equal to one minus the probability that it occurs. Mathematically:
\[ P(A^c) = 1 - P(A) \] This is particularly useful when you know the probability of an event but need to find out the likelihood of it not happening.

• For instance, if you know \( P(A) = 0.7 \), then \( P(A^c) = 0.3 \).
• In our exercise, we apply this rule to the intersection of sets \(A\) and \(B\), specifically finding \( P((A \cap B)^c) \), which means the probability that an element is part of either \(A\) or \(B\) but not both.

By calculating \(1 - P(A \cap B)\), the exercise employs the complement rule to deduce that the final probability we need is 0.8.

Intersection of Sets

The intersection of sets refers to the common elements shared between the sets. It can be thought of as an overlap of two or more sets. In this problem, the interest lies in the intersection between sets \(A\) and \(B\).
Set \(A = \{2, 3\}\) and Set \(B = \{3, 4\}\) have a single common element, which is 3. Thus, \( A \cap B = \{3\} \).
The calculation of probabilities involving intersections often requires knowing the probabilities assigned to these intersections, although, sometimes like in this situation, it can be found indirectly using other given data and probability rules.
Understanding the intersection is crucial for applications in probability, as it allows us to focus precisely on overlapping cases within different events.

Calculus Applications

Calculus applications in probability theory often deal with continuous cases, but understanding calculus concepts can benefit even simple discrete problems. Calculus provides techniques that help in evaluating more complex probability scenarios, such as calculating areas under curves for probability distributions.
While this specific problem involves basic arithmetic without the direct use of calculus, appreciating these deeper methods provides a strong foundation for tackling extensive and nuanced probability problems. Calculus primarily comes into play in the probability when:

• Determining probabilities over continuous intervals
• Handling probability density functions
• Finding cumulative distribution functions

In summary, learning about calculus applications broadens the scope of tackling probability problems, allowing for solutions to a vast array of real-world and theoretical questions.