Problem 11 - Calculus for Biology and Medicine

Question

Assume that \(\Omega=\\{1,2,3,4,5\\}\) \(P(1)=0.1, P(2)=0.2\), and \(P(3)=P(4)=0.05 .\) Furthermore, assume that \(A=\\{1,3,5\\}\) and \(B=\\{2,3,4\\}\) Find \(P\left(A^{c}\right)\).

Step-by-Step Solution

Answer

The probability of the complement set, \(P(A^c)\), is 0.25.

Step 1: Understand the Complement of a Set

The complement of a set \(A\), denoted as \(A^c\), consists of all elements in the universal set \(\Omega\) that are not in \(A\). So, if \(A = \{1,3,5\}\), then \(A^c = \Omega - A = \{2,4\}\).

Step 2: Recognize Probability Assignment

Recall from the exercise that each element of \(\Omega\) is assigned a probability: \(P(1) = 0.1, P(2) = 0.2, P(3) = 0.05, P(4) = 0.05\). The probability of any element not listed (like 5) can be calculated as the remainder up to 1, where \(P(5) = 1 - (0.1 + 0.2 + 0.05 + 0.05) = 0.6\).

Step 3: Find the Probability of the Complement

Calculate \(P(A^c)\) by summing the probabilities of elements in \(A^c\). Since \(A^c = \{2,4\}\), then \(P(A^c) = P(2) + P(4) = 0.2 + 0.05 = 0.25\).

Key Concepts

Complement of a Set

In probability theory, the complement of a set is an essential concept. It involves defining elements that are not part of a particular set but remain within a larger collection known as the universal set. When discussing the complement of a set \(A\), denoted as \(A^c\), we are considering all elements that are not included in set \(A\). For example, if our universal set \(\Omega = \{1, 2, 3, 4, 5\}\) and set \(A = \{1, 3, 5\}\), then the complement of \(A\) would be \(A^c = \{2, 4\}\).

This concept helps us in identifying the elements that are excluded from set \(A\), which is pivotal in calculating probabilities for scenarios not captured by the set \(A\). The complement allows you to see the full picture by focusing on what is left out. It's like looking at the inverse side of a coin—one side shows heads (set \(A\)) and the other tails (complement \(A^c\)).

Universal Set

The universal set, often symbolized as \(\Omega\), is the complete set that encompasses all objects or elements within a specific discussion or problem domain. In mathematical discussions and particularly in probability, the universal set forms the basis of talks about sets and complements.

For the problem at hand, the universal set is \(\Omega = \{1, 2, 3, 4, 5\}\). It includes all the possible outcomes that our discussion encompasses. Each subset, such as \(A\) or \(B\), contains elements from this universal set. When considering probability, the universal set provides the total set of possible events, and subsets represent particular outcomes for which we may calculate probabilities.

• It's the stage on which every probabilistic scenario plays out.
• Understanding \(\Omega\) ensures no possibility is overlooked when calculating probabilities or complements.

Recognizing the universal set enhances our ability to handle subsets and their complements efficiently, ensuring that our probability calculations remain accurate and comprehensive.

Probability Assignment

Probability assignment is the method of associating a probability value with each possible outcome in the universal set \(\Omega\). These values must satisfy the basic axioms of probability: each probability is between 0 and 1, and the sum of all probabilities is 1.

In the exercise, the probability assignments are as follows: \(P(1) = 0.1\), \(P(2) = 0.2\), \(P(3) = 0.05\), and \(P(4) = 0.05\). The probability for \(P(5)\) is calculated by ensuring the total sums up to 1: \(P(5) = 1 - (0.1 + 0.2 + 0.05 + 0.05) = 0.6\).

• Each probability depicts the chance of that particular element occurring in a random selection.
• The sum being 1 reaffirms that all possibilities have been accounted for.

This clear and methodical approach ensures that each outcome is given a realistic chance of occurrence, making further calculations, such as finding the probability of a set or its complement, straightforward and reliable.