Problem 13 - 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 \(P(3)\).

Step-by-Step Solution

Answer

The probability of 3, \( P(3) \), is 0.3.

Step 1: Understand the Given Information

We are given a sample space \( \Omega=\{1,2,3,4\} \) and several probabilities: \( P(1) = 0.1 \), \( P(A) = 0.7 \) where \( A = \{2,3\} \), and \( P(B) = 0.5 \) where \( B = \{3,4\} \). We need to find \( P(3) \).

Step 2: Use the Total Probability Rule

We know that the probabilities of all outcomes in the sample space must sum up to 1. Thus, \( P(1) + P(2) + P(3) + P(4) = 1 \). Since \( P(1) = 0.1 \), we have \[0.1 + P(2) + P(3) + P(4) = 1.\]

Step 3: Express P(2) and P(3) using P(A)

Given \( P(A) = 0.7 \) and \( A = \{2,3\} \), we have \[P(2) + P(3) = 0.7.\]

Step 4: Express P(3) and P(4) using P(B)

Given \( P(B) = 0.5 \) and \( B = \{3,4\} \), we have \[P(3) + P(4) = 0.5.\]

Step 5: Solve the System of Equations

We now have two equations:1. \( P(2) + P(3) = 0.7 \)2. \( P(3) + P(4) = 0.5 \).Using the total probability equation, \[0.1 + P(2) + P(3) + P(4) = 1,\]we can substitute the values for \( P(4) \) from the second equation,\[P(4) = 0.5 - P(3).\]Substitute \( P(4) = 0.5 - P(3) \) into the total probability equation,\[0.1 + P(2) + P(3) + (0.5 - P(3)) = 1,\]which simplifies to\[0.6 + P(2) = 1,\]giving \( P(2) = 0.4 \). Then, using \( P(2) + P(3) = 0.7 \), find\[0.4 + P(3) = 0.7,\]which gives \( P(3) = 0.3 \).

Key Concepts

Sample Space

In probability, the term "sample space" refers to the set of all possible outcomes of a particular experiment. Imagine rolling a die; the sample space is the set of results you might get, which is 1, 2, 3, 4, 5, and 6. However, in our exercise, we consider a smaller set: \(\Omega=\{1,2,3,4\}\). This means when we perform our experiment, any outcome must be one of these numbers.

To understand probabilities within this framework, each outcome in the sample space has a certain likelihood associated with it. All those probabilities must add up to 1, representing certainty that one of these outcomes indeed occurs. For example, if \( P(1) = 0.1 \), then the remaining outcomes \( \{2, 3, 4\} \) must share the remaining probability of 0.9. Understanding a sample space helps frame our thinking about possible outcomes and the probability distributions involved.

Total Probability Rule

The Total Probability Rule is a fundamental principle in probability theory. It essentially states that the sum of the probabilities of all individual outcomes within a sample space must equal 1. This rule ensures that we have accounted for all possible outcomes.

In our exercise scenario, the sample space consists of the elements \(\{1, 2, 3, 4\}\). According to the Total Probability Rule, we express this as:
\[P(1) + P(2) + P(3) + P(4) = 1.\]
Given that \(P(1) = 0.1\), we are left with
\[0.1 + P(2) + P(3) + P(4) = 1.\]
This equation is crucial for solving problems involving unknown probabilities and ensures that fractional parts account for a complete whole.

System of Equations

To solve probability problems with multiple conditions, we often use a system of equations. This technique allows us to find unknown probabilities by linking them through known relationships. In our exercise, we encountered a situation with two such known relationships.

First, we have the equation from event \(A\):
\[P(2) + P(3) = 0.7.\]
Second, we use the equation derived from event \(B\):
\[P(3) + P(4) = 0.5.\]
Additionally, the total probability equation reinforces these relationships:
\[0.1 + P(2) + P(3) + P(4) = 1.\]
By solving these equations step-by-step, we identify the individual probabilities. For instance, substituting to find \(P(4)\) and utilizing known sums like \(P(2) + P(3)\), leads us to \(P(3) = 0.3\).

Systems of equations can seem dense, but they provide a systematic approach to solving complex probability puzzles. By strategically substituting known values, we can unlock unknown probabilities.