Problem 17 - Calculus for Biology and Medicine

Question

Assume that \(P(A \cap B)=0.1, P(A)=0.4\), and \(P\left(A^{c} \cap B^{c}\right)=\) 0.2. Find \(P(B)\).

Step-by-Step Solution

Answer

P(B) = 0.5.

Step 1: Understand the Given Probabilities

We are given that the probability of intersection of events A and B, \( P(A \cap B) = 0.1 \), and the probability of event A, \( P(A) = 0.4 \). Additionally, the probability of neither A nor B occurring, \( P(A^c \cap B^c) = 0.2 \). We need to find \( P(B) \), the probability of event B occurring.

Step 2: Calculate the Complement Probabilities

Using the complementary probability formula, we know \( P(A^c) = 1 - P(A) = 1 - 0.4 = 0.6 \). Since \( P(A^c \cap B^c) = 0.2 \), the remaining probability must cover \( P(A^c \cap B) \) and \( P(A \cap B) \).

Step 3: Use the Total Probability Rule

The total probability can be expressed as \( P(A \cap B) + P(A \cap B^c) + P(A^c \cap B) + P(A^c \cap B^c) = 1 \). We substitute the known values: \( 0.1 + P(A \cap B^c) + P(A^c \cap B) + 0.2 = 1 \).

Step 4: Express Unknowns in Terms of Known Values

Rearrange to solve for \( P(A \cap B^c) + P(A^c \cap B) \): \[ P(A \cap B^c) + P(A^c \cap B) = 1 - 0.1 - 0.2 = 0.7 \].

Step 5: Calculate the Relationship Between Unknowns

Using \( P(A) = P(A \cap B) + P(A \cap B^c) \), we know \( 0.4 = 0.1 + P(A \cap B^c) \), thus \( P(A \cap B^c) = 0.3 \).

Step 6: Determine \(P(B)\) by Substitution

Now substitute into the remaining equation: \( 0.7 = 0.3 + P(A^c \cap B) \). Solve for \( P(A^c \cap B) \): \( P(A^c \cap B) = 0.4 \).

Step 7: Combine Probabilities to Find \(P(B)\)

Since \( P(B) = P(A \cap B) + P(A^c \cap B) \), substitute the values: \( P(B) = 0.1 + 0.4 = 0.5 \).

Key Concepts

Intersection of Events

In probability, the intersection of events refers to the probability that two or more events occur simultaneously. It is denoted by the symbol \( \cap \). In mathematical terms, if we have events \( A \) and \( B \), the intersection is represented as \( P(A \cap B) \).
For example, if \( A \) is 'it rains' and \( B \) is 'the temperature drops below freezing', \( P(A \cap B) \) represents the probability that both it rains and the temperature goes below freezing at the same time.

• The formula for the intersection of two independent events \( A \) and \( B \) is: \[ P(A \cap B) = P(A) \times P(B) \]
• For non-independent events, you can't simply multiply individual probabilities; you must use additional given data, like in our exercise.

In our problem, we are given \( P(A \cap B) = 0.1 \), which is used as part of the solution.
This value is crucial, as it affects the total probability and its components.

Complementary Probability

Complementary probability considers the likelihood of an event not occurring. If \( A \) is an event, the complement, denoted \( A^c \), is the scenario where \( A \) does not happen. The complementary probability is calculated as \( P(A^c) = 1 - P(A) \).
Understanding complements helps in determining the likelihood of the opposite outcome when you know one. For instance, if the probability of it raining (event \( A \)) is 0.4, the complementary probability of it not raining is \( P(A^c) = 0.6 \).
In terms of logical probabilities, calculations involving complements help fill in missing parts of probability evaluations, as was required in our given exercise.
Given \( P(A^c \cap B^c) = 0.2 \) in the problem, it represents the probability that neither event \( A \) nor \( B \) occurs. This value is essential for solving the puzzle of finding other probabilities.

Total Probability Rule

The Total Probability Rule is a fundamental concept that provides a way to account for all possible outcomes of random events. It states that the sum of probabilities of all possible outcomes should always be 1.
For two events, the total probability can be represented as:
\[ P(A \cap B) + P(A \cap B^c) + P(A^c \cap B) + P(A^c \cap B^c) = 1 \]
This formula considers all scenarios where events \( A \) and \( B \) might or might not occur. By knowing this principle, you can solve for unknown probabilities if three of them are known.
In the exercise, this rule was crucial in the step where the equation included known probabilities and sought the unknowns. With this, values like \( P(A \cap B^c) \) and \( P(A^c \cap B) \) were calculated to eventually find \( P(B) \).
This representation helps ensure that no probability is left out, maintaining the balance of possible outcomes.

Probability Calculation

Probability calculation involves the methods and formulas used to determine the likelihood of events. To calculate the probability of an event \( B \), you might need to use provided information about intersections or complements.
From the problem, we ultimately aimed to find \( P(B) \) using known values. Using the formula:
\[ P(B) = P(A \cap B) + P(A^c \cap B) \]
You substitute the previously solved values into this equation. This allows for a deeper understanding of how separate probabilities relate and combine to give the complete probability for \( B \).

• Consider the relationships between \( P(A) \), \( P(A^c) \), and \( P(B) \).
• Use known probabilities like \( P(A \cap B) \) to help calculate \( P(A^c \cap B) \).

Remember, organizing your steps methodically, using all given values, and applying rules like intersections and complements can help tackle intricate probability problems.