Problem 12 - A Modern Introduction to Probability and Statistics: Understanding Why and How

Question

The events \(A, B\), and \(C\) satisfy: \(\mathrm{P}(A \mid B \cap C)=1 / 4, \mathrm{P}(B \mid C)=1 / 3\), and \(\mathrm{P}(C)=1 / 2\). Calculate \(\mathrm{P}\left(A^{c} \cap B \cap C\right)\).

Step-by-Step Solution

Answer

\( \mathrm{P}(A^c \cap B \cap C) = \frac{1}{8} \).

Step 1: Express the Unknown Probability

We need to find \( \mathrm{P}(A^c \cap B \cap C) \). We can express it using the complement rule: \( \mathrm{P}(A^c \cap B \cap C) = \mathrm{P}(B \cap C) - \mathrm{P}(A \cap B \cap C) \).

Step 2: Use Conditional Probability to Find P(A ∩ B ∩ C)

Given \( \mathrm{P}(A \mid B \cap C) = \frac{1}{4} \), we know \( \mathrm{P}(A \mid B \cap C) = \frac{\mathrm{P}(A \cap B \cap C)}{\mathrm{P}(B \cap C)} \). Let \( x = \mathrm{P}(B \cap C) \); then \( \mathrm{P}(A \cap B \cap C) = \frac{1}{4}x \).

Step 3: Find P(B ∩ C) Using Conditional Probability

Given \( \mathrm{P}(B \mid C) = \frac{1}{3} \), we have \( \mathrm{P}(B \mid C) = \frac{\mathrm{P}(B \cap C)}{\mathrm{P}(C)} \). Using \( \mathrm{P}(C) = \frac{1}{2} \), \( \mathrm{P}(B \cap C) = \frac{1}{3} \times \frac{1}{2} = \frac{1}{6} \).

Step 4: Calculate P(A ∩ B ∩ C)

From Step 2, since \( x = \frac{1}{6} \), we have \( \mathrm{P}(A \cap B \cap C) = \frac{1}{4} \times \frac{1}{6} = \frac{1}{24} \).

Step 5: Substitute Values to Find P(A^c ∩ B ∩ C)

Substitute into the expression from Step 1: \( \mathrm{P}(A^c \cap B \cap C) = \mathrm{P}(B \cap C) - \mathrm{P}(A \cap B \cap C) = \frac{1}{6} - \frac{1}{24} = \frac{4}{24} - \frac{1}{24} = \frac{3}{24} = \frac{1}{8} \).

Key Concepts

Complement Rule

When we talk about the complement rule in probability, we're actually working with the idea of all possible outcomes. In simple terms, the complement of an event is everything that can happen, except what we defined as our event. Let's denote an event by the letter \( A \). The complement of \( A \), noted as \( A^c \), includes all the outcomes that are not part of \( A \).

The complement rule is a handy shortcut in probability. It tells us that the probability of \( A^c \) is the same as 1 minus the probability of \( A \). Mathematically, it is expressed as:

• \( \mathrm{P}(A^c) = 1 - \mathrm{P}(A) \)

In the exercise you worked on, you had to find \( P(A^c \cap B \cap C) \). By knowing \( P(A \cap B \cap C) \) and \( P(B \cap C) \), you can apply the complement rule as shown initially to simplify the calculation.

This type of rule is especially useful because instead of calculating every possible outcome separately, you can focus on the total minus the outcomes you don't want. 📚

Events Intersection

In probability, the intersection of events refers to occurrences that happen simultaneously. Imagine two events \( A \) and \( B \). The intersection of these events, denoted by \( A \cap B \), represents all the outcomes that belong to both \( A \) and \( B \). In layman terms, it's like finding overlapping interests between two groups of people.

When you calculate the probability of intersections, you are essentially figuring out the chance of both events happening together. This is often fundamental in solving problems where multiple conditions must be met simultaneously. In the exercise, the intersection \( B \cap C \) was used as a condition for further analysis.

• The probability formula for intersection when dependencies like conditional probabilities are present is:
• \( \mathrm{P}(A \cap B) = \mathrm{P}(A \mid B) \times \mathrm{P}(B) \)

Understanding this overlap and the role of intersections helps deepen your grasp on how events might influence each other or combine, which is key in solving complicated probability puzzles.🔍

Probability Calculations

Probability calculations, at their core, involve determining the likelihood of certain outcomes given specific conditions. These calculations are often premised on basic rules and conditional probabilities, which you might already be familiar with.

In the exercise, several calculations were needed to derive the final probability. Conditional probability allows you to find out the likelihood of an event occurring given that another event has already occurred. It's expressed as \( \mathrm{P}(A \mid B) = \frac{\mathrm{P}(A \cap B)}{\mathrm{P}(B)} \). Using this, one can backtrack to find specific probabilities when combined events are involved. For instance, the exercise used given probabilities \( \mathrm{P}(B \mid C) \) and \( \mathrm{P}(A \mid B \cap C) \) to navigate through the solution.

• Start by identifying known probabilities and relations.
• Apply conditional probability to find missing probabilities.
• Combine these to find the probability of the event you are interested in.

By mastering these calculation techniques, you streamline the process of solving probability problems, making it more intuitive and less intimidating.🙌