91Ó°ÊÓ

For customers purchasing a refrigerator at a certain appliance store, let \(A\) be the event that the refrigerator was manufactured in the U.S., \(B\) be the event that the refrigerator had an icemaker, and \(C\) be the event that the customer purchased an extended warranty. Relevant probabilities are. $$ \begin{aligned} &P(A)=.75 \quad P(B \mid A)=.9 \quad P\left(B \mid A^{\prime}\right)=.8 \\ &P(C \mid A \cap B)=.8 \quad P\left(C \mid A \cap B^{\prime}\right)=.6 \\ &P\left(C \mid A^{\prime} \cap B\right)=.7 \quad P\left(C \mid A^{\prime} \cap B^{\prime}\right)=.3 \end{aligned} $$ a. Construct a tree diagram consisting of first-, second-, and third- generation branches and place an event label and appropriate probability next to each branch. b. Compute \(P(A \cap B \cap C)\). c. Compute \(P(B \cap C)\). d. Compute \(P(C)\). e. Compute \(P(A \mid B \cap C)\), the probability of a U.S. purchase given that an icemaker and extended warranty are also purchased.

Step by step solution

01

Understanding the Event Probabilities

1. Event \( A \) represents refrigerators made in the U.S. with \( P(A) = 0.75 \).
2. Event \( B \) represents refrigerators with an icemaker. The conditional probabilities are \( P(B \mid A) = 0.9 \) and \( P(B \mid A') = 0.8 \), where \( A' \) is the complement of \( A \).
3. Event \( C \) represents the purchase of an extended warranty. The conditional probabilities given combinations of \( A \) and \( B \) are provided: \( P(C \mid A \cap B) = 0.8 \), \( P(C \mid A \cap B') = 0.6 \), \( P(C \mid A' \cap B) = 0.7 \), \( P(C \mid A' \cap B') = 0.3 \).

02

Constructing the Tree Diagram

- Start with event \( A \): Branch into \( A \) (0.75) and \( A' \) (0.25).
- From \( A \), branch into \( B \) (0.9) and \( B' \) (0.1). From \( A' \), branch into \( B \) (0.8) and \( B' \) (0.2).
- Next generation for each path, branch into \( C \) and \( C' \) using the given conditional probabilities for \( C \).
- Each path represents a unique intersection of \( A \ B \ C \) or their complements.

03

Calculating P(A ∩ B ∩ C)

Use the tree path probability: For \( A \cap B \cap C \):
\[ P(A \cap B \cap C) = P(A) \times P(B \mid A) \times P(C \mid A \cap B) = 0.75 \times 0.9 \times 0.8 = 0.54 \]

04

Calculating P(B ∩ C)

Calculate the probabilities across all combinations of \( A \) and \( A' \):
\[ P(B \cap C) = P(A \cap B \cap C) + P(A' \cap B \cap C) \]Where:
\[ P(A' \cap B \cap C) = P(A') \times P(B \mid A') \times P(C \mid A' \cap B) = 0.25 \times 0.8 \times 0.7 = 0.14 \]
Thus:
\[ P(B \cap C) = 0.54 + 0.14 = 0.68 \]

05

Calculating P(C)

Sum over all paths that end in \( C \):
\[ P(C) = P(A \cap B \cap C) + P(A \cap B' \cap C) + P(A' \cap B \cap C) + P(A' \cap B' \cap C) \]Calculate each probability path:
\[ P(A \cap B' \cap C) = 0.75 \times 0.1 \times 0.6 = 0.045 \]\[ P(A' \cap B' \cap C) = 0.25 \times 0.2 \times 0.3 = 0.015 \]
Combine:
\[ P(C) = 0.54 + 0.045 + 0.14 + 0.015 = 0.74 \]

06

Calculating P(A | B ∩ C)

Use the conditional probability formula:
\[ P(A \mid B \cap C) = \frac{P(A \cap B \cap C)}{P(B \cap C)} = \frac{0.54}{0.68} \approx 0.794 \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Conditional Probability

Conditional probability is a way to determine the likelihood of an event occurring, given that another event has already taken place. It's a fundamental concept in probability theory and helps us make sense of how events are interconnected.
For example, consider events such as buying refrigerators or accessories. If we already know a refrigerator is made in the U.S. (event A), we can then calculate the probability it has an icemaker (event B). This is given by the conditional probability \( P(B \mid A) \). Here, it represents the chance an icemaker is present, given it's a U.S. made refrigerator.
To calculate it, you'd use the formula:

• \( P(B \mid A) = \frac{P(A \cap B)}{P(A)} \)

This formula shows how often both events happen together relative to the occurrence of the first event.
In this exercise, conditional probabilities are provided for combinations of purchases, which can greatly enhance decision-making in real-world scenarios.

Tree Diagram

A tree diagram is a visual representation of all possible outcomes of a series of events. It's especially useful when you need to understand complex multi-step probabilities.
Each branch of the tree represents an event and its probability. As you move from left to right in the diagram, you encounter different layers or branches that represent different events in sequence.
In our original exercise:

• First, refrigerators are categorized as being made in the U.S. or not (event A).
• The next branches show whether they have icemakers (event B) based on whether they are made in the U.S. or not.
• The final branches depict whether an extended warranty is purchased (event C) given the other two events.

Every complete path from the start to an endpoint on the tree signifies a unique combination of events, and the probability at the end of each branch is the product of the probabilities along that path. This step-by-step visual breakdown makes complicated probability tasks easier to comprehend.

Intersection of Events

The intersection of events in probability is all about finding the probability that two or more events happen at the same time. This is known as "joint probability."
For instance, if you want to know the probability of both events A and B occurring together, you'd refer to \( P(A \cap B) \). This concept answers questions like, "What's the chance that a fridge is U.S. made and has an icemaker?"
To calculate it, you multiply the probabilities of each event happening when considering their interdependencies. In structured exercises like this:

• Follow the tree path for which both events happen together and multiply the branch probabilities.

It's crucial for solving questions that deal with simultaneous events, like determining the probability of buying a fridge that's both U.S. made and comes with additional features, ensuring you're capturing every part of your query's conditions.

Complementary Events

Complementary events encompass the idea of one event happening or not. In simpler terms, if event A occurs, its complement \( A' \) does not, and vice versa. These two outcomes cover every possibility between them.
In probability theory, the probability of an event plus the probability of its complement is always 1. That can be expressed mathematically as:

• \( P(A) + P(A') = 1 \).

This relationship helps in calculating unknown probabilities if their complements are known.
In real-life scenarios, consider an appliance store where \( A \) is the event "a refrigerator is made in the U.S." Thus, \( A' \) would be "a refrigerator is not made in the U.S." Such simple yet powerful concepts assist in understanding full scenarios where certain conditions might alternatively occur, which is significant for holistic evaluations in probability studies and exercises.