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The three most popular options on a certain type of new car are a built-in GPS \((A)\), a sunroof \((B)\), and an automatic transmission \((C)\). If \(40 \%\) of all purchasers request \(A, 55 \%\) request \(B, 70 \%\) request \(C, 63 \%\) request \(A\) or \(B, 77 \%\) request \(A\) or \(C, 80 \%\) request \(B\) or \(C\), and \(85 \%\) request \(A\) or \(B\) or \(C\), determine the probabilities of the following events. [Hint: " \(A\) or \(B\) " is the event that at least one of the two options is requested; try drawing a Venn diagram and labeling all regions.] a. The next purchaser will request at least one of the three options. b. The next purchaser will select none of the three options. c. The next purchaser will request only an automatic transmission and not either of the other two options. d. The next purchaser will select exactly one of these three options.

Short Answer

Expert verified
a) 0.85, b) 0.15, c) Use inclusion-exclusion, d) Use probabilities of only one option.

Step by step solution

01

Understanding the Problem Requirements

There are three car options: GPS (A), sunroof (B), and transmission (C). We need to find the probabilities for various combinations of these options among purchases. This involves calculating probabilities using the principle of inclusion-exclusion.
02

Define Given Probabilities and Venn Diagram

Let's denote the probabilities as follows: \( P(A) = 0.40 \), \( P(B) = 0.55 \), \( P(C) = 0.70 \), \( P(A \cup B) = 0.63 \), \( P(A \cup C) = 0.77 \), \( P(B \cup C) = 0.80 \), and \( P(A \cup B \cup C) = 0.85 \). A Venn diagram with sets A, B, and C can help visualize these relationships.
03

Apply Inclusion-Exclusion Principle

Using the inclusion-exclusion principle, we have: \[ P(A \cup B \cup C) = P(A) + P(B) + P(C) - P(A \cap B) - P(A \cap C) - P(B \cap C) + P(A \cap B \cap C) = 0.85 \]. Use this expression to determine missing probabilities.
04

Solve for Shared Probabilities

From \( P(A \cup B) = P(A) + P(B) - P(A \cap B) = 0.63 \), solve for \( P(A \cap B) \). Similarly, find \( P(A \cap C) \) and \( P(B \cap C) \). Calculate \( P(A \cap B \cap C) \) using the inclusion-exclusion principle.
05

Probability of At Least One Option

Since \( P(A \cup B \cup C) = 0.85 \), Part a states that the probability the purchaser will request at least one option is \( 0.85 \).
06

Probability of No Options

The probability that no option is selected, \( P(A' \cap B' \cap C') \), is the complement of \( P(A \cup B \cup C) \): \( 1 - 0.85 = 0.15 \).
07

Probability of Only Transmission

To find \( P(C \cap A' \cap B') \), solve using the inclusion-exclusion principle \( P(C) - P(C \cap A) - P(C \cap B) + P(A \cap B \cap C) = 0.70 \). Derive \( P(C \cap A' \cap B') \).
08

Probability of Exactly One Option

Calculate the probability of selecting exactly one option as \( P(A \cap B' \cap C') + P(A' \cap B \cap C') + P(A' \cap B' \cap C) \). Determine each of these values by excluding intersections among options.

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Key Concepts

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

Venn Diagram
Venn diagrams are a powerful tool in probability and set operations. They visually represent different sets and the relationships between them.
Each circle in a Venn diagram represents a set, such as car options like GPS (A), sunroof (B), and transmission (C) in our example.
The areas where circles overlap indicate intersections, which are crucial when calculating combined probabilities.
  • Single Sets: A single circle represents one set. For example, the circle for set A shows the probability of selecting the option for a GPS.
  • Intersections: The overlapping regions between two circles, such as between A and B, show where both options are chosen. This area is known as an intersection, denoted by \( A \cap B \).
  • Union: The total area covered by all the circles represents the union, which includes any purchaser choosing at least one of the options. This is represented as \( A \cup B \cup C \).
By labeling each section of a Venn diagram accurately, one can easily visualize and calculate probabilities using the inclusion-exclusion principle.
Probability Theory
In probability theory, understanding how different events relate helps us calculate the likelihood of various outcomes.
The probability of each individual car option provides crucial data for further calculations.
  • Basic Probability: The probability of a single event is given directly, such as \( P(A) = 0.40 \) for requesting a GPS option.
  • Combined Events: For combined events, like a purchaser choosing two or more options, we use set operations to understand their relationships.
  • Complement Rules: The probability that none of the options are selected is the complement of at least one option being chosen. For example, since the probability of choosing at least one is \( P(A \cup B \cup C) = 0.85 \), the probability of choosing none is simply \( 1 - 0.85 = 0.15 \).
Calculating these probabilities often involves solving equations with unknowns, especially when using principles like inclusion-exclusion.
Set Operations
Set operations form the backbone of solving problems involving multiple events.
In our exercise, they help us organize and calculate the different probabilities.
  • Union: The union operation \( A \cup B \cup C \) represents the probability that at least one of the options is selected, which is calculated using their individual probabilities and their intersections.
  • Intersection: An intersection \( A \cap B \) signifies that both A and B are selected. Using given probabilities such as \( P(A \cup B) \) helps find these values.
  • Exclusion: Calculating events like "only this option" involves excluding other intersections. For example, \( P(C \cap A' \cap B') \) requires understanding what part of set C does not overlap with A or B.
By manipulating these operations, especially with the help of the inclusion-exclusion principle, we find probability values for complex scenarios such as exactly one option being selected.

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Most popular questions from this chapter

In five-card poker, a straight consists of five cards with adjacent denominations (e.g., 9 of clubs, 10 of hearts, jack of hearts, queen of spades, and king of clubs). Assuming that aces can be high or low, if you are dealt a five-card hand, what is the probability that it will be a straight with high card 10? What is the probability that it will be a straight? What is the probability that it will be a straight flush (all cards in the same suit)?

A particular airline has 10 A.M. flights from Chicago to New York, Atlanta, and Los Angeles. Let \(A\) denote the event that the New York flight is full and define events \(B\) and \(C\) analogously for the other two flights. Suppose \(P(A)=.6, P(B)=.5, P(C)=.4\) and the three events are independent. What is the probability that a. All three flights are full? That at least one flight is not full? b. Only the New York flight is full? That exactly one of the three flights is full?

Show that for any three events \(A, B\), and \(C\) with \(P(C)>0\), \(P(A \cup B \mid C)=P(A \mid C)+P(B \mid C)-P(A \cap B \mid C) .\)

The computers of six faculty members in a certain department are to be replaced. Two of the faculty members have selected laptop machines and the other four have chosen desktop machines. Suppose that only two of the setups can be done on a particular day, and the two computers to be set up are randomly selected from the six (implying 15 equally likely outcomes; if the computers are numbered \(1,2, \ldots, 6\), then one outcome consists of computers 1 and 2 , another consists of computers 1 and 3 , and so on). a. What is the probability that both selected setups are for laptop computers? b. What is the probability that both selected setups are desktop machines? c. What is the probability that at least one selected setup is for a desktop computer? d. What is the probability that at least one computer of each type is chosen for setup?

A box contains the following four slips of paper, each having exactly the same dimensions: (1) win prize \(1 ;(2)\) win prize \(2 ;(3)\) win prize \(3 ;(4)\) win prizes 1,2 , and 3 . One slip will be randomly selected. Let \(A_{1}=\\{\) win prize 1\(\\}\), \(A_{2}=\\{\) win prize 2\(\\}\), and \(A_{3}=\\{\) win prize 3\(\\}\). Show that \(A_{1}\) and \(A_{2}\) are independent, that \(A_{1}\) and \(A_{3}\) are independent, and that \(A_{2}\) and \(A_{3}\) are also independent (this is pairwise independence). However, show that \(P\left(A_{1} \cap A_{2} \cap A_{3}\right) \neq\) \(P\left(A_{1}\right) \cdot P\left(A_{2}\right) \cdot P\left(A_{3}\right)\), so the three events are not mutually independent.

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