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Chapter 2: Problem 25

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

Suppose identical tags are placed on both the left ear and the right ear of a fox. The fox is then let loose for a period of time. Consider the two events \(C_{1}=\\{\) left ear tag is lost \(\\}\) and \(C_{2}=\\{\) right ear tag is lost \(\\}\). Let \(\pi=P\left(C_{1}\right)=P\left(C_{2}\right)\), and assume \(C_{1}\) and \(C_{2}\) are independent events. Derive an expression (involving \(\pi\) ) for the probability that exactly one tag is lost, given that at most one is lost ("Ear Tag Loss in Red Foxes," J. Wildlife Mgmt., 1976: 164-167). [Hint: Draw a tree diagram in which the two initial branches refer to whether the left ear tag was lost.]

Computer keyboard failures can be attributed to electrical defects or mechanical defects. A repair facility currently has 25 failed keyboards, 6 of which have electrical defects and 19 of which have mechanical defects. a. How many ways are there to randomly select 5 of these keyboards for a thorough inspection (without regard to order)? b. In how many ways can a sample of 5 keyboards be selected so that exactly two have an electrical defect? c. If a sample of 5 keyboards is randomly selected, what is the probability that at least 4 of these will have a mechanical defect?

Individual A has a circle of five close friends (B, C, D, E, and F). A has heard a certain rumor from outside the circle and has invited the five friends to a party to circulate the rumor. To begin, A selects one of the five at random and tells the rumor to the chosen individual. That individual then selects at random one of the four remaining individuals and repeats the rumor. Continuing, a new individual is selected from those not already having heard the rumor by the individual who has just heard it, until everyone has been told. a. What is the probability that the rumor is repeated in the order \(\mathrm{B}, \mathrm{C}, \mathrm{D}, \mathrm{E}\), and \(\mathrm{F}\) ? b. What is the probability that \(\mathrm{F}\) is the third person at the party to be told the rumor? c. What is the probability that \(\mathrm{F}\) is the last person to hear the rumor? d. If at each stage the person who currently "has" the rumor does not know who has already heard it and selects the next recipient at random from all five possible individuals, what is the probability that \(\mathrm{F}\) has still not heard the rumor after it has been told ten times at the party?

Consider randomly selecting a student at a certain university, and let \(A\) denote the event that the selected individual has a Visa credit card and \(B\) be the analogous event for a MasterCard. Suppose that \(P(A)=.5, P(B)=.4\), and \(P(A \cap B)=.25\). a. Compute the probability that the selected individual has at least one of the two types of cards (i.e., the probability of the event \(A \cup B\) ). b. What is the probability that the selected individual has neither type of card? c. Describe, in terms of \(A\) and \(B\), the event that the selected student has a Visa card but not a MasterCard, and then calculate the probability of this event.

Show that if one event \(A\) is contained in another event \(B\) (i.e., \(A\) is a subset of \(B\) ), then \(P(A) \leq P(B)\). [Hint: For such \(A\) and \(B, A\) and \(B \cap A^{\prime}\) are disjoint and \(B=A \cup\left(B \cap A^{\prime}\right)\), as can be seen from a Venn diagram.] For general \(A\) and \(B\), what does this imply about the relationship among \(P(A \cap B), P(A)\) and \(P(A \cup B)\) ?
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