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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

Professor Stan der Deviation can take one of two routes on his way home from work. On the first route, there are four railroad crossings. The probability that he will be stopped by a train at any particular one of the crossings is .1, and trains operate independently at the four crossings. The other route is longer but there are only two crossings, independent of one another, with the same stoppage probability for each as on the first route. On a particular day, Professor Deviation has a meeting scheduled at home for a certain time. Whichever route he takes, he calculates that he will be late if he is stopped by trains at at least half the crossings encountered. a. Which route should he take to minimize the probability of being late to the meeting? b. If he tosses a fair coin to decide on a route and he is late, what is the probability that he took the four-crossing route?

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 \(P(A \cup B)\) ?

According to the article "Optimization of Distribution Parameters for Estimating Probability of Crack Detection" (J. of Aircraft, 2009: 2090-2097), the following "Palmberg" equation is commonly used to determine the probability \(P_{d}(c)\) of detecting a crack of size \(c\) in an aircraft structure: $$ P_{d}(c)=\frac{\left(c / c^{*}\right)^{\beta}}{1+\left(c / c^{*}\right)^{\beta}} $$ where \(c^{*}\) is the crack size that corresponds to a \(.5\) detection probability (and thus is an assessment of the quality of the inspection process). a. Verify that \(P_{d}\left(c^{*}\right)=.5\) b. What is \(P_{d}\left(2 c^{*}\right)\) when \(\beta=4\) ? c. Suppose an inspector inspects two different panels, one with a crack size of \(c^{*}\) and the other with a crack size of \(2 c^{*}\). Again assuming \(\beta=4\) and also that the results of the two inspections are independent of one another, what is the probability that exactly one of the two cracks will be detected? d. What happens to \(P_{d}(c)\) as \(\beta \rightarrow \infty\) ?

Consider randomly selecting a single individual and having that person test drive 3 different vehicles. Define events \(A_{1}\), \(\mathrm{A}_{2}\), and \(\mathrm{A}_{3}\) by \(A_{1}=\) likes vehicle \(\\# 1 \quad A_{2}=\) likes vehicle \(\\# 2\) \(A_{3}=\) likes vehicle \(\\# 3\) Suppose that \(P\left(A_{1}\right)=.55, P\left(A_{2}\right)=.65, P\left(A_{3}\right)=.70\), \(P\left(A_{1} \cup A_{2}\right)=.80, P\left(A_{2} \cap A_{3}\right)=.40\), and \(P\left(A_{1} \cup A_{2} \cup A_{3}\right)=.88\). a. What is the probability that the individual likes both vehicle #1 and vehicle #2? b. Determine and interpret \(P\left(A_{2} \mid A_{3}\right)\). c. Are \(\mathrm{A}_{2}\) and \(\mathrm{A}_{3}\) independent events? Answer in two different ways. d. If you learn that the individual did not like vehicle #1, what now is the probability that he/she liked at least one of the other two vehicles?

A friend of mine is giving a dinner party. His current wine supply includes 8 bottles of zinfandel, 10 of merlot, and 12 of cabernet (he only drinks red wine), all from different wineries. a. If he wants to serve 3 bottles of zinfandel and serving order is important, how many ways are there to do this? b. If 6 bottles of wine are to be randomly selected from the 30 for serving, how many ways are there to do this? c. If 6 bottles are randomly selected, how many ways are there to obtain two bottles of each variety? d. If 6 bottles are randomly selected, what is the probability that this results in two bottles of each variety being chosen? e. If 6 bottles are randomly selected, what is the probability that all of them are the same variety?
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