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91Ó°ÊÓ

The three major options on a car model are an automatic transmission \((A)\), a sunroof \((B)\), and an upgraded stereo \((C)\). If \(70 \%\) of all purchasers request \(A, 80 \%\) request \(B, 75 \%\) request \(C, 85 \%\) request \(A\) or \(B, 90 \%\) request \(A\) or \(C, 95 \%\) request \(B\) or \(C\), and \(98 \%\) request \(A\) or \(B\) or \(C\), compute the probabilities of the following events. [Hint: "A or \(B^{\prime \prime}\) 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 neither of the other two options. d. The next purchaser will select exactly one of these three options.

Short Answer

Expert verified
a. 0.98, b. 0.02, c. 0.03, d. 0.09.

Step by step solution

01

Define the Probabilities

Let's define the probabilities for each event given: \(P(A) = 0.70\), \(P(B) = 0.80\), \(P(C) = 0.75\). We are also given the following combined probabilities: \(P(A \cup B) = 0.85\), \(P(A \cup C) = 0.90\), \(P(B \cup C) = 0.95\), and \(P(A \cup B \cup C) = 0.98\).
02

Calculate Probability for Part a

For part (a), we need the probability that at least one of the options is requested, which is given as \(P(A \cup B \cup C) = 0.98\). Thus, the probability that at least one option is requested is \(0.98\).
03

Determine Probability for Part b

For part (b), we need the probability that none of the options are requested, which is the complement of at least one being requested: \(1 - P(A \cup B \cup C) = 1 - 0.98 = 0.02\).
04

Use Inclusion-Exclusion Principle

To proceed with parts (c) and (d), apply the inclusion-exclusion principle: \[P(A \cap B \cap C) = P(A) + P(B) + P(C) - P(A \cup B) - P(A \cup C) - P(B \cup C) + P(A \cup B \cup C)\].
05

Solve for \(P(A \cap B \cap C)\)

Plug the values into the formula: \[P(A \cap B \cap C) = 0.70 + 0.80 + 0.75 - 0.85 - 0.90 - 0.95 + 0.98 = 0.53\].
06

Probability Only Auto Transmission (Part c)

Calculate \(P(A \cap B^c \cap C^c)\) using: \[P(A) - P(A \cap B) - P(A \cap C) + P(A \cap B \cap C) \]. With symmetry and known values, \(P(A \cap B) = P(A) + P(B) - P(A \cup B) = 0.65\), similarly \(P(A \cap C) = 0.55\). Thus, \(P(A \cap B^c \cap C^c) = 0.70 - 0.65 - 0.55 + 0.53 = 0.03\).
07

Probability Exactly One Option (Part d)

Compute: \[P( ext{exactly one}) = (P(A \cap B^c \cap C^c) + P(B \cap A^c \cap C^c) + P(C \cap A^c \cap B^c))\]. Set up equations using symmetry and solve, finding \(P(B \cap A^c \cap C^c)\) and \(P(C \cap A^c \cap B^c)\), resulting similarly in \(0.03\) each.Add: \(0.03 + 0.03 + 0.03 = 0.09\).

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Key Concepts

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

Inclusion-Exclusion Principle
The Inclusion-Exclusion Principle is a fundamental concept in probability theory, used to find the probability of the union of multiple events. It helps in avoiding overcounting when more than one event can overlap.

In mathematical terms, to calculate the probability of the union of three events, such as in this exercise with automatic transmission \(A\), sunroof \(B\), and upgraded stereo \(C\), the formula is:\[P(A \cup B \cup C) = P(A) + P(B) + P(C) - P(A \cap B) - P(A \cap C) - P(B \cap C) + P(A \cap B \cap C)\]

This formula breaks down the problem into individual probabilities and then subtracts the probabilities of pairs of events to remove overcounting. Finally, it adds back the probability of all three events happening at the same time, which was subtracted thrice in total among the pairs. Applying this principle helps to systematically handle complex probability problems.
Venn Diagram
A Venn Diagram is a visual tool used in probability to show the relationships between different sets of data. They are particularly useful for illustrating the Inclusion-Exclusion Principle.

In the context of this problem, imagine three overlapping circles, each representing one of the car options: automatic transmission, sunroof, and upgraded stereo. Where these circles overlap, different combinations of chosen options are represented.

Using a Venn Diagram allows you to visually break down who chose multiple options and where those overlaps occur. It's a helpful technique for visual learners and those tackling questions involving unions and intersections of sets.

Creating a Venn Diagram for our problem can make it easier to see where the calculations for \(P(A \cup B)\), \(P(A \cap B \cap C)\), and other combined probabilities stem from.
Complement Rule
The Complement Rule is a convenient method to find the probability of an event not happening by subtracting the probability of the event happening from 1. This rule is very useful when calculating the probability of 'none' or 'at least one'.

For example, in this exercise, if we want to find the probability that none of the car options are selected (part b of the exercise), we use the complement rule. Since we know that the probability of at least one option being selected is \(P(A \cup B \cup C) = 0.98\), the probability that none are selected is:\[1 - P(A \cup B \cup C) = 1 - 0.98 = 0.02\]

This rule provides a straightforward way to compute probabilities associated with the "not happening" scenarios by reversing our perspective on the problem.
Basic Probability Rules
Basic probability rules form the foundation of probability theory. Understanding these rules aids in solving even complex problems involving multiple events and their combinations.

These rules include:
  • **Addition Rule**: Used to calculate the probability of the union of two events. For two events A and B, it is given by \(P(A \cup B) = P(A) + P(B) - P(A \cap B)\).
  • **Multiplication Rule**: Used to calculate the probability of the intersection of independent events. For independent events A and B, \(P(A \cap B) = P(A) \times P(B)\).
  • **Complement Rule**: As discussed, allows us to find the probability of an event not occurring by using \(1 - P(Event)\).


In the exercise, these probabilities are key to solving parts like finding exactly one option, which requires understanding intersections (using multiplication for dependent events like \(P(A \cap B^c \cap C^c)\)) and additions for combinations.

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