/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 98 The general addition rule for th... [FREE SOLUTION] | 91Ó°ÊÓ

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The general addition rule for three events states that $$ \begin{aligned} &P(A \text { or } B \text { or } C)=P(A)+P(B)+P(C) \\ &\quad-P(A \text { and } B)-P(A \text { and } C) \\ &\quad-P(B \text { and } C)+P(A \text { and } B \text { and } C) \end{aligned} $$ A new magazine publishes columns entitled "Art" (A), "Books" (B), and "Cinema" (C). Suppose that \(14 \%\) of all subscribers read \(\mathrm{A}, 23 \%\) read \(\mathrm{B}, 37 \%\) read \(\mathrm{C}, 8 \%\) read \(\mathrm{A}\) and \(\mathrm{B}, 9 \%\) read \(\mathrm{A}\) and \(\mathrm{C}, 13 \%\) read \(\mathrm{B}\) and \(\mathrm{C}\), and \(5 \%\) read all three columns. What is the probability that a randomly selected subscriber reads at least one of these three columns?

Short Answer

Expert verified
The probability that a randomly selected subscriber reads at least one of the columns is \(0.49\)

Step by step solution

01

- Identify the given probabilities

The problem provides several probabilities that we need to plug into the general addition rule for three events. These probabilities are as follows: \(P(A) = 0.14, P(B) = 0.23, P(C) = 0.37, P(A \text { and } B) = 0.08, P(A \text { and } C) = 0.09, P(B \text { and } C) = 0.13, P(A \text { and } B \text { and } C) = 0.05.\)
02

- Apply the general addition rule

Express the probabilities of the three events A, B, and C (reading Art, Books, or Cinema) using the general addition rule. That is: \[ P(A \text { or } B \text { or } C) = P(A) + P(B) + P(C) - P(A \text { and } B) - P(A \text { and } C) - P(B \text { and } C) + P(A \text { and } B \text { and } C)\]
03

- Compute the desired probability

Next, substitute the given probabilities into the formular: \[ P(A \text { or } B \text { or } C) = 0.14 + 0.23 + 0.37 - 0.08 - 0.09 - 0.13 + 0.05\]. Complete the computation to find the value of \(P(A \text { or } B \text { or } C)\).

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Key Concepts

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

Probability Theory
Probability theory is an essential branch of mathematics that deals with the likelihood of different outcomes or events. It allows us to analyze and make predictions about random occurrences based on known probabilities. These probabilities are numerical values between 0 and 1, where 0 indicates impossibility and 1 signifies certainty.

In this context, probability is often used to understand complex systems by considering individual possibilities and their interactions. The foundational principles of probability include events, sample spaces, and the basic operations on these sets, such as union, intersection, and complement. The goal is to assess how likely different combinations of these events are to occur by using probability rules and theorems.

Understanding probability theory provides powerful tools in fields like statistics, finance, science, and engineering, where decision-making and risk assessment are crucial. By using various probability rules, such as the general addition rule, we can find the likelihood of specific outcomes when dealing with multiple events.
Three Events
When dealing with probability, analyzing three events simultaneously can become complex due to the various combinations that need to be considered. In our scenario, these three events are the readers' engagement with different magazine columns: Art (A), Books (B), and Cinema (C).

The complexity arises because we must evaluate not only the probability of each event occurring individually but also the likelihood of two or more events overlapping. This includes:
  • Readers who are interested in both Art and Books (A and B).
  • Readers who enjoy both Art and Cinema (A and C).
  • Readers who like both Books and Cinema (B and C).
  • And finally, those who appreciate all three sections (A, B, and C).

By assessing each of these possibilities, we can apply comprehensive rules, such as the general addition rule, to determine the overall probability of a subscriber engaging with at least one column. These calculations help us understand patterns in reader preferences and tailor content accordingly.
Inclusion-Exclusion Principle
The inclusion-exclusion principle is a fundamental concept in combinatorics and probability theory that helps us find the probability of the union of several events. It corrects the over-counting that occurs when simply summing up individual probabilities of multiple events.

When analyzing the probability of at least one of the three events (A, B, or C) occurring, adding up probabilities directly might lead to an inaccurate result due to shared probabilities in overlapping areas. The inclusion-exclusion principle systematically subtracts and adds intersecting terms to find the precise probability:
  • Add each event separately: \( P(A) + P(B) + P(C) \)
  • Subtract probabilities for all pairwise intersections: \(- P(A \text { and } B) - P(A \text { and } C) - P(B \text { and } C) \)
  • Add back the probability of the triple intersection (if applicable): \(+ P(A \text { and } B \text { and } C) \)

This principle ensures no double-counting while including all necessary components to find the true probability. It beautifully simplifies complex event interactions into a comprehensible calculation framework.

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Most popular questions from this chapter

Consider a Venn diagram picturing two events \(A\) and \(B\) that are not disjoint. a. Shade the event \((A \cup B)^{C} .\) On a separate Venn diagram shade the event \(A^{C} \cap B^{C} .\) How are these two events related? b. Shade the event \((A \cap B)^{C} .\) On a separate Venn diagram shade the event \(A^{C} \cup B^{C} .\) How are these two events related? (Note: These two relationships together are called DeMorgan's laws.)

Two different airlines have a flight from Los Angeles to New York that departs each weekday morning at a certain time. Let \(E\) denote the event that the first airline's flight is fully booked on a particular day, and let \(F\) denote the event that the second airline's flight is fully booked on that same day. Suppose that \(P(E)=.7\), \(P(F)=.6\), and \(P(E \cap F)=.54 .\) a. Calculate \(P(E \mid F)\), the probability that the first airline's flight is fully booked given that the second airline's flight is fully booked. b. Calculate \(P(F \mid E)\).

A college library has four copies of a certain book; the copies are numbered \(1,2,3\), and 4 . Two of these are selected at random. The first selected book is placed on 2-hour reserve, and the second book can be checked out overnight. a. Construct a tree diagram to display the 12 outcomes in the sample space. b. Let \(A\) denote the event that at least one of the books selected is an even-numbered copy. What outcomes are in \(A\) ? c. Suppose that copies 1 and 2 are first printings, whereas copies 3 and 4 are second printings. Let \(B\) denote the event that exactly one of the copies selected is a first printing. What outcomes are contained in \(B\) ?

The Associated Press (San Luis Obispo Telegram-Tribune, August 23,1995 ) reported on the results of mass screening of schoolchildren for tuberculosis (TB). For Santa Clara County, California, the proportion of all tested kindergartners who were found to have TB was .0006. The corresponding proportion for recent immigrants (thought to be a high-risk group) was. 0075 . Suppose that a Santa Clara County kindergartner is selected at random. Are the events selected student is a recent immigrant and selected student has \(T B\) independent or dependent events? Justify your answer using the given information.

In a school machine shop, \(60 \%\) of all machine breakdowns occur on lathes and \(15 \%\) occur on drill presses. Let \(E\) denote the event that the next machine breakdown is on a lathe, and let \(F\) denote the event that a drill press is the next machine to break down. With \(P(E)=.60\) and \(P(F)=.15\), calculate: a. \(P\left(E^{C}\right)\) b. \(P(E \cup F)\) c. \(P\left(E^{C} \cap F^{C}\right)\)

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