/*! 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

A library has five copies of a certain textbook on reserve of which two copies ( 1 and 2 ) are first printings and the other three \((3,4\), and 5\()\) are second printings. \(\mathrm{A}\) student examines these books in random order, stopping only when a second printing has been selected. a. Display the possible outcomes in a tree diagram. b. What outcomes are contained in the event \(A\), that exactly one book is examined before the chance experiment terminates? c. What outcomes are contained in the event \(C\), that the chance experiment terminates with the examination of book 5 ?

The National Public Radio show Car Talk has a feature called "The Puzzler." Listeners are asked to send in answers to some puzzling questions-usually about cars but sometimes about probability (which, of course, must account for the incredible popularity of the program!). Suppose that for a car question, 800 answers are submitted, of which 50 are correct. a. Suppose that the hosts randomly select two answers from those submitted with replacement. Calculate the probability that both selected answers are correct. (For purposes of this problem, keep at least five digits to the right of the decimal.) b. Suppose now that the hosts select the answers at random but without replacement. Use conditional probability to evaluate the probability that both answers selected are correct. How does this probability compare to the one computed in Part (a)?

A shipment of 5000 printed circuit boards contains 40 that are defective. Two boards will be chosen at random, without replacement. Consider the two events \(E_{1}=\) event that the first board selected is defective and \(E_{2}=\) event that the second board selected is defective. a. Are \(E_{1}\) and \(E_{2}\) dependent events? Explain in words. b. Let \(n o t E_{1}\) be the event that the first board selected is not defective (the event \(E_{1}^{C}\) ). What is \(P\left(\right.\) not \(E_{1}\) )? c. How do the two probabilities \(P\left(E_{2} \mid E_{1}\right)\) and \(P\left(E_{2} \mid\right.\) not \(\left.E_{1}\right)\) compare? d. Based on your answer to Part (c), would it be reasonable to view \(E_{1}\) and \(E_{2}\) as approximately independent?

Two individuals, \(A\) and \(B\), are finalists for a chess championship. They will play a sequence of games, each of which can result in a win for \(A\), a win for \(\mathrm{B}\), or a draw. Suppose that the outcomes of successive games are independent, with \(P(\) A wins game \()=.3\), \(P(\) B wins game \()=.2\), and \(P(\) draw \()=.5\). Each time a player wins a game, he earns 1 point and his opponent earns no points. The first player to win 5 points wins the championship. For the sake of simplicity, assume that the championship will end in a draw if both players obtain 5 points at the same time. a. What is the probability that \(A\) wins the championship in just five games? b. What is the probability that it takes just five games to obtain a champion? c. If a draw earns a half-point for each player, describe how you would perform a simulation to estimate \(\mathrm{P}(\mathrm{A}\) wins the championship). d. If neither player earns any points from a draw, would the simulation in Part (c) take longer to perform? Explain your reasoning.

The article "Chances Are You Know Someone with a Tattoo, and He's Not a Sailor" (Associated Press, June 11, 2006) included results from a survey of adults aged 18 to 50 . The accompanying data are consistent with summary values given in the article. $$ \begin{array}{l|cc} & \text { At Least One Tattoo } & \text { No Tattoo } \\ \hline \text { Age 18-29 } & 18 & 32 \\ \text { Age 30-50 } & 6 & 44 \\ \hline \end{array} $$ Assuming these data are representative of adult Americans and that an adult American is selected at random, use the given information to estimate the following probabilities. a. \(P(\) tattoo \()\) b. \(P(\) tattoo \(\mid\) age \(18-29)\) c. \(P(\) tattoo \(\mid\) age \(30-50\) ) d. \(P(\) age \(18-29 \mid\) tattoo \()\)

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