/*! 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 92 The general addition rule for th... [FREE SOLUTION] | 91Ó°ÊÓ

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The general addition rule for three events states that $$ \begin{aligned} P(A \text { or } B \text { or } C)=& P(A)+P(B)+P(C) \\ &-P(A \text { and } B)-P(A \text { and } C) \\ &-P(B \text { and } C)+P(A \text { and } B \text { and } C) \end{aligned} $$ A new magazine publishes columns entitled "Art" (A), "Books" (B), and "Cinema" (C). Suppose that \(14 \%\) of all subscribers read 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

convert percentages into probabilities

First, convert the given percentages into probabilities. For instance, 14% becomes \(0.14\), 23% becomes \(0.23\), and so forth. The reason being that probabilities range between 0 (representing an impossible event) and 1 (representing a certain event), and percentages need to be divided by 100 to fit within this range. Hence, \( P(A)=0.14, P(B)=0.23, P(C)=0.37, P(A \, and \, B)=0.08, P(A \, and \, C)=0.09, P(B \, and \, C)=0.13\) and \( P(A \, and \, B \, and \, C)=0.05\).
02

apply the general addition rule

The next step is to substitute these probabilities into the general addition rule for three events. From the problem, it is understood that the formula can be written as:\[ P(A \, or \, B \, or \, C) = P(A) + P(B) + P(C) - P(A \, and \, B) - P(A \, and \, C) - P(B \, and \, C) + P(A \, and \, B \, and \, C) \] Substitute the obtained probabilities into this formula to get the result.
03

substitute and solve

Substitute the probabilities from step 1 into the formula:\[ P(A \, or \, B \, or \, C) = 0.14 + 0.23 + 0.37 - 0.08 - 0.09 - 0.13 + 0.05 = 0.49 \] Hence, the probability that a randomly selected subscriber reads at least one of these three columns is 0.49.

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Key Concepts

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

General Addition Rule
The General Addition Rule is an essential concept in probability theory, especially when dealing with multiple events. This rule allows us to find the probability that at least one of several events occurs. When applied to three events A, B, and C, the formula accounts for the individual probabilities of each event, the intersections of pairs of events, and the intersection of all three events.

To recall, the General Addition Rule for three events is: \[ P(A \text{ or } B \text{ or } C) = P(A) + P(B) + P(C) - P(A \text{ and } B) - P(A \text{ and } C) - P(B \text{ and } C) + P(A \text{ and } B \text{ and } C) \] This formula might seem daunting at first, but it thoughtfully includes every possibility of overlaps correctly. These overlaps are the intersections, and each must be adjusted for in the calculation to ensure we avoid double-counting any probabilities.
  • Individual events: The probability of just A, B, or C happening alone.
  • Pairs intersections: Represents when two of the events occur together, such as A and B, B and C, or A and C.
  • Triple intersection: All three events occur together, which needs to be added back as it was subtracted out with the pairs.
Probability Calculation
Probability calculation is crucial in determining the likelihood of various outcomes. It includes converting percentages into decimals, substituting values into formulas, and understanding theoretical outcomes. As per the problem involving magazine subscribers, we calculate the probability by converting given percentages to probabilities and substituting them into the General Addition Rule.

Here's how the conversion and calculation work:
  • Convert percentages: Each percentage is divided by 100 to transform it into a probability value. For instance, 14% converts to 0.14, representing the probability of event A.
  • Substitute into the formula: Once all the probabilities are ready, they are placed into the formula to see how often at least one event occurs.
Probability provides a structured approach to quantify uncertainty. In practice, it helps us make informed decisions, forecast, and interpret results in countless scenarios.
Statistics in Media Studies
Statistics in media studies play a vital role in understanding audience behavior, preferences, and media consumption. In the context of the exercise, statistics allow a magazine publisher to understand what percentage of their subscribers engage with various sections like Art, Books, or Cinema. By leveraging probabilities and statistical calculations, publishers can tailor content to better meet the needs of their audience.

Understanding these statistics helps to make informed decisions on which sections might need more emphasis based on readership preferences. It also highlights the importance of statistical literacy for interpreting and making decisions based on data results. In media studies, statistical analysis of how people interact with content provides invaluable insights.
Applications of these statistics could mean:
  • Recognizing reader trends to predict future interests.
  • Strategizing advertising efforts to align with areas of high reader engagement.
  • Assessing the potential for new content areas by examining the probabilities of current interest overlap.
By integrating these insights, media companies can enhance their content strategies, ensuring they provide valuable and relevant material to their audience.

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Most popular questions from this chapter

A radio station that plays classical music has a "by request" program each Saturday evening. The percentages of requests for composers on a particular night are as follows: \(\begin{array}{lr}\text { Bach } & 5 \% \\ \text { Beethoven } & 26 \% \\\ \text { Brahms } & 9 \% \\ \text { Dvorak } & 2 \% \\ \text { Mendelssohn } & 3 \% \\ \text { Mozart } & 21 \% \\ \text { Schubert } & 12 \% \\ \text { Schumann } & 7 \% \\ \text { Tchaikovsky } & 14 \% \\ \text { Wagner } & 1 \%\end{array}\) Suppose that one of these requests is to be selected at random. a. What is the probability that the request is for one of the three \(\mathrm{B}^{\prime}\) s? b. What is the probability that the request is not for one of the two S's? c. Neither Bach nor Wagner wrote any symphonies. What is the probability that the request is for a composer who wrote at least one symphony?

The USA Today article referenced in Exercise \(6.37\) also gave information on seat belt usage by age, which is summarized in the following table of counts: $$ \begin{array}{lcc} & \begin{array}{c} \text { Does Not Use } \\ \text { Seat Belt } \\ \text { Regularly } \end{array} & \begin{array}{c} \text { Uses } \\ \text { Seat Belt } \\ \text { Regularly } \end{array} \\ \hline 18-24 & 59 & 41 \\ 25-34 & 73 & 27 \\ 35-44 & 74 & 26 \\ 45-54 & 70 & 30 \\ 55-64 & 70 & 30 \\ 65 \text { and older } & 82 & 18 \\ \hline \end{array} $$ Consider the following events: \(S=\) event that a randomly selected individual uses a seat belt regularly, \(A_{1}=\) event that a randomly selected individual is in age group \(18-24\), and \(A_{6}=\) event that a randomly selected individual is in age group 65 and older. a. Convert the counts to proportions and then use them to compute the following probabilities: i. \(P\left(A_{1}\right)\) ii. \(P\left(A_{1} \cap S\right)\) iii. \(P\left(A_{1} \mid S\right)\) iv. \(P\left(\right.\) not \(\left.A_{1}\right) \quad\) v. \(P\left(S \mid A_{1}\right) \quad\) vi. \(P\left(S \mid A_{6}\right)\) b. Using the probabilities \(P\left(S \mid A_{1}\right)\) and \(P\left(S \mid A_{6}\right)\) computed in Part (a), comment on how \(18-24\) -year-olds and seniors differ with respect to seat belt usage.

A construction firm bids on two different contracts. Let \(E_{1}\) be the event that the bid on the first contract is successful, and define \(E_{2}\) analogously for the second contract. Suppose that \(P\left(E_{1}\right)=.4\) and \(P\left(E_{2}\right)=.2\) and that \(E_{1}\) and \(E_{2}\) are independent events. a. Calculate the probability that both bids are successful (the probability of the event \(E_{1}\) and \(E_{2}\) ). b. Calculate the probability that neither bid is successful (the probability of the event \(\left(\right.\) not \(\left.E_{1}\right)\) and \(\left(\right.\) not \(\left.E_{2}\right)\) ). c. What is the probability that the firm is successful in at least one of the two bids?

Consider a system consisting of four components, as pictured in the following diagram: Components 1 and 2 form a series subsystem, as do Components 3 and 4 . The two subsystems are connected in parallel. Suppose that \(P(1\) works \()=.9, P(2\) works \()=.9\), \(P(3\) works \()=.9\), and \(P(4\) works \()=.9\) and that the four components work independently of one another. a. The \(1-2\) subsystem works only if both components work. What is the probability of this happening? b. What is the probability that the \(1-2\) subsystem doesn't work? that the \(3-4\) subsystem doesn't work? c. The system won't work if the \(1-2\) subsystem doesn't work and if the \(3-4\) subsystem also doesn't work. What is the probability that the system won't work? that it will work? d. How would the probability of the system working change if a \(5-6\) subsystem were added in parallel with the other two subsystems? e. How would the probability that the system works change if there were three components in series in each of the two subsystems?

USA Today (June 6,2000 ) gave information on seat belt usage by gender. The proportions in the following table are based on a survey of a large number of adult men and women in the United States: $$ \begin{array}{l|cc} \hline & \text { Male } & \text { Female } \\ \hline \text { Uses Seat Belts Regularly } & .10 & .175 \\ \begin{array}{l} \text { Does Not Use Seat Belts } \\ \text { Regularly } \end{array} & .40 & .325 \\ \hline \end{array} $$ Assume that these proportions are representative of adults in the United States and that a U.S. adult is selected at random. a. What is the probability that the selected adult regularly uses a seat belt? b. What is the probability that the selected adult regularly uses a seat belt given that the individual selected is male? c. What is the probability that the selected adult does not use a seat belt regularly given that the selected individual is female? d. What is the probability that the selected individual is female given that the selected individual does not use a seat belt regularly? e. Are the probabilities from Parts (c) and (d) equal? Write a couple of sentences explaining why this is so.

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