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Chapter 6: Problem 92

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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Most popular questions from this chapter

A Gallup survey of 2002 adults found that \(46 \%\) of women and \(37 \%\) of men experience pain daily (San Luis Obispo Tribune, April 6, 2000). Suppose that this information is representative of U.S. adults. If a U.S. adult is selected at random, are the events selected adult is male and selected adult experiences pain daily independent or dependent? Explain.

Insurance status - covered (C) or not covered (N) \- is determined for each individual arriving for treatment at a hospital's emergency room. Consider the chance experiment in which this determination is made for two randomly selected patients. The simple events are \(O_{1}=(\mathrm{C}, \mathrm{C})\) \(O_{2}=(\mathrm{C}, \mathrm{N}), O_{3}=(\mathrm{N}, \mathrm{C})\), and \(O_{4}=(\mathrm{N}, \mathrm{N}) .\) Suppose that probabilities are \(P\left(O_{1}\right)=.81, P\left(O_{2}\right)=.09, P\left(O_{3}\right)=.09\), and \(P\left(O_{4}\right)=.01\). a. What outcomes are contained in \(A\), the event that at most one patient is covered, and what is \(P(A)\) ? b. What outcomes are contained in \(B\), the event that the two patients have the same status with respect to coverage, and what is \(P(B)\) ?

The article "Doctors Misdiagnose More Women, Blacks" (San Luis Obispo Tribune, April 20, 2000) gave the following information, which is based on a large study of more than 10,000 patients treated in emergency rooms in the eastern and midwestern United States: 1\. Doctors misdiagnosed heart attacks in \(2.1 \%\) of all patients. 2\. Doctors misdiagnosed heart attacks in \(4.3 \%\) of black patients. 3\. Doctors misdiagnosed heart attacks in \(7 \%\) of women under 55 years old. Use the following event definitions: \(M=\) event that a heart attack is misdiagnosed, \(B=\) event that a patient is black, and \(W=\) event that a patient is a woman under 55 years old. Translate each of the three statements into probability notation.

The Los Angeles Times (June 14,1995 ) reported that the U.S. Postal Service is getting speedier, with higher overnight on-time delivery rates than in the past. The Price Waterhouse accounting firm conducted an independent audit by seeding the mail with letters and recording ontime delivery rates for these letters. Suppose that the results were as follows (these numhers are fictitions hut are compatible with summary values given in the article): $$ \begin{array}{lcc} & \begin{array}{l} \text { Number } \\ \text { of Letters } \\ \text { Mailed } \end{array} & \begin{array}{l} \text { Number of } \\ \text { Lefters Arriving } \\ \text { on Time } \end{array} \\ \hline \text { Los Angeles } & 500 & 425 \\ \text { New York } & 500 & 415 \\ \text { Washington, D.C. } & 500 & 405 \\ \text { Nationwide } & 6000 & 5220 \\ & & \\ \hline \end{array} $$ Use the given information to estimate the following probabilities: a. The probability of an on-time delivery in Los Angeles b. The probability of late delivery in Washington, D.C. c. The probability that two letters mailed in New York are both delivered on time d. The probability of on-time delivery nationwide

Only \(0.1 \%\) of the individuals in a certain population have a particular disease (an incidence rate of .001). Of those who have the disease, \(95 \%\) test positive when a certain diagnostic test is applied. Of those who do not have the disease, \(90 \%\) test negative when the test is applied. Suppose that an individual from this population is randomly selected and given the test. a. Construct a tree diagram having two first-generation branches, for has disease and doesn't have disease, and two second-generation branches leading out from each of these, for positive test and negative test. Then enter appropriate probabilities on the four branches. b. Use the general multiplication rule to calculate \(P(\) has disease and positive test). c. Calculate \(P\) (positive test). d. Calculate \(P\) (has disease \(\mid\) positive test). Does the result surprise you? Give an intuitive explanation for the size of this probability.
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