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Chapter 6: Problem 24

An individual is presented with three different glasses of cola, labeled C, D, and P. He is asked to taste all three and then list them in order of preference. Suppose that the same cola has actually been put into all three glasses. a. What are the simple events in this chance experiment, and what probability would you assign to each one? b. What is the probability that \(\mathrm{C}\) is ranked first? c. What is the probability that \(\mathrm{C}\) is ranked first \(a n d \mathrm{D}\) is ranked last?

Short Answer

Expert verified
a. The simple events and their probabilities are: (CDP,1/6), (CPD,1/6), (DCP,1/6), (DPC,1/6), (PCD,1/6), (PDC,1/6). \nb. The probability that C is ranked first is 1/3. \nc. The probability that C is ranked first and D is ranked last is 1/6.

Step by step solution

01

Identify the Simple Events

The simple events in this experiment are the different possible orderings of the glasses C, D, and P. These are all possible permutations of three distinct items, which amounts to \(3!=3*2*1=6\) different outcomes. The possible outcomes are: CDP, CPD, DC, DPC, PCD, PD. Since he doesn't know that all the glasses are filled with the same Cola, he might randomly select any of these combinations.
02

Assign Probabilities

There are 6 equally likely possibilities or outcomes that the individual can choose. Therefore, the probability for each outcome is \(1/6\). Because all orderings are equally likely, it is reasonable to assign each of these 6 simple events the same probability of \(1/6\).
03

Calculate the Probability of C being Ranked First

Looking at the 6 possible outcomes, 2 out of 6 have C ranked first: CDP and CPD. Therefore, the probability of Cola C being ranked first is \(2/6=1/3\).
04

Calculate the Probability of Both C being Ranked First and D being Ranked Last

Looking at the outcomes, only one outcome has C ranked first and D ranked last: CPD. Therefore, the probability of C being first and D being last is \(1/6\).

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Key Concepts

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

Chance Experiments
A chance experiment is any process that generates a well-defined outcome that cannot be predicted with certainty before the experiment is carried out. In the problem at hand, the experiment involves tasting three identical glasses of cola labeled C, D, and P, and ranking them in order of preference. The outcome in this situation is the order of preference.

Each possible outcome of a chance experiment is known as a simple event. The simple events here are the different possible orderings the individual could list the glasses, given the tastes are indistinguishable. It is crucial to ensure that these events are well-defined and mutually exclusive for correct probability assignment.
Probability Assignment
In probability assignment, each simple event from a chance experiment is allocated a probability value, which is a measure of the likelihood of that event occurring. These values should always sum up to 1, because the certainty of some event happening is absolute. In the cola tasting experiment, the probability assignment is straightforward since there is no bias in ranking preferences and the individual is equally likely to choose any order.

This assignment assumes all outcomes are equally probable, known as a uniform probability distribution. Each of the 6 possible permutations (listed in step 1 of the solution) are assigned a probability of \(1/6\), reflecting an equal chance of any sequence being the chosen one.
Permutations
A permutation is an arrangement of a set of items in a specific order. In the provided problem, we deal with arranging three items (C, D, and P). The number of permutations of 'n' distinct items is given by the factorial of 'n', represented as 'n!'.

For example, with three items (C, D, and P), there are 3! (which is 3*2*1) permutations, giving us a total of 6 possible different orderings. It's these permutations that represent the simple events in our chance experiment. Understanding the concept of permutations is essential for correctly identifying potential outcomes in similar probability problems.
Outcome Probabilities
Determining outcome probabilities is the essence of solving any probability problem. It involves figuring out the likelihood of each individual outcome within a set of possible outcomes. Once we have our list of simple events from a chance experiment, like the different orderings of cola glasses C, D, and P, we must calculate the probability of each event.

To improve the comprehensibility of outcome probabilities, consider visual aids like tree diagrams or lists. In our problem, the specific outcome probabilities were calculated in two parts: the probability of C being ranked first (\(1/3\)), and the probability of C being first and D being last (\(1/6\)). These calculations depend on counting the favorable outcomes against the total number of possible outcomes.

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Most popular questions from this chapter

Suppose that, starting at a certain time, batteries coming off an assembly line are examined one by one to see whether they are defective (let \(\mathrm{D}=\) defective and \(\mathrm{N}=\) not defective). The chance experiment terminates as soon as a nondefective battery is obtained. a. Give five possible experimental outcomes. b. What can be said about the number of outcomes in the sample space? c. What outcomes are in the event \(E\), that the number of batteries examined is an even number?

Consider the chance experiment in which both tennis racket head size and grip size are noted for a randomly selected customer at a particular store. The six possible outcomes (simple events) and their probabilities are displayed in the following table: a. The probability that grip size is \(4 \frac{1}{2}\) in. (event \(\mathrm{A}\) ) is $$ P(A)=P\left(O_{2} \text { or } O_{5}\right)=.20+.15=.35 $$ How would you interpret this probability? b. Use the result of Part (a) to calculate the probability that grip size is not \(4 \frac{1}{2}\) in. c. What is the probability that the racket purchased has an oversize head (event \(B\) ), and how would you interpret this probability? d. What is the probability that grip size is at least \(4 \frac{1}{2}\) in.?

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.

Refer to Exercise 6.18. Adding probabilities in the first row of the given table yields \(P(\) midsize \()=.45\), whereas from the first column, \(\mathrm{P}\left(4 \frac{3}{8}\right.\) in. grip) \(=.30\). Is the following true? $$ P\left(\text { midsize } \text { or } 4 \frac{3}{8} \text { in. grip }\right)=.45+.30=.75 $$ Explain.

Consider the following information about travelers on vacation: \(40 \%\) check work email, \(30 \%\) use a cell phone to stay connected to work, \(25 \%\) bring a laptop with them on vacation, \(23 \%\) both check work email and use a cell phone to stay connected, and \(51 \%\) neither check work email nor use a cell phone to stay connected nor bring a laptop. In addition \(88 \%\) of those who bring a laptop also check work email and \(70 \%\) of those who use a cell phone to stay connected also bring a laptop. With \(E=\) event that a traveler on vacation checks work email, \(C=\) event that a traveler on vacation uses a cell phone to stay connected, and \(L=\) event that a traveler on vacation brought a laptop, use the given information to determine the following probabilities. A Venn diagram may help. a. \(P(E)\) b. \(P(C)\) c. \(P(L)\) d. \(P(E\) and \(C)\) e. \(P\left(E^{C}\right.\) and \(C^{C}\) and \(L^{C}\) ) f. \(P(\) Eor C or \(L\) ) g. \(P(E \mid L)\) j. \(P(E\) and \(L)\) h. \(P(L \mid C)\) k. \(P(C\) and \(L)\) i. \(P(E\) and \(C\) and \(L)\) 1\. \(P(C \mid E\) and \(L)\)
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