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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

The newspaper article "Folic Acid Might Reduce Risk of Down Syndrome" (USA Today, September 29 , 1999) makes the following statement: "Older women are at a greater risk of giving birth to a baby with Down Syndrome than are younger women. But younger women are more fertile, so most children with Down Syndrome are born to mothers under \(30 .\) " Let \(D=\) event that a randomly selected baby is born with Down Syndrome and \(Y=\) event that a randomly selected baby is born to a young mother (under age 30 ). For each of the following probability statements, indicate whether the statement is consistent with the quote from the article, and if not, explain why not. a. \(P(D \mid Y)=.001, \quad P\left(D \mid Y^{C}\right)=.004, \quad P(Y)=.7\) b. \(P(D \mid Y)=.001, \quad P\left(D \mid Y^{C}\right)=.001, \quad P(Y)=.7\) c. \(P(D \mid Y)=.004, \quad P\left(D \mid Y^{C}\right)=.004, \quad P(Y)=.7\) d. \(P(D \mid Y)=.001, \quad P\left(D \mid Y^{C}\right)=.004, \quad P(Y)=.4\) e. \(P(D \mid Y)=.001, \quad P\left(D \mid Y^{C}\right)=.001, \quad P(Y)=.4\) f. \(P(D \mid Y)=.004, \quad P\left(D \mid Y^{C}\right)=.004, \quad P(Y)=.4\)

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 not \(E_{1}\) be the event that the first board selected is not defective (the event \(E_{1}^{C}\) ). What is \(P\left(\right.\) not \(\left.E_{1}\right)\) ? 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?

Suppose that a six-sided die is "loaded" so that any particular even-numbered face is twice as likely to be observed as any particular odd-numbered face. a. What are the probabilities of the six simple events? (Hint: Denote these events by \(O_{1}, \ldots, O_{6}\). Then \(P\left(O_{1}\right)=p\), \(P\left(O_{2}\right)=2 p, P\left(O_{3}\right)=p, \ldots, P\left(O_{6}\right)=2 p .\) Now use a condi- tion on the sum of these probabilities to determine \(p .\) ) b. What is the probability that the number showing is an odd number? at most \(3 ?\) c. Now suppose that the die is loaded so that the probability of any particular simple event is proportional to the number showing on the corresponding upturned face; that is, \(P\left(O_{1}\right)=c, P\left(O_{2}\right)=2 c, \ldots, P\left(O_{6}\right)=6 c .\) What are the probabilities of the six simple events? Calculate the probabilities of Part (b) for this die.

An article in the New York Times (March 2, 1994) reported that people who suffer cardiac arrest in New York City have only a 1 in 100 chance of survival. Using probability notation, an equivalent statement would be \(P\) (survival \()=.01\) for people who suffer a cardiac arrest in New York City. (The article attributed this poor survival rate to factors common in large cities: traffic congestion and the difficulty of finding victims in large buildings.) a. Give a relative frequency interpretation of the given probability. b. The research that was the basis for the New York Times article was a study of 2329 consecutive cardiac arrests in New York City. To justify the " 1 in 100 chance of survival" statement, how many of the 2329 cardiac arrest sufferers do you think survived? Explain.

A certain company sends \(40 \%\) of its overnight mail parcels by means of express mail service \(A_{1}\). Of these parcels, \(2 \%\) arrive after the guaranteed delivery time (use \(L\) to denote the event late delivery). If a record of an overnight mailing is randomly selected from the company's files, what is the probability that the parcel went by means of \(A_{1}\) and was late?
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