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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

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 \(P(B)\) ?

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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.

In an article that appears on the web site of the American Statistical Association (www.amstat.org), Carlton Gunn, a public defender in Seattle, Washington, wrote about how he uses statistics in his work as an attorney. He states: I personally have used statistics in trying to challenge the reliability of drug testing results. Suppose the chance of a mistake in the taking and processing of a urine sample for a drug test is just 1 in 100 . And your client has a "dirty" (i.e., positive) test result. Only a 1 in 100 chance that it could be wrong? Not necessarily. If the vast majority of all tests given - say 99 in 100 \- are truly clean, then you get one false dirty and one true dirty in every 100 tests, so that half of the dirty tests are false. Define the following events as \(T D=\) event that the test result is dirty, \(T C=\) event that the test result is clean, \(D=\) event that the person tested is actually dirty, and \(C=\) event that the person tested is actually clean. a. Using the information in the quote, what are the values of \mathbf{i} . ~ \(P(T D \mid D)\) iii. \(P(C)\) ii. \(P(T D \mid C) \quad\) iv. \(P(D)\) b. Use the law of total probability to find \(P(T D)\). c. Use Bayes' rule to evaluate \(P(C \mid T D)\). Is this value consistent with the argument given in the quote? Explain.

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