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Chapter 3: Q3-51E (page 183)

Encoding variability in software. At the 2012 Gulf Petrochemicals and Chemicals Association (GPCA) Forum, Oregon State University software engineers presented a paper on modelling and implementing variation in computer software. The researchers employed the compositional choice calculus (CCC)—a formal language for representing, generating, and organizing variation in tree-structured artefacts. The CCC language was compared to two other coding languages—the annotative choice calculus (ACC) and the computational feature algebra (CFA). Their research revealed the following: Any type of expression (e.g., plain expressions, dimension declarations, or lambda abstractions) found in either ACC or CFA can be found in CCC; plain expressions exist in both ACC and CFA; dimension declarations exist in ACC, but not CFA; lambda abstractions exist in CFA, but not ACC. Based on this information, draw a Venn diagram illustrating the relationships among the three languages. (Hint: An expression represents a sample point in the Venn diagram.)

Short Answer

Expert verified

Answer

Fig.1 Venn diagram (shown in step 2)

Step by step solution

01

Step-by-Step SolutionStep 1: Introduction

A Venn diagram is a probability diagram with one or more circles inside a rectangle and demonstrates logical relationships between occurrences. In a diagram, the rectangle symbolizes the sample space or the universal set, which collects all possible outcomes. A circle within a rectangle symbolizes an event, i.e., a subset of the sample space.

02

Venn diagram shows the relationship among the languages

The above Venn diagram shows that the CCC (Compositional Choice Calculus) language was compared to two other coding languages: ACC and CFA. They exist in:

  1. ACC (Annotative Choice Calculus) exist in dimension declaration.
  2. CFA (Computational Feature Algebra) exist in lambda abstraction.
  3. Both the languages ACC and CFA have existed in plain expression.

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Most popular questions from this chapter

Randomization in a study of TV commercials. Gonzaga University professors conducted a study of more than 1,500 television commercials and published their results in the Journal of Sociology, Social Work, and Social Welfare (Vol. 2, 2008). Commercials from eight networks—ABC, FAM, FOX, MTV, ESPN, CBS, CNN, and NBC—were sampled for 8 days, with one network randomly selected each day. The table below shows the actual order determined by random draw:

ABC—July 6 (Wed)

FAM—July 7 (Thr)

FOX—July 9 (Sat)

MTV—July 10 (Sun)

ESPN—July 11 (Mon)

CBS—July 12 (Tue)

CNN—July 16 (Sat)

NBC—July 17 (Sun)

a. What is the probability that ESPN was selected on Monday, July 11?

b. Consider the four networks chosen for the weekends (Saturday and Sunday). How many ways could the researchers select four networks from the eight for the weekend analysis of commercials? (Assume that the assignment order for the four weekend days was immaterial to the analysis.)

c. Knowing that the networks were selected at random, what is the probability that ESPN was one of the four networks selected for the weekend analysis of commercials?

An experiment results in one of the following sample points: E1,E2,E3 orE4 . Find PE4for each of the following cases.

  1. PE1=0.1,PE2=0.2,PE3=0.3
  2. PE1=PE2=PE3=PE4
  3. PE1=PE2=0.1andPE3=PE4

The three-dice gambling problem. According toSignificance(December 2015), the 16th-century mathematician Jerome Cardan was addicted to a gambling game involving tossing three fair dice. One outcome of interest— which Cardan called a “Fratilli”—is when any subset of the three dice sums to 3. For example, the outcome {1, 1, 1} results in 3 when you sum all three dice. Another possible outcome that results in a “Fratilli” is {1, 2, 5}, since the first two dice sum to 3. Likewise, {2, 3, 6} is a “Fratilli,” since the second die is a 3. Cardan was an excellent mathematician but calculated the probability of a “Fratilli” incorrectly as 115/216 = .532.

a. Show that the denominator of Cardan’s calculation, 216, is correct. [Hint: Knowing that there are 6 possible outcomes for each die, show that the total number of possible outcomes from tossing three fair dice is 216.]

b. One way to obtain a “Fratilli” is with the outcome {1,1, 1}. How many possible ways can this outcome be obtained?

c. Another way to obtain a “Fratilli” is with an outcome that includes at least one die with a 3. First, find the number of outcomes that do not result in a 3 on any of the dice. [Hint: If none of the dice can result in a 3, then there are only 5 possible outcomes for each die.] Now subtract this result from 216 to find the number of outcomes that include at least one 3.

d. A third way to obtain a “Fratilli” is with the outcome {1, 2, 1}, where the order of the individual die outcomes does not matter. How many possible ways can this outcome be obtained?

e. A fourth way to obtain a “Fratilli” is with the outcome {1, 2, 2}, where the order of the individual die outcomes does not matter. How many possible ways can this outcome be obtained?

f. A fifth way to obtain a “Fratilli” is with the outcome {1, 2, 4}, where the order of the individual die outcomes does not matter. How many possible ways can this outcome be obtained? [Hint:There are 3 choices for the first die, 2 for the second, and only 1 for the third.]

g. A sixth way to obtain a “Fratilli” is with the outcome {1, 2, 5}, where the order of the individual die outcomes does not matter. How many possible ways can this outcome be obtained? [See Hintfor part f.]

h. A final way to obtain a “Fratilli” is with the outcome {1, 2, 6}, where the order of the individual die outcomes does not matter. How many possible ways can this outcome be obtained? [See Hintfor part f.]

i. Sum the results for parts b–h to obtain the total number of possible “Fratilli” outcomes.

j. Compute the probability of obtaining a “Fratilli” outcome. Compare your answer with Cardan’s.

Appeals of federal civil trials. The Journal of the American Law and Economics Association (Vol. 3, 2001) publishedthe results of a study of appeals of federal civil trials. Thefollowing table, extracted from the article, gives a breakdownof 2,143 civil cases that were appealed by either theplaintiff or the defendant. The outcome of the appeal, aswell as the type of trial (judge or jury), was determined foreach civil case. Suppose one of the 2,143 cases is selected

at random and both the outcome of the appeal and type of trial are observed.

Jury

Judge

Totals

Plaintiff trial win-reserved

194

71

265

Plaintiff trial win-affirmed/dismissed

429

240

669

Defendant trial win-reserved

111

68

179

Defendant trial win- affirmed/dismissed

731

678

1030

Total

1465

678

2143

a. Find P (A), where A = {jury trial}.

b. Find P (B), where B = {plaintiff trial win is reversed}.

c. Are A and B mutually exclusive events?

d. FindP(AC)

e. FindP(A∪B)

f. FindP(A∩B)

Management system failures. Refer to the Process Safety Progress (December 2004) study of 83 industrial accidents caused by management system failures, Exercise 2.150(p. 142). A summary of the root causes of these 83 incidents is reproduced in the following table. One of the 83 incidents is randomly selected and the root cause is determined.

Management system cause category

Number of incidents

Engineering and design

27

Procedures and practices

24

Management and oversight

22

Training and communication

10

Total

83

a. List the sample points for this problem and assign reasonable probabilities to them.

b. Find and interpret the probability that an industrial accident is caused by faulty engineering and design.

c. Find and interpret the probability that an industrial accident is caused by something other than faulty procedures and practices.
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