Statistical Techniques in Business and Economics Douglas Lind 17e - Test Bank Instant Download - Complete Test Bank With Answers Sample Questions Are Posted Below Statistical Techniques in Business and Economics, 17e (Lind) Chapter 5 A Survey of Probability Concepts 1) The probability of rolling a 3 or 2 on …
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Statistical Techniques in Business and Economics Douglas Lind 17e – Test Bank
Instant Download – Complete Test Bank With Answers
Sample Questions Are Posted Below
Statistical Techniques in Business and Economics, 17e (Lind)
Chapter 5 A Survey of Probability Concepts
1) The probability of rolling a 3 or 2 on a single die is an example of conditional probability.
Answer: FALSE
Explanation: This is classical probability.
Difficulty: 1 Easy
Topic: Rules of Multiplication to Calculate Probability
Learning Objective: 05-02 Assign probabilities using a classical, empirical, or subjective approach.
Bloom’s: Understand
AACSB: Communication
Accessibility: Keyboard Navigation
2) The probability of rolling a 3 or 2 on a single die is an example of mutually exclusive events.
Answer: TRUE
Explanation: This is mutually exclusive as you cannot roll both a 2 and a 3 at the same time. Only one of these events can happen on a roll of a single die.
Difficulty: 1 Easy
Topic: Approaches to Assigning Probabilities
Learning Objective: 05-02 Assign probabilities using a classical, empirical, or subjective approach.
Bloom’s: Understand
AACSB: Communication
Accessibility: Keyboard Navigation
3) An individual can assign a subjective probability to an event based on the individual’s knowledge about the event.
Answer: TRUE
Explanation: When someone uses available knowledge to assign a probability to an event, this is subjective probability as it is based on an opinion.
Difficulty: 1 Easy
Topic: Approaches to Assigning Probabilities
Learning Objective: 05-02 Assign probabilities using a classical, empirical, or subjective approach.
Bloom’s: Remember
AACSB: Communication
Accessibility: Keyboard Navigation
4) To apply the special rule of addition, the events must be mutually exclusive.
Answer: TRUE
Explanation: The special rule of addition assumes there is no intersection or joint probability between events, so the events cannot occur concurrently (i.e., they are mutually exclusive so the joint probability is zero).
Difficulty: 1 Easy
Topic: Rules of Addition for Computing Probabilities
Learning Objective: 05-03 Calculate probabilities using the rules of addition.
Bloom’s: Understand
AACSB: Communication
Accessibility: Keyboard Navigation
5) A joint probability measures the likelihood that two or more events will happen concurrently.
Answer: TRUE
Explanation: A joint probability measures the chance that several events can happen at the same time. If the joint probability is zero, then the events are mutually exclusive.
Difficulty: 1 Easy
Topic: Rules of Addition for Computing Probabilities
Learning Objective: 05-03 Calculate probabilities using the rules of addition.
Bloom’s: Remember
AACSB: Communication
Accessibility: Keyboard Navigation
6) The joint probability of two independent events, A and B, is computed as P(A and B) = P(A) × P(B).
Answer: TRUE
Explanation: If Events A and B are independent, then the P(A) = P(A|B) and P(B) = P(B|A). In other words, P(A and B) = P(A) × P(B) as the conditional probability is zero.
Difficulty: 1 Easy
Topic: Rules of Multiplication to Calculate Probability
Learning Objective: 05-04 Calculate probabilities using the rules of multiplication.
Bloom’s: Remember
AACSB: Communication
7) The joint probability of two events, A and B, that are NOT independent is computed as
P(A and B) = P(A) × P(B|A).
Answer: TRUE
Explanation: When two events are not independent, then P(B) ≠ P(B|A). In other words, the probability of B depends on the other event A, so we must use P(B|A) in calculating the joint probability of A and B rather than just P(B).
Difficulty: 1 Easy
Topic: Rules of Multiplication to Calculate Probability
Learning Objective: 05-04 Calculate probabilities using the rules of multiplication.
Bloom’s: Remember
AACSB: Communication
8) A coin is tossed four times. The joint probability that all four tosses will result in a head is 1/4 or 0.25.
Answer: FALSE
Explanation: Each outcome’s probability is 0.5. The joint probability is (0.5)(0.5)(0.5)(0.5) = 0.0625.
Difficulty: 2 Medium
Topic: Rules of Multiplication to Calculate Probability
Learning Objective: 05-04 Calculate probabilities using the rules of multiplication.
Bloom’s: Apply
AACSB: Analytic
9) If there are “m” ways of doing one thing, and “n” ways of doing another thing, the multiplication formula states that there are (m) × (n) ways of doing both.
Answer: TRUE
Explanation: The multiplication formula states the number of possible arrangements that are possible for two or more events.
Difficulty: 2 Medium
Topic: Principles of Counting
Learning Objective: 05-07 Determine the number of outcomes using principles of counting.
Bloom’s: Remember
AACSB: Communication
10) A combination of a set of objects is defined by the order of the objects.
Answer: FALSE
Explanation: The order of objects is important for permutations but not combinations.
Difficulty: 1 Easy
Topic: Principles of Counting
Learning Objective: 05-07 Determine the number of outcomes using principles of counting.
Bloom’s: Remember
AACSB: Communication
Accessibility: Keyboard Navigation
11) The complement rule states that the probability of an event not occurring is equal to 1 minus the probability of its occurrence.
Answer: TRUE
Explanation: If P(A) and P(~A) are complements, then P(A) = 1 − P(~A) and P(~A) = 1 − P(A).
Difficulty: 1 Easy
Topic: Approaches to Assigning Probabilities
Learning Objective: 05-03 Calculate probabilities using the rules of addition.
Bloom’s: Remember
AACSB: Communication
12) If two events are mutually exclusive, then P(A and B) = P(A) × P(B).
Answer: FALSE
Explanation: If two events are mutually exclusive, then P(A and B) = 0
Difficulty: 1 Easy
Topic: Rules of Addition for Computing Probabilities
Learning Objective: 05-03 Calculate probabilities using the rules of addition.
Bloom’s: Understand
AACSB: Communication
13) An illustration of an experiment is turning the ignition key of an automobile as it comes off the assembly line to determine whether or not the engine will start.
Answer: TRUE
Explanation: An experiment is a process that leads to the occurrence of one and only one of several possible outcomes. In this case, the experiment is turning the key (the process) and the possible outcomes are (1) the car starts, or (2) the car doesn’t start.
Difficulty: 2 Medium
Topic: What is a Probability?
Learning Objective: 05-01 Define the terms probability, experiment, event, and outcome.
Bloom’s: Understand
AACSB: Communication
Accessibility: Keyboard Navigation
14) Bayes’ theorem is a method to revise the probability of an event given additional information.
Answer: TRUE
Explanation: In Bayes’ theorem, we take initial probabilities known on the basis of current information (prior probabilities) and revise them based on new information (posterior probabilities).
Difficulty: 1 Easy
Topic: Bayes Theorem
Learning Objective: 05-06 Calculate probabilities using Bayes theorem.
Bloom’s: Remember
AACSB: Communication
Accessibility: Keyboard Navigation
15) Bayes’ theorem is used to calculate a subjective probability.
Answer: FALSE
Explanation: Bayes’ theorem is used to calculate posterior probabilities (a revised probability based on new information). Subjective probability is a method used to assign a probability based on the opinion of someone using available information.
Difficulty: 1 Easy
Topic: Bayes Theorem
Learning Objective: 05-06 Calculate probabilities using Bayes theorem.
Bloom’s: Remember
AACSB: Communication
Accessibility: Keyboard Navigation
16) The National Center for Health Statistics reported that of every 883 deaths in recent years, 24 resulted from an automobile accident, 182 from cancer, and 333 from heart disease. What is the probability that a particular death is due to an automobile accident?
Answer: A
Explanation: Based on the empirical approach to probability, 24/883 = 0.027.
Difficulty: 2 Medium
Topic: Approaches to Assigning Probabilities
Learning Objective: 05-02 Assign probabilities using a classical, empirical, or subjective approach.
Bloom’s: Apply
AACSB: Analytic
Accessibility: Keyboard Navigation
17) If two events A and B are mutually exclusive, what does the special rule of addition state?
Answer: A
Explanation: By definition, P(A or B) = P(A) + P(B) is the special rule of addition.
Difficulty: 1 Easy
Topic: Rules of Addition for Computing Probabilities
Learning Objective: 05-03 Calculate probabilities using the rules of addition.
Bloom’s: Remember
AACSB: Communication
18) What does the complement rule state?
Answer: B
Explanation: By definition, P(A) plus its complement, P(not A) must equal 1.
Difficulty: 1 Easy
Topic: Rules of Addition for Computing Probabilities
Learning Objective: 05-03 Calculate probabilities using the rules of addition.
Bloom’s: Remember
AACSB: Communication
19) Which approach to probability is exemplified by the following formula?
Probability of an event =
Answer: B
Explanation: The empirical rule is based on the frequency of observed experimental outcomes.
Difficulty: 1 Easy
Topic: Approaches to Assigning Probabilities
Learning Objective: 05-02 Assign probabilities using a classical, empirical, or subjective approach.
Bloom’s: Remember
AACSB: Communication
20) A study of 200 computer service firms revealed these incomes after taxes:
| Income After Taxes | Number of Firms |
| Under $1 million | 102 |
| $1 million up to $20 million | 61 |
| $20 million or more | 37 |
What is the probability that a particular firm selected has $1 million or more in income after taxes?
Answer: C
Explanation: A couple of approaches can be used to answer the question. Using the complement rule, the probability of firms less than $1 million is 102/200 = 0.51. The complement or firms with $1 million or more income is 1.0 − 0.51 = 0.49.
Difficulty: 2 Medium
Topic: Approaches to Assigning Probabilities
Learning Objective: 05-03 Calculate probabilities using the rules of addition.
Bloom’s: Apply
AACSB: Analytic
21) A firm offers routine physical examinations as part of a health service program for its employees. The exams showed that 8% of the employees needed corrective shoes, 15% needed major dental work, and 3% needed both corrective shoes and major dental work. What is the probability that an employee selected at random will need either corrective shoes or major dental work?
Answer: A
Explanation: The two events, corrective shoes and dental work, are not mutually exclusive because an employee can need both. The P(corrective shoes or dental work) = P(corrective shoes) + P(dental work) − P(corrective shoes and dental work) = 0.8 + 0.15 − 0.03 = 0.20.
Difficulty: 2 Medium
Topic: Rules of Addition for Computing Probabilities
Learning Objective: 05-03 Calculate probabilities using the rules of addition.
Bloom’s: Apply
AACSB: Analytic
22) A survey of top executives revealed that 35% of them regularly read Time magazine, 20% read Newsweek, and 40% read U.S. News & World Report. A total of 10% read both Time and U.S. News & World Report. What is the probability that a particular top executive reads either Time or U.S. News & World Report regularly?
Answer: D
Explanation: The three events—reading Time, Newsweek, or U.S. News & World Report—are not mutually exclusive because executives can read more than one of the magazines. The P(Time or U.S. News) = P(Time) + P(U.S. News) − P(Time and U.S. News) = 0.35 + 0.40 − 0.10 = 0.65.
Difficulty: 2 Medium
Topic: Rules of Addition for Computing Probabilities
Learning Objective: 05-03 Calculate probabilities using the rules of addition.
Bloom’s: Apply
AACSB: Analytic
23) A study by the National Park Service revealed that 50% of the vacationers going to the Rocky Mountain region visit Yellowstone Park, 40% visit the Grand Tetons, and 35% visit both. What is the probability that a vacationer will visit at least one of these magnificent attractions?
Answer: C
Explanation: The two events, visiting Yellowstone and visiting the Grand Tetons, are not mutually exclusive because vacationers can visit both locations. The P(Yellowstone or Grand Tetons) = P(Yellowstone) + P(Grand Tetons) − P(Yellowstone and Grand Tetons) = 0.50 + 0.40 − 0.35 = 0.55.
Difficulty: 2 Medium
Topic: Rules of Addition for Computing Probabilities
Learning Objective: 05-03 Calculate probabilities using the rules of addition.
Bloom’s: Apply
AACSB: Analytic
24) A tire manufacturer advertises, “The median life of our new all-season radial tire is 50,000 miles. An immediate adjustment will be made on any tire that does not last 50,000 miles.” You purchased four of these tires. What is the probability that all four tires will wear out before traveling 50,000 miles?
Answer: D
Explanation: The median corresponds with the 50th percentile. So the probability that a tire wears out before 50,000 miles is 0.5. Each outcome’s probability is 0.5. The joint probability is (0.5)(0.5)(0.5)(0.5) = 0.0625.
Difficulty: 2 Medium
Topic: Rules of Multiplication to Calculate Probability
Learning Objective: 05-04 Calculate probabilities using the rules of multiplication.
Bloom’s: Apply
AACSB: Analytic
25) A sales representative calls on four hospitals in Westchester County. It is immaterial what order he calls on them. How many ways can he organize his calls?
Answer: B
Explanation: Use the multiplication formula: (4)(3)(2)(1) = 24.
Difficulty: 2 Medium
Topic: Principles of Counting
Learning Objective: 05-07 Determine the number of outcomes using principles of counting.
Bloom’s: Apply
AACSB: Analytic
Accessibility: Keyboard Navigation
26) There are 10 AAA batteries in a box and 3 are defective. Two batteries are selected without replacement. What is the probability of selecting a defective battery followed by another defective battery?
Answer: D
Explanation: The probability of a defective battery on the first selection is 3/10 = 0.3. The probability of selecting a second defective battery is a conditional probability that assumes the first selection was defective, so the probability of a second defective battery is 2/9. The joint probability is (3/10)(2/9) = 0.066667.
Difficulty: 2 Medium
Topic: Rules of Multiplication to Calculate Probability
Learning Objective: 05-04 Calculate probabilities using the rules of multiplication.
Bloom’s: Apply
AACSB: Reflective Thinking
27) Giorgio offers the person who purchases an 8-ounce bottle of Allure two free gifts, chosen from the following: an umbrella, a 1-ounce bottle of Midnight, a feminine shaving kit, a raincoat, or a pair of rain boots. If you purchased Allure, what is the probability you randomly select an umbrella and a shaving kit in that order?
Answer: C
Explanation: There are five different gifts. Therefore, the probability of any gift is 1/5 = 0.2. The probability of selecting a second gift is a conditional probability that assumes the first selection was an umbrella, so the probability of a second gift, a shaving kit, is 1/4 = 0.25. The joint probability is (1/5)(1/4) = 1/20 = 0.05.
Difficulty: 2 Medium
Topic: Rules of Multiplication to Calculate Probability
Learning Objective: 05-04 Calculate probabilities using the rules of multiplication.
Bloom’s: Apply
AACSB: Analytic
Accessibility: Keyboard Navigation
28) A board of directors consists of eight men and four women. A four-member search committee is randomly chosen to recommend a new company president. What is the probability that all four members of the search committee will be women?
Answer: D
Explanation: There are four women in a group of 12 individuals. Therefore, the probability of picking a woman on the first selection is 4/12, the second selection is 3/11, the third selection is 2/10, and the fourth is 1/9. This is an application of the multiplication rule for events that are not independent. The joint probability is (4/12)(3/11)(2/10)(1/9) = 0.002.
Difficulty: 2 Medium
Topic: Rules of Multiplication to Calculate Probability
Learning Objective: 05-04 Calculate probabilities using the rules of multiplication.
Bloom’s: Apply
AACSB: Analytic
29) A lamp manufacturer designed five lamp bases and four lampshades that could be used together. How many different arrangements of base and shade can be offered?
Answer: D
Explanation: Using the multiplication formula, (5)(4) = 20.
Difficulty: 2 Medium
Topic: Principles of Counting
Learning Objective: 05-07 Determine the number of outcomes using principles of counting.
Bloom’s: Apply
AACSB: Analytic
Accessibility: Keyboard Navigation
30) A gumball machine has just been filled with 50 black, 150 white, 100 red, and 100 yellow gumballs that have been thoroughly mixed. Sue and Jim each purchase one gumball. What is the likelihood that both Sue and Jim will get red gumballs?
Answer: B
Explanation: The probability of a red gumball on the first selection is 100/400 = 0.25. The probability of selecting a second red gumball is a conditional probability that assumes the first selection was a red gumball, so the probability of a second red gumball is 99/399. The joint probability is (100/400)(99/399) = 0.062.
Difficulty: 2 Medium
Topic: Rules of Multiplication to Calculate Probability
Learning Objective: 05-04 Calculate probabilities using the rules of multiplication.
Bloom’s: Apply
AACSB: Analytic
Accessibility: Keyboard Navigation
31) What does equal?
Answer: C
Explanation: (6*5*4*3*2*1)(2*1)/(4*3*2*1)(3*2*1) = 10
Difficulty: 2 Medium
Topic: Principles of Counting
Learning Objective: 05-07 Determine the number of outcomes using principles of counting.
Bloom’s: Apply
AACSB: Analytic; Reflective Thinking
32) In a management trainee program, 80% of the trainees are female, while 20% are male. Ninety percent of the females attended college; 78% of the males attended college. A management trainee is selected at random. What is the probability that the person selected is a female who did NOT attend college?
Answer: B
Explanation: First, the conditional probability that a person attended college given the person is female is P(college | female) = 0.9. The complement is P(no college | female) = 0.1. Now the joint probability of selecting a female who did not attend college is P(female and no college) = P(female) P(no college | female) = (0.8)(0.1) = 0.08.
Difficulty: 3 Hard
Topic: Rules of Multiplication to Calculate Probability
Learning Objective: 05-04 Calculate probabilities using the rules of multiplication.
Bloom’s: Apply
AACSB: Analytic
Accessibility: Keyboard Navigation
33) In a management trainee program, 80% of the trainees are female, while 20% are male. Ninety percent of the females attended college; 78% of the males attended college. A management trainee is selected at random. What is the probability that the person selected is a female who attended college?
Answer: D
Explanation: First, the conditional probability that a person attended college given the person is female is P(college | female) = 0.9. Now the joint probability of selecting a female who did attend college is P(female and college) = P(female) P(college | female) = (0.8)(0.9) = 0.72.
Difficulty: 2 Medium
Topic: Rules of Multiplication to Calculate Probability
Learning Objective: 05-04 Calculate probabilities using the rules of multiplication.
Bloom’s: Apply
AACSB: Analytic
Accessibility: Keyboard Navigation
34) In a management trainee program, 80% of the trainees are female, while 20% are male. Ninety percent of the females attended college; 78% of the males attended college. A management trainee is selected at random. What is the probability that the person selected is a male who did NOT attend college?
Answer: A
Explanation: First, the conditional probability that a person attended college given the person is male is P(college | male) = 0.78. The complement is P(no college | male) = 0.22. Now the joint probability of selecting a male who did not attend college is P(male and no college) = P(male) P(no college | male) = (0.2)(0.22) = 0.044.
Difficulty: 2 Medium
Topic: Rules of Multiplication to Calculate Probability
Learning Objective: 05-04 Calculate probabilities using the rules of multiplication.
Bloom’s: Apply
AACSB: Analytic
Accessibility: Keyboard Navigation
35) In a management trainee program, 80% of the trainees are female, while 20% are male. Ninety percent of the females attended college; 78% of the males attended college. A management trainee is selected at random. What is the probability that the person selected is a male who did NOT attend college?
Answer: A
Explanation: First, the conditional probability that a person attended college given the person is male is P(college | male). The complement is P(no college | male). Now the joint probability of selecting a male who did not attend college is P(male and no college) = P(male) P(no college | male).
Difficulty: 2 Medium
Topic: Rules of Multiplication to Calculate Probability
Learning Objective: 05-04 Calculate probabilities using the rules of multiplication.
Bloom’s: Apply
AACSB: Analytic
Accessibility: Keyboard Navigation
36) In a management trainee program, 80% of the trainees are female, while 20% are male. Ninety percent of the females attended college; 78% of the males attended college. A management trainee is selected at random. What is the probability that the person selected is a female who attended college?
Answer: C
Explanation: First, the conditional probability that a person attended college given the person is female is P(college | female). The complement is P(no college | female). Now the joint probability of selecting a female who did not attend college is P(female and no college) = P(female) P(no college | female).
Difficulty: 2 Medium
Topic: Rules of Multiplication to Calculate Probability
Learning Objective: 05-04 Calculate probabilities using the rules of multiplication.
Bloom’s: Apply
AACSB: Analytic
Accessibility: Keyboard Navigation
37) A supplier delivers an order for 20 electric toothbrushes to a store. By accident, three of the electric toothbrushes are defective. What is the probability that the first two electric toothbrushes sold are defective?
Answer: D
Explanation: The probability of a defective unit on the first selection is 3/20 = 0.15. The probability of selecting a second unit is a conditional probability that assumes the first selection was defective, so the probability of a second defective toothbrush is 2/19. The joint probability is (3/20)(2/19) = 0.01579.
Difficulty: 2 Medium
Topic: Rules of Multiplication to Calculate Probability
Learning Objective: 05-04 Calculate probabilities using the rules of multiplication.
Bloom’s: Apply
AACSB: Analytic
Accessibility: Keyboard Navigation
38) An electronics firm sells four models of stereo receivers, three amplifiers, and six speaker brands. When the four types of components are sold together, they form a “system.” How many different systems can the electronics firm offer?
Answer: C
Explanation: Using the multiplication formula, (4)(3)(6) = 72.
Difficulty: 2 Medium
Topic: Principles of Counting
Learning Objective: 05-07 Determine the number of outcomes using principles of counting.
Bloom’s: Apply
AACSB: Analytic
Accessibility: Keyboard Navigation
39) The numbers 0 through 9 are used in code groups of four to identify an item of clothing. Code 1083 might identify a blue blouse, size medium. The code group 2031 might identify a pair of pants, size 18, and so on. Repetitions of numbers are not permitted—in other words, the same number cannot be used more than once in a total sequence. As examples, 2,256, 2,562, or 5,559 would not be permitted. How many different code groups can be designed?
Answer: A
Explanation: Using the multiplication formula, (10)(9)(8)(7) = 5,040 different codes without repeated digits.
Difficulty: 2 Medium
Topic: Principles of Counting
Learning Objective: 05-07 Determine the number of outcomes using principles of counting.
Bloom’s: Apply
AACSB: Analytic
Accessibility: Keyboard Navigation
40) How many permutations of the three letters C, D, and E are possible?
Answer: C
Explanation: Using the permutation formula 3!/0! = (3)(2)(1) = 6.
Difficulty: 2 Medium
Topic: Principles of Counting
Learning Objective: 05-07 Determine the number of outcomes using principles of counting.
Bloom’s: Apply
AACSB: Analytic
Accessibility: Keyboard Navigation
41) You are assigned to design color codes for different parts. Three colors are used to code on each part. Once a combination of three colors is used—such as green, yellow, and red—these three colors cannot be rearranged to use as a code for another part. If there are 35 combinations, how many colors are available?
Answer: B
Explanation: Using the combination formula, x!/(3!(x − 3)!) or trial and error, there are 7 colors.
Difficulty: 2 Medium
Topic: Principles of Counting
Learning Objective: 05-07 Determine the number of outcomes using principles of counting.
Bloom’s: Apply
AACSB: Analytic
42) A developer of a new subdivision wants to build homes that are all different. There are three different interior plans that can be combined with any of five different home exteriors. How many different homes can be built?
Answer: C
Explanation: Using the multiplication formula, (3)(5) = 15.
Difficulty: 2 Medium
Topic: Principles of Counting
Learning Objective: 05-07 Determine the number of outcomes using principles of counting.
Bloom’s: Apply
AACSB: Analytic
Accessibility: Keyboard Navigation
43) Six basic colors are used in decorating a new condominium. They are applied to a unit in groups of four colors. One unit might have gold as the principal color, blue as a complementary color, red as the accent color, and touches of white. Another unit might have blue as the principal color, white as the complementary color, gold as the accent color, and touches of red. If repetitions are permitted, how many different units can be decorated?
Answer: D
Explanation: Using the multiplication formula, (6)(6)(6)(6) = 1,296.
Difficulty: 2 Medium
Topic: Principles of Counting
Learning Objective: 05-07 Determine the number of outcomes using principles of counting.
Bloom’s: Apply
AACSB: Analytic
Accessibility: Keyboard Navigation
44) The ABCD football association is considering a Super Ten Football Conference. The top 10 football teams in the country, based on past records, would be members of the Super Ten Conference. Each team would play every other team in the conference during the season and the team winning the most games would be declared the national champion. How many games would the conference commissioner have to schedule each year? (Remember, Oklahoma versus Michigan is the same as Michigan versus Oklahoma.)
Answer: A
Explanation: Using the combination formula, n = 10 and r = 2, 10!/2!(10 − 2)! = 90/2 = 45.
Difficulty: 2 Medium
Topic: Principles of Counting
Learning Objective: 05-07 Determine the number of outcomes using principles of counting.
Bloom’s: Apply
AACSB: Analytic
45) A rug manufacturer has decided to use seven compatible colors in her rugs. However, in weaving a rug, only five spindles can be used. In advertising, the rug manufacturer wants to indicate the number of different color groupings for sale. How many color groupings using the seven colors taken five at a time are there? (This assumes that five different colors will go into each rug—in other words, there are no repetitions of color.)
Answer: B
Explanation: Using the combination formula, n = 7 and r = 5; 7!/5!(7 − 5)! = 42/2 = 21.
Difficulty: 2 Medium
Topic: Principles of Counting
Learning Objective: 05-07 Determine the number of outcomes using principles of counting.
Bloom’s: Apply
AACSB: Analytic
46) The first card selected from a standard 52-card deck was a king. If it is returned to the deck, what is the probability that a king will be drawn on the second selection?
Answer: B
Explanation: The probability is 4/52 = 1/13, or 0.077.
Difficulty: 2 Medium
Topic: Approaches to Assigning Probabilities
Learning Objective: 05-04 Calculate probabilities using the rules of multiplication.
Bloom’s: Apply
AACSB: Analytic
Accessibility: Keyboard Navigation
47) The first card selected from a standard 52-card deck was a king. If it is NOT returned to the deck, what is the probability that a king will be drawn on the second selection?
Answer: C
Explanation: The probability of a king on the first selection is 4/52 = 1/13. The probability of selecting a second king is a conditional probability that assumes the first selection was a king, so the probability of a second king is 3/51, or 0.0588.
Difficulty: 2 Medium
Topic: Approaches to Assigning Probabilities
Learning Objective: 05-04 Calculate probabilities using the rules of multiplication.
Bloom’s: Apply
AACSB: Analytic
Accessibility: Keyboard Navigation
48) Which approach to probability assumes that events are equally likely?
Answer: A
Explanation: By definition, the classical approach assumes that events are equally likely.
Difficulty: 1 Easy
Topic: Approaches to Assigning Probabilities
Learning Objective: 05-02 Assign probabilities using a classical, empirical, or subjective approach.
Bloom’s: Remember
AACSB: Communication
Accessibility: Keyboard Navigation
49) An experiment may have ________.
Answer: C
Explanation: By definition, an experiment results in one or more outcomes.
Difficulty: 1 Easy
Topic: What is a Probability?
Learning Objective: 05-01 Define the terms probability, experiment, event, and outcome.
Bloom’s: Remember
AACSB: Communication
Accessibility: Keyboard Navigation
50) When are two experimental outcomes mutually exclusive?
Answer: B
Explanation: By example, an experiment measures the variable gender. Therefore, there are two possible outcomes: male or female. The outcome is mutually exclusive since a person can be classified as either male or female, not both.
Difficulty: 1 Easy
Topic: Approaches to Assigning Probabilities
Learning Objective: 05-02 Assign probabilities using a classical, empirical, or subjective approach.
Bloom’s: Understand
AACSB: Communication
Accessibility: Keyboard Navigation
51) Probabilities are important information when ________.
Answer: D
Explanation: Inferential statistics uses sample data to make decisions with a stated probability of making an error.
Difficulty: 1 Easy
Topic: What is a Probability?
Learning Objective: 05-01 Define the terms probability, experiment, event, and outcome.
Bloom’s: Understand
AACSB: Communication
Accessibility: Keyboard Navigation
52) The result of a particular experiment is called a(n) ________.
Answer: D
Explanation: By definition, experiments result in a set of observable outcomes.
Difficulty: 1 Easy
Topic: What is a Probability?
Learning Objective: 05-01 Define the terms probability, experiment, event, and outcome.
Bloom’s: Remember
AACSB: Communication
Accessibility: Keyboard Navigation
53) The probability of two or more events occurring concurrently is called a(n) ________.
Answer: C
Explanation: The P(A and B and C) is a joint probability.
Difficulty: 1 Easy
Topic: Rules of Multiplication to Calculate Probability
Learning Objective: 05-04 Calculate probabilities using the rules of multiplication.
Bloom’s: Remember
AACSB: Communication
Accessibility: Keyboard Navigation
54) The probability of a particular event occurring, given that another event has occurred, is known as a(n) ________.
Answer: A
Explanation: P(A | B) is a conditional probability indicating that the probability of the event A is affected by event B.
Difficulty: 1 Easy
Topic: Rules of Multiplication to Calculate Probability
Learning Objective: 05-04 Calculate probabilities using the rules of multiplication.
Bloom’s: Remember
AACSB: Communication
Accessibility: Keyboard Navigation
55) A graphical method used to calculate joint and conditional probabilities is ________.
Answer: A
Difficulty: 1 Easy
Topic: Rules of Multiplication to Calculate Probability
Learning Objective: 05-05 Compute probabilities using a contingency table.
Bloom’s: Remember
AACSB: Communication
Accessibility: Keyboard Navigation
56) When an experiment is conducted “without replacement,” ________.
Answer: A
Explanation: “Without replacement” means that when an individual or object is observed or measured, it is not returned to the population. So on the next selection from the population, each individual or object has a higher probability of selection because the population is now N − 1.
Difficulty: 1 Easy
Topic: Rules of Multiplication to Calculate Probability
Learning Objective: 05-04 Calculate probabilities using the rules of multiplication.
Bloom’s: Understand
AACSB: Communication
Accessibility: Keyboard Navigation
57) If two events are independent, then their joint probability is computed with ________.
Answer: B
Explanation: The special rule of multiplication is P(A and B) = P(A)P(B) for independent events.
Difficulty: 1 Easy
Topic: Rules of Multiplication to Calculate Probability
Learning Objective: 05-04 Calculate probabilities using the rules of multiplication.
Bloom’s: Understand
AACSB: Communication
Accessibility: Keyboard Navigation
58) When applying the special rule of addition for mutually exclusive events, the joint probability is ________.
Answer: C
Explanation: For mutually exclusive events, the joint probability is P(A and B) = 0.
Difficulty: 1 Easy
Topic: Rules of Addition for Computing Probabilities
Learning Objective: 05-03 Calculate probabilities using the rules of addition.
Bloom’s: Understand
AACSB: Communication
Accessibility: Keyboard Navigation
59) When an event’s probability depends on the occurrence of another event, the probability is a(n) ________.
Answer: A
Explanation: It is a conditional probability, expressed as P(B | A), or the probability of B given A.
Difficulty: 1 Easy
Topic: Rules of Multiplication to Calculate Probability
Learning Objective: 05-04 Calculate probabilities using the rules of multiplication.
Bloom’s: Understand
AACSB: Communication
Accessibility: Keyboard Navigation
60) A group of employees of Unique Services will be surveyed about a new pension plan. In-depth interviews with each employee selected in the sample will be conducted. The employees are classified as follows:
| Classification | Event | Number of Employees |
| Supervisors | A | 120 |
| Maintenance | B | 50 |
| Production | C | 1,460 |
| Management | D | 302 |
| Secretarial | E | 68 |
What is the probability that the first person selected is classified as a maintenance employee?
Answer: C
Explanation: Applying the empirical probability approach, there are 50 maintenance employees out of a total of 2,000 employees. The probability is 50/2,000 = 0.025.
Difficulty: 2 Medium
Topic: Approaches to Assigning Probabilities
Learning Objective: 05-02 Assign probabilities using a classical, empirical, or subjective approach.
Bloom’s: Apply
AACSB: Analytic
61) A group of employees of Unique Services will be surveyed about a new pension plan. In-depth interviews with each employee selected in the sample will be conducted. The employees are classified as follows:
| Classification | Event | Number of Employees |
| Supervisors | A | 120 |
| Maintenance | B | 50 |
| Production | C | 1,460 |
| Management | D | 302 |
| Secretarial | E | 68 |
What is the probability that the first person selected is either in maintenance or in secretarial?
Answer: C
Explanation: Given mutually exclusive classes, P(maintenance or secretarial) = P(maintenance) + P(secretarial) = 50/2,000 + 68/2,000 = 0.059.
Difficulty: 2 Medium
Topic: Rules of Addition for Computing Probabilities
Learning Objective: 05-03 Calculate probabilities using the rules of addition.
Bloom’s: Apply
AACSB: Analytic
62) A group of employees of Unique Services will be surveyed about a new pension plan. In-depth interviews with each employee selected in the sample will be conducted. The employees are classified as follows:
| Classification | Event | Number of Employees |
| Supervisors | A | 120 |
| Maintenance | B | 50 |
| Production | C | 1,460 |
| Management | D | 302 |
| Secretarial | E | 68 |
What is the probability that the first person selected is either in management or in supervision?
Answer: D
Explanation: For mutually exclusive classes, P(management or supervision) = P(management) + P(supervision) = 302/2,000 + 120/2,000 = 0.21.
Difficulty: 2 Medium
Topic: Rules of Addition for Computing Probabilities
Learning Objective: 05-03 Calculate probabilities using the rules of addition.
Bloom’s: Apply
AACSB: Analytic
63) The process used to calculate the revised probability of an event given additional information can be obtained through ________.
Answer: A
Explanation: Bayes’ theorem is used to calculate posterior probabilities, or revised probabilities based on additional information added to the prior probabilities we have presently.
Difficulty: 1 Easy
Topic: Bayes Theorem
Learning Objective: 05-06 Calculate probabilities using Bayes theorem.
Bloom’s: Understand
AACSB: Communication
Accessibility: Keyboard Navigation
64) Each salesperson in a large department store chain is rated on their sales ability and their potential for advancement. The data for the 500 sampled salespeople are summarized in the following table.
| Potential for Advancement | ||||
| Fair | Good | Excellent | ||
| Sales ability | Below average | 16 | 12 | 22 |
| Average | 45 | 60 | 45 | |
| Above average | 93 | 72 | 135 | |
What is the probability that a salesperson selected at random has above-average sales ability and has excellent potential for advancement?
Answer: C
Explanation: The events are not independent. P(above-average ability and excellent potential) = P(above-average ability) P(excellent potential | above-average ability) = (300/500) (135/300) = 135/500 = 0.27.
Difficulty: 2 Medium
Topic: Contingency Tables
Learning Objective: 05-05 Compute probabilities using a contingency table.
Bloom’s: Apply
AACSB: Analytic
65) Each salesperson in a large department store chain is rated on his or her sales ability and potential for advancement. The data for the 500 sampled salespeople are summarized in the following table.
| Potential for Advancement | ||||
| Fair | Good | Excellent | ||
| Sales ability | Below average | 16 | 12 | 22 |
| Average | 45 | 60 | 45 | |
| Above average | 93 | 72 | 135 | |
What is the probability that a salesperson selected at random will have average sales ability and good potential for advancement?
Answer: B
Explanation: The events are not independent. P(average ability and good potential) = P(average ability) P(good potential | average ability) = (150/500)(60/150) = 60/500 = 0.12.
Difficulty: 2 Medium
Topic: Contingency Tables
Learning Objective: 05-05 Compute probabilities using a contingency table.
Bloom’s: Apply
AACSB: Analytic
66) Each salesperson in a large department store chain is rated on his or her sales ability and potential for advancement. The data for the 500 sampled salespeople are summarized in the following table.
| Potential for Advancement | ||||
| Fair | Good | Excellent | ||
| Sales ability | Below average | 16 | 12 | 22 |
| Average | 45 | 60 | 45 | |
| Above average | 93 | 72 | 135 | |
What is the probability that a salesperson selected at random will have below-average sales ability and fair potential for advancement?
Answer: A
Explanation: The events are not independent. P(below average ability and fair potential) = P(below average ability) P(fair potential | below average ability) = (60/500)(16/60) = 16/500 = 0.032.
Difficulty: 2 Medium
Topic: Contingency Tables
Learning Objective: 05-05 Compute probabilities using a contingency table.
Bloom’s: Apply
AACSB: Analytic
67) Each salesperson in a large department store chain is rated on his or her sales ability and potential for advancement. The data for the 500 sampled salespeople are summarized in the following table.
| Potential for Advancement | ||||
| Fair | Good | Excellent | ||
| Sales ability | Below average | 16 | 12 | 22 |
| Average | 45 | 60 | 45 | |
| Above average | 93 | 72 | 135 | |
What is the probability that a salesperson selected at random will have an excellent potential for advancement given he or she also has above-average sales ability?
Answer: D
Explanation: This is a conditional probability: P(excellent potential | above-average ability)(135/300) = 0.45. There are 300 salespeople who are above average; 135 have excellent potential.
Difficulty: 2 Medium
Topic: Contingency Tables
Learning Objective: 05-05 Compute probabilities using a contingency table.
Bloom’s: Apply
AACSB: Analytic
68) Each salesperson in a large department store chain is rated on his or her sales ability and potential for advancement. The data for the 500 sampled salespeople are summarized in the following table.
| Potential for Advancement | ||||
| Fair | Good | Excellent | ||
| Sales ability | Below average | 16 | 12 | 22 |
| Average | 45 | 60 | 45 | |
| Above average | 93 | 72 | 135 | |
What is the probability that a salesperson selected at random will have an excellent potential for advancement given he or she also has average sales ability?
Answer: B
Explanation: This is a conditional probability: P(excellent potential | average ability)(45/150) = 0.30. There are 150 salespeople who are average; 45 have excellent potential.
Difficulty: 2 Medium
Topic: Contingency Tables
Learning Objective: 05-05 Compute probabilities using a contingency table.
Bloom’s: Apply
AACSB: Analytic
69) A study of interior designers’ opinions with respect to the most desirable primary color for executive offices showed the following:
| Primary Color | Number of Opinions |
| Red | 92 |
| Orange | 86 |
| Yellow | 46 |
| Green | 91 |
| Blue | 37 |
| Indigo | 46 |
| Violet | 2 |
What is the probability that a designer does NOT prefer red?
Answer: B
Explanation: Using the complement rule, P(red) = 92/400 = 0.23. Therefore, P(not red) = 1.00 − 0.23 = 0.77.
Difficulty: 2 Medium
Topic: Rules of Addition for Computing Probabilities
Learning Objective: 05-03 Calculate probabilities using the rules of addition.
Bloom’s: Apply
AACSB: Analytic
70) A study of interior designers’ opinions with respect to the most desirable primary color for executive offices showed the following:
| Primary Color | Number of Opinions |
| Red | 92 |
| Orange | 86 |
| Yellow | 46 |
| Green | 91 |
| Blue | 37 |
| Indigo | 46 |
| Violet | 2 |
What is the probability that a designer does NOT prefer yellow?
Answer: C
Explanation: Using the complement rule, P(yellow) = 46/400 = 0.115. Therefore, P(not yellow) = 1.00 − 0.115 = 0.885.
Difficulty: 2 Medium
Topic: Rules of Addition for Computing Probabilities
Learning Objective: 05-03 Calculate probabilities using the rules of addition.
Bloom’s: Apply
AACSB: Analytic
71) A study of interior designers’ opinions with respect to the most desirable primary color for executive offices showed the following:
| Primary Color | Number of Opinions |
| Red | 92 |
| Orange | 86 |
| Yellow | 46 |
| Green | 91 |
| Blue | 37 |
| Indigo | 46 |
| Violet | 2 |
What is the probability that a designer does NOT prefer blue?
Answer: B
Explanation: Using the complement rule, P(blue) = 37/400 = 0.0925. Therefore, P(not blue) = 1.00 − 0.0925 = 0.9075.
Difficulty: 2 Medium
Topic: Rules of Addition for Computing Probabilities
Learning Objective: 05-03 Calculate probabilities using the rules of addition.
Bloom’s: Apply
AACSB: Analytic
72) An automatic machine inserts mixed vegetables into a plastic bag. Past experience revealed that some packages were underweight and some were overweight, but most of them had satisfactory weight.
| Weight | % of Total |
| Underweight | 2.5 |
| Satisfactory | 90.0 |
| Overweight | 7.5 |
What is the probability of selecting three packages that are overweight?
Answer: B
Explanation: Apply the multiplication rule: (0.075)(0.075)(0.075) = 0.0004219.
Difficulty: 2 Medium
Topic: Rules of Multiplication to Calculate Probability
Learning Objective: 05-04 Calculate probabilities using the rules of multiplication.
Bloom’s: Apply
AACSB: Analytic
73) An automatic machine inserts mixed vegetables into a plastic bag. Past experience revealed that some packages were underweight and some were overweight, but most of them had satisfactory weight.
| Weight | % of Total |
| Underweight | 2.5 |
| Satisfactory | 90.0 |
| Overweight | 7.5 |
What is the probability of selecting three packages that are satisfactory?
Answer: C
Explanation: Apply the multiplication rule: (0.90)(0.90)(0.90) = 0.729.
Difficulty: 2 Medium
Topic: Rules of Multiplication to Calculate Probability
Learning Objective: 05-04 Calculate probabilities using the rules of multiplication.
Bloom’s: Apply
AACSB: Analytic
74) In a finance class, the final grade is based on three tests. Historically, the instructor tells the class that the joint probability of scoring As on the first two tests is 0.5. A student assigns a probability of 0.9 that she will get an A on the first test. What is the probability that the student will score an A on the second test given that she scored an A on the first test?
Answer: C
Explanation: The general rule of multiplication states that P(A and B) = P(A|B) P(B), so P(A|B) = P(A and B)/P(B). Therefore, P(A on second test| A on first test) = P(A on both tests) / P(A on first test) = 0.5/0.9 = 0.56.
Difficulty: 3 Hard
Topic: Bayes Theorem
Learning Objective: 05-06 Calculate probabilities using Bayes theorem.
Bloom’s: Apply
AACSB: Analytic
Accessibility: Keyboard Navigation
75) Using the terminology of Bayes’ theorem, a posterior probability can also be defined as a
Answer: A
Explanation: A posterior probability is a revised probability based on additional information.
Difficulty: 1 Easy
Topic: Bayes Theorem
Learning Objective: 05-06 Calculate probabilities using Bayes theorem.
Bloom’s: Understand
AACSB: Communication
Accessibility: Keyboard Navigation
76) If A and B are mutually exclusive events with P(A) = 0.2 and P(B) = 0.6, then P(A or B) = ________.
Answer: C
Explanation: Use formula [5-4]: P(A or B) = P(A) + P(B) − P(A and B). Since the events are mutually exclusive, P(A and B) = 0 and P(A or B) = (0.20) + (0.60) = 0.80.
Difficulty: 2 Medium
Topic: Rules of Addition for Computing Probabilities
Learning Objective: 05-03 Calculate probabilities using the rules of addition.
Bloom’s: Apply
AACSB: Analytic
77) If P(A) = 0.62, P(B) = 0.47, and P(A or B) = 0.88, then P(A and B) = ________.
Answer: D
Explanation: Use formula [5-4]: P(A or B) = P(A) + P(B) − P(A and B). Rearrange formula [5-4]: P(A and B) = P(A) + P(B) − P(A or B) = 0.62 + 0.47 − 0.88 = 0.21. P(A and B) = 0.21.
Difficulty: 2 Medium
Topic: Rules of Addition for Computing Probabilities
Learning Objective: 05-03 Calculate probabilities using the rules of addition.
Bloom’s: Apply
AACSB: Analytic
78) A method of assigning probabilities based upon judgment, opinions, and available information is referred to as the ________.
Answer: D
Difficulty: 1 Easy
Topic: Approaches to Assigning Probabilities
Learning Objective: 05-02 Assign probabilities using a classical, empirical, or subjective approach.
Bloom’s: Remember
AACSB: Communication
Accessibility: Keyboard Navigation
79) Your favorite soccer team has two remaining matches to complete the season. The possible outcomes of a soccer match are win, lose, or tie. What is the possible number of outcomes for the season?
Answer: D
Explanation: Use the tree diagram to determine the possible outcomes, which are: {W,W}, {W,L}, {W,T}, {L,W}, {L,L}, {L,T}, {T,W}, {T,L}, {T,T}. There are 9 possible outcomes.
Difficulty: 2 Medium
Topic: Principles of Counting
Learning Objective: 05-07 Determine the number of outcomes using principles of counting.
Bloom’s: Apply
AACSB: Analytic
Accessibility: Keyboard Navigation
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Original price was: $30.00.$20.00Current price is: $20.00.
Original price was: $30.00.$20.00Current price is: $20.00.
Original price was: $30.00.$20.00Current price is: $20.00.
Original price was: $30.00.$20.00Current price is: $20.00.
Original price was: $30.00.$20.00Current price is: $20.00.
