Applied Statistics in Business And Economics 5th Edition by Doane -Test Bank Instant Download - Complete Test Bank With Answers Sample Questions Are Posted Below Chapter 05 Probability True / False Questions 1. A sample space is the set of all possible outcomes in an experiment. True False 2. The …
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Applied Statistics in Business And Economics 5th Edition by Doane -Test Bank
Instant Download – Complete Test Bank With Answers
Sample Questions Are Posted Below
Chapter 05
Probability
True / False Questions
| 1. | A sample space is the set of all possible outcomes in an experiment.
True False |
| 2. | The sum of all the probabilities of simple events in a sample space equals one.
True False |
| 3. | The sum of the probabilities of all compound events in a sample space equals one.
True False |
| 4. | Probability is the measure of the relative likelihood that an event will occur.
True False |
| 5. | The probability of the union of two events P(A or B) can exceed one.
True False |
| 6. | The empirical view of probability is based on relative frequencies.
True False |
| 7. | Grandma’s predicting rain based on how much her arthritis is acting up is an example of the classical view of probability.
True False |
| 8. | The odds of an event can be calculated by dividing the event’s probability by the probability of its complement.
True False |
| 9. | The union of two events A and B is the event consisting of all outcomes in the sample space that are contained in both event A and event B.
True False |
| 10. | The general law of addition for probabilities says P(A or B) = P(A) + P(B) – P(A∩B).
True False |
| 11. | Events A and B are mutually exclusive if P(A∩B) = 0.
True False |
| 12. | Independent events are mutually exclusive.
True False |
| 13. | If events A and B are mutually exclusive, then P(A) + P(B) = 0.
True False |
| 14. | P(A | B) is the joint probability of events A and B divided by the probability of A.
True False |
| 15. | Two events A and B are independent if P(A | B) is the same as P(A).
True False |
| 16. | If events A and B are dependent, it can be concluded that one event causes the other.
True False |
| 17. | For any event A, the probability of A is always 0 ≤ P(A) ≤ 1.
True False |
| 18. | If events A and B are mutually exclusive, the joint probability of the events is zero.
True False |
| 19. | When the outcome of a random experiment is a continuous measurement, the sample space is described by a rule instead of listing the possible simple events.
True False |
| 20. | If A and B are independent events, then P(A or B) = P(A) P(B).
True False |
| 21. | If A and B are mutually exclusive events, then P(A∩B) = P(A) + P(B).
True False |
| 22. | The probability of A and its complement (A´) will always sum to one.
True False |
| 23. | If event A occurs, then its complement (A´) will also occur.
True False |
| 24. | The sum of the probabilities of two mutually exclusive events is one.
True False |
| 25. | P(A∩B) = .50 is an example of a joint probability.
True False |
| 26. | The general law of addition for probabilities says P(A or B) = P(A) P(B).
True False |
| 27. | If P(A) = 0.50, P(B) = 0.30, and P(A∩B) = 0.15, then A and B are independent events.
True False |
| 28. | Insurance company life tables are an example of the classical (a priori) approach to probability.
True False |
| 29. | When two events cannot occur at the same time, they are said to be mutually exclusive.
True False |
| 30. | The probability of events A or B occurring can be found by summing their probabilities.
True False |
| 31. | When two events A and B are independent, the probability of their intersection can be found by multiplying their probabilities.
True False |
| 32. | Two events are mutually exclusive when they contain no outcomes in common.
True False |
| 33. | In a contingency table, the probability of the union of two events is found by taking the frequency of the intersection of the two events and dividing by the total.
True False |
| 34. | Bayes’ Theorem shows how to revise a prior probability to obtain a conditional or posterior probability when another event’s occurrence is known.
True False |
| 35. | The union of two events is all outcomes in either or both, while the intersection is only those events in both.
True False |
| 36. | A contingency table is a cross-tabulation of frequencies for two categorical variables.
True False |
| 37. | The number of arrangements of sampled items drawn from a population is found with the formula for permutations (if order is important) or combinations (if order does not matter).
True False |
| 38. | If P(A) = .20 then the odds against event A‘s occurrence are 4 to 1.
True False |
| 39. | The value of 7! is 5040.
True False |
Multiple Choice Questions
| 40. | Events A and B are mutually exclusive when:
|
| 41. | If two events are complementary, then we know that:
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| 42. | Regarding probability, which of the following is correct?
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| 43. | Independent events A and B would be consistent with which of the following statements:
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| 44. | Find the probability that either event A or B occurs if the chance of A occurring is .5, the chance of B occurring is .3, and events A and B are independent.
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| 45. | Regarding the rules of probability, which of the following statements is correct?
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| 46. | Within a given population, 22 percent of the people are smokers, 57 percent of the people are males, and 12 percent are males who smoke. If a person is chosen at random from the population, what is the probability that the selected person is either a male or a smoker?
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| 47. | Information was collected on those who attended the opening of a new movie. The analysis found that 56 percent of the moviegoers were female, 26 percent were under age 25, and 17 percent were females under the age of 25. Find the probability that a moviegoer is either female or under age 25.
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| 48. | Given the contingency table shown here, find P(V).
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| 49. | Given the contingency table shown here, find P(V | W).
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| 50. | Given the contingency table shown here, find the probability P(V´), that is, the probability of the complement of V.
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| 51. | Given the contingency table shown here, find P(W∩S).
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| 52. | Given the contingency table shown here, find P(A or M).
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| 53. | Given the contingency table shown here, find P(A2).
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| 54. | Given the contingency table shown here, find P(A3∩B2).
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| 55. | Given the contingency table shown here, find P(A2 | B3).
|
| 56. | Given the contingency table shown here, find P(A1 or B2).
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| 57. | Given the contingency table shown here, find P(A1∩A2).
|
| 58. | Given the contingency table shown here, find the probability that either event A2 or event B2 will occur.
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| 59. | Given the contingency table shown here, find P(B).
|
| 60. | Given the contingency table shown here, find P(A or B).
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| 61. | Given the contingency table shown here, find P(B | A).
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| 62. | Given the contingency table shown here, what is the probability that a randomly chosen employee who is under age 25 would be absent 2 or more days?
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| 63. | Oxnard Casualty wants to ensure that their e-mail server has 99.98 percent reliability. They will use several independent servers in parallel, each of which is 95 percent reliable. What is the smallest number of independent file servers that will accomplish the goal?
|
| 64. | Given the contingency table shown here, does the decision to retire appear independent of the employee type?
Survey question: Do you plan on retiring or keep working when you turn 65?
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| 65. | Given the contingency table shown here, find the probability that a randomly chosen employee is a line worker who plans to retire at age 65.
Survey question: Do you plan on retiring or keep working when you turn 65?
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| 66. | Given the contingency table shown here, find P(R∩L).
Survey question: Do you plan on retiring or keep working when you turn 65?
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| 67. | Given the contingency table shown here, find P(W | M).
Survey question: Do you plan on retiring or keep working when you turn 65?
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| 68. | Given the contingency table shown here, find P(L or W).
Survey question: Do you plan on retiring or keep working when you turn 65?
|
| 69. | Ramjac Company wants to set up k independent file servers, each capable of running the company’s intranet. Each server has average “uptime” of 98 percent. What must k be to achieve 99.999 percent probability that the intranet will be “up”?
|
| 70. | Given the contingency table shown here, what is the probability that a mother in the study smoked during pregnancy?
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| 71. | Given the contingency table shown here, what is the probability that a mother smoked during pregnancy if her education level was below high school?
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| 72. | Given the contingency table shown here, what is the probability that a mother smoked during pregnancy and had a college degree?
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| 73. | Given the contingency table shown here, what is the probability that a mother smoked during pregnancy or that she graduated from college?
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| 74. | Given the contingency table shown here, if a mother attended some college but did not have a degree, what is the probability that she did not smoke during her pregnancy?
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| 75. | Given the contingency table shown here, find the probability that a mother with some college smoked during pregnancy.
|
| 76. | Given the contingency table shown here, if a survey participant is selected at random, what is the probability he/she is an undergrad who favors the change to a quarter system?
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| 77. | Given the contingency table shown here, if a faculty member is chosen at random, what is the probability he/she opposes the change to a quarter system?
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| 78. | Given the contingency table shown here, what is the probability that a participant selected at random is a graduate student who opposes the change to a quarter system?
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| 79. | Given the contingency table shown here, what is the probability that a student attends a public school in a rural area?
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| 80. | Given the contingency table shown here, if a randomly chosen student attends a religious school, what is the probability the location is rural?
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| 81. | Given the contingency table shown here, if a randomly chosen student attends school in an inner-city location, what is the probability that it is a public school?
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| 82. | Given the contingency table shown here, find P(E).
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| 83. | Given the contingency table shown here, find P(E | F).
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| 84. | Given the contingency table shown here, find P(A∩M).
|
| 85. | Given the contingency table shown here, find P(F or G).
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| 86. | Given the contingency table shown here, find the probability that a randomly chosen individual is a female economics major.
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| 87. | Debbie has two stocks, X and Y. Consider the following events:
X = the event that the price of stock X has increased
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| 88. | If P(A | B) = 0.40 and P(B) = 0.30, find P(A∩B).
|
| 89. | A company is producing two types of ski goggles. Thirty percent of the production is of type A, and the rest is of type B. Five percent of all type A goggles are returned within 10 days after the sale, whereas only two percent of type B are returned. If a pair of goggles is returned within the first 10 days after the sale, the probability that the goggles returned are of type B is:
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| 90. | Given the contingency table shown here, find the joint probability that a call sampled at random out of this population is local and 2-5 minutes long.
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| 91. | Given the contingency table shown here, if a call is sampled at random, find the marginal probability that the call is long distance.
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| 92. | If a call is sampled at random, the conditional probability that the call is not “6+” minutes long given that it is a long distance call is:
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| 93. | The following table gives a classification of the 10,000 shareholders of Oxnard Xylophone Distributors, Inc. A few numbers are missing from the table. Given that a shareholder holding 500-999 shares is picked, there is a 0.625 probability that the shareholder will be a woman. Consequently, what is the number of men holding 1000 or more shares?
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| 94. | In any sample space P(A | B) and P(B | A):
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| 95. | If P(A∩B) = 0.50, can P(A) = 0.20?
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| 96. | The following relationship always holds true for events A and B in a sample space.
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| 97. | The following probabilities are given about events A and B in a sample space: P(A) = 0.30, P(B) = 0.40, P(A or B) = 0.60. We can say that:
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| 98. | If P(A) = 0.35, P(B) = 0.60, and P(A or B) = 0.70, then:
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| 99. | The following table shows the survival experience of 1,000 males who retire at age 65:
Based on these data, the probability that a 75-year-old male will survive to age 80 is:
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| 100. | Given the contingency table shown here, find P(G | M).
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| 101. | Given the contingency table shown here, find P(V or S).
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| 102. | Given the contingency table shown here, find P(V | S).
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| 103. | The manager of Ardmore Pharmacy knows that 25 percent of the customers entering the store buy prescription drugs, 65 percent buy over-the-counter drugs, and 18 percent buy both types of drugs. What is the probability that a randomly selected customer will buy at least one of these two types of drugs?
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| 104. | Two events are complementary (i.e., they are complements) if:
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| 105. | Which statement is false?
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| 106. | The number of unique orders in which five items (A, B, C, D, E) can be arranged is:
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| 107. | If four items are chosen at random without replacement from seven items, in how many ways can the four items be arranged, treating each arrangement as a different event (i.e., if order is important)?
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| 108. | How many ways can we choose three items at random without replacement from five items (A, B, C, D, E) if the order of the selected items is not important?
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| 109. | The value of 6C2 is:
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| 110. | The value of 4P2 is:
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| 111. | The probability that event A occurs, given that event B has occurred, is an example of:
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| 112. | If each of two independent file servers has a reliability of 93 percent and either alone can run the website, then the overall website availability is:
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| 113. | In a certain city, 5 percent of all drivers have expired licenses, 10 percent have an unpaid parking ticket, and 1 percent have both an expired license and an unpaid parking ticket. Are these events independent?
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| 114. | In a certain city, 5 percent of all drivers have expired licenses and 10 percent have an unpaid parking ticket. If these events are independent, what is the probability that a driver has both an expired license and an unpaid parking ticket?
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| 115. | If two events are collectively exhaustive, what is the probability that one or the other will occur?
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| 116. | Which best exemplifies a subjective probability?
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| 117. | Which best exemplifies the classical definition of probability?
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| 118. | Which best exemplifies the empirical definition of probability?
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| 119. | From the following tree, find the probability that a randomly chosen person will get the flu vaccine and will also get the flu.
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| 120. | From the following tree, find the probability that a randomly chosen person will not get a vaccination and will not get the flu.
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| 121. | From the following tree, find the probability that a randomly chosen person will get the flu.
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| 122. | At Joe’s Restaurant, 80 percent of the diners are new customers (N), while 20 percent are returning customers (R). Fifty percent of the new customers pay by credit card, compared with 70 percent of the regular customers. If a customer pays by credit card, what is the probability that the customer is a new customer?
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| 123. | At Dolon General Hospital, 30 percent of the patients have Medicare insurance (M) while 70 percent do not have Medicare insurance (M´). Twenty percent of the Medicare patients arrive by ambulance, compared with 10 percent of the non-Medicare patients. If a patient arrives by ambulance, what is the probability that the patient has Medicare insurance?
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Short Answer Questions
| 124. | Axolotl, Inc. decided to double the number of independent network file servers, each server being capable of running its entire internal corporate network. Bob suggested that “doubling the number of file servers in our corporate network will just double the chances of failure.” Is Bob right? Has the company made an error?
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| 125. | A baseball player said, “Yes, I’m in a batting slump. But the law of averages has got to catch up sooner or later. I’m bound to improve.” Discuss the logic of this statement.
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| 126. | Recently, a few companies have been accused of issuing stock options to their top executives at the lowest stock price of the year. Executives can later sell these stock shares at a higher price and make a large profit. In some companies, this happened several years in a row. An executive argued, “It’s just a coincidence. There are about 250 trading days in a year, so there is a reasonable chance that my option price just happened to be lowest of the year.” Discuss.
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Chapter 05 Probability Answer Key
True / False Questions
| 1. | A sample space is the set of all possible outcomes in an experiment.
TRUE Review the definition of sample space. |
| AACSB: Analytical Thinking Accessibility: Keyboard Navigation Blooms: Remember Difficulty: 1 Easy Learning Objective: 05-01 Describe the sample space of a random experiment. Topic: Random Experiments |
| 2. | The sum of all the probabilities of simple events in a sample space equals one.
TRUE Simple events are nonoverlapping, so they sum to unity. |
| AACSB: Analytical Thinking Accessibility: Keyboard Navigation Blooms: Understand Difficulty: 1 Easy Learning Objective: 05-03 Apply the definitions and rules of probability. Topic: Rules of Probability |
| 3. | The sum of the probabilities of all compound events in a sample space equals one.
FALSE Compound events may overlap, so you can’t simply add their probabilities. |
| AACSB: Analytical Thinking Accessibility: Keyboard Navigation Blooms: Understand Difficulty: 1 Easy Learning Objective: 05-03 Apply the definitions and rules of probability. Topic: Probability |
| 4. | Probability is the measure of the relative likelihood that an event will occur.
TRUE This is one of three ways to view probability. |
| AACSB: Analytical Thinking Accessibility: Keyboard Navigation Blooms: Remember Difficulty: 1 Easy Learning Objective: 05-03 Apply the definitions and rules of probability. Topic: Probability |
| 5. | The probability of the union of two events P(A or B) can exceed one.
FALSE Review the General Rule of Addition. No event probability can exceed one. |
| AACSB: Analytical Thinking Accessibility: Keyboard Navigation Blooms: Understand Difficulty: 2 Medium Learning Objective: 05-03 Apply the definitions and rules of probability. Topic: Rules of Probability |
| 6. | The empirical view of probability is based on relative frequencies.
TRUE This is one view of probability, based on experience. |
| AACSB: Analytical Thinking Accessibility: Keyboard Navigation Blooms: Remember Difficulty: 1 Easy Learning Objective: 05-02 Distinguish among the three views of probability. Topic: Probability |
| 7. | Grandma’s predicting rain based on how much her arthritis is acting up is an example of the classical view of probability.
FALSE This would be a subjective probability, where no data are available. |
| AACSB: Analytical Thinking Accessibility: Keyboard Navigation Blooms: Apply Difficulty: 2 Medium Learning Objective: 05-02 Distinguish among the three views of probability. Topic: Probability |
| 8. | The odds of an event can be calculated by dividing the event’s probability by the probability of its complement.
TRUE Review the definition of odds. |
| AACSB: Analytical Thinking Accessibility: Keyboard Navigation Blooms: Remember Difficulty: 2 Medium Learning Objective: 05-04 Calculate odds from given probabilities. Topic: Probability |
| 9. | The union of two events A and B is the event consisting of all outcomes in the sample space that are contained in both event A and event B.
FALSE The union is all outcomes in either A or B. |
| AACSB: Analytical Thinking Accessibility: Keyboard Navigation Blooms: Remember Difficulty: 2 Medium Learning Objective: 05-03 Apply the definitions and rules of probability. Topic: Rules of Probability |
| 10. | The general law of addition for probabilities says P(A or B) = P(A) + P(B) – P(A∩B).
TRUE We must subtract the probability of the event intersection from the sum. |
| AACSB: Analytical Thinking Accessibility: Keyboard Navigation Blooms: Remember Difficulty: 1 Easy Learning Objective: 05-03 Apply the definitions and rules of probability. Topic: Rules of Probability |
| 11. | Events A and B are mutually exclusive if P(A∩B) = 0.
TRUE Nonoverlapping events cannot both occur (intersection is the empty set). |
| AACSB: Analytical Thinking Accessibility: Keyboard Navigation Blooms: Remember Difficulty: 2 Medium Learning Objective: 05-03 Apply the definitions and rules of probability. Topic: Rules of Probability |
| 12. | Independent events are mutually exclusive.
FALSE Only if P(A∩B) = 0 would they be mutually exclusive. |
| AACSB: Analytical Thinking Accessibility: Keyboard Navigation Blooms: Remember Difficulty: 2 Medium Learning Objective: 05-05 Determine when events are independent. Topic: Independent Events |
| 13. | If events A and B are mutually exclusive, then P(A) + P(B) = 0.
FALSE Their intersection would have zero probability, but not their sum. |
| AACSB: Analytical Thinking Accessibility: Keyboard Navigation Blooms: Remember Difficulty: 2 Medium Learning Objective: 05-03 Apply the definitions and rules of probability. Topic: Rules of Probability |
| 14. | P(A | B) is the joint probability of events A and B divided by the probability of A.
FALSE You would divide the joint probability by P(B) rather than by P(A). |
| AACSB: Analytical Thinking Accessibility: Keyboard Navigation Blooms: Remember Difficulty: 2 Medium Learning Objective: 05-03 Apply the definitions and rules of probability. Topic: Rules of Probability |
| 15. | Two events A and B are independent if P(A | B) is the same as P(A).
TRUE This is the definition of independence. Event B does not affect event A. |
| AACSB: Analytical Thinking Accessibility: Keyboard Navigation Blooms: Remember Difficulty: 1 Easy Learning Objective: 05-05 Determine when events are independent. Topic: Independent Events |
| 16. | If events A and B are dependent, it can be concluded that one event causes the other.
FALSE Maybe there is causation, but both events could be affected by something else. |
| AACSB: Analytical Thinking Accessibility: Keyboard Navigation Blooms: Understand Difficulty: 2 Medium Learning Objective: 05-05 Determine when events are independent. Topic: Independent Events |
| 17. | For any event A, the probability of A is always 0 ≤ P(A) ≤ 1.
TRUE Any probability must be between 0 and 1. |
| AACSB: Analytical Thinking Accessibility: Keyboard Navigation Blooms: Remember Difficulty: 1 Easy Learning Objective: 05-03 Apply the definitions and rules of probability. Topic: Probability |
| 18. | If events A and B are mutually exclusive, the joint probability of the events is zero.
TRUE The intersection of nonoverlapping events is the empty set. |
| AACSB: Analytical Thinking Accessibility: Keyboard Navigation Blooms: Remember Difficulty: 2 Medium Learning Objective: 05-03 Apply the definitions and rules of probability. Topic: Rules of Probability |
| 19. | When the outcome of a random experiment is a continuous measurement, the sample space is described by a rule instead of listing the possible simple events.
TRUE For example, speeds of vehicles on a freeway are not listable (nonintegers). |
| AACSB: Analytical Thinking Accessibility: Keyboard Navigation Blooms: Understand Difficulty: 1 Easy Learning Objective: 05-01 Describe the sample space of a random experiment. Topic: Random Experiments |
| 20. | If A and B are independent events, then P(A or B) = P(A) P(B).
FALSE Review the definition of independence. The proffered statement would have been correct if “and” were substituted for “or.” The correct statement would be P(A or B) = P(A) + P(B) – P(A∩B). |
| AACSB: Analytical Thinking Accessibility: Keyboard Navigation Blooms: Remember Difficulty: 2 Medium Learning Objective: 05-05 Determine when events are independent. Topic: Independent Events |
| 21. | If A and B are mutually exclusive events, then P(A∩B) = P(A) + P(B).
FALSE Their intersection is empty so it has zero probability. |
| AACSB: Analytical Thinking Accessibility: Keyboard Navigation Blooms: Remember Difficulty: 2 Medium Learning Objective: 05-03 Apply the definitions and rules of probability. Topic: Rules of Probability |
| 22. | The probability of A and its complement (A´) will always sum to one.
TRUE An event and its complement comprise the sample space. |
| AACSB: Analytical Thinking Accessibility: Keyboard Navigation Blooms: Remember Difficulty: 2 Medium Learning Objective: 05-03 Apply the definitions and rules of probability. Topic: Rules of Probability |
| 23. | If event A occurs, then its complement (A´) will also occur.
FALSE An event and its complement are nonoverlapping. |
| AACSB: Analytical Thinking Accessibility: Keyboard Navigation Blooms: Remember Difficulty: 1 Easy Learning Objective: 05-03 Apply the definitions and rules of probability. Topic: Rules of Probability |
| 24. | The sum of the probabilities of two mutually exclusive events is one.
FALSE Review definitions for event probabilities. No probability can exceed one. |
| AACSB: Analytical Thinking Accessibility: Keyboard Navigation Blooms: Remember Difficulty: 2 Medium Learning Objective: 05-03 Apply the definitions and rules of probability. Topic: Rules of Probability |
| 25. | P(A∩B) = .50 is an example of a joint probability.
TRUE Intersection of two events yields a joint probability. |
| AACSB: Analytical Thinking Accessibility: Keyboard Navigation Blooms: Remember Difficulty: 1 Easy Learning Objective: 05-03 Apply the definitions and rules of probability. Topic: Rules of Probability |
| 26. | The general law of addition for probabilities says P(A or B) = P(A) P(B).
FALSE Review the General Law of Addition: P(A or B) = P(A) + P(B) – P(A∩B). |
| AACSB: Analytical Thinking Accessibility: Keyboard Navigation Blooms: Remember Difficulty: 1 Easy Learning Objective: 05-03 Apply the definitions and rules of probability. Topic: Rules of Probability |
| 27. | If P(A) = 0.50, P(B) = 0.30, and P(A∩B) = 0.15, then A and B are independent events.
TRUE This is true because for independent events: P(A∩B) = P(A) P(B) and in this example P(A) P(B) = (.50)(.30) = .15 = P(A∩B). |
| AACSB: Analytical Thinking Accessibility: Keyboard Navigation Blooms: Apply Difficulty: 2 Medium Learning Objective: 05-05 Determine when events are independent. Topic: Independent Events |
| 28. | Insurance company life tables are an example of the classical (a priori) approach to probability.
FALSE Actuarial data are based on relative frequencies (empirical probability). |
| AACSB: Analytical Thinking Accessibility: Keyboard Navigation Blooms: Understand Difficulty: 2 Medium Learning Objective: 05-02 Distinguish among the three views of probability. Topic: Probability |
| 29. | When two events cannot occur at the same time, they are said to be mutually exclusive.
TRUE This is the definition of mutually exclusive events. |
| AACSB: Analytical Thinking Accessibility: Keyboard Navigation Blooms: Remember Difficulty: 2 Medium Learning Objective: 05-03 Apply the definitions and rules of probability. Topic: Rules of Probability |
| 30. | The probability of events A or B occurring can be found by summing their probabilities.
FALSE Event probabilities cannot be summed unless their intersection is the empty set. |
| AACSB: Analytical Thinking Accessibility: Keyboard Navigation Blooms: Understand Difficulty: 2 Medium Learning Objective: 05-03 Apply the definitions and rules of probability. Topic: Rules of Probability |
| 31. | When two events A and B are independent, the probability of their intersection can be found by multiplying their probabilities.
TRUE This is true by the definition of independent events. |
| AACSB: Analytical Thinking Accessibility: Keyboard Navigation Blooms: Remember Difficulty: 2 Medium Learning Objective: 05-05 Determine when events are independent. Topic: Independent Events |
| 32. | Two events are mutually exclusive when they contain no outcomes in common.
TRUE This is the definition of mutually exclusive events. |
| AACSB: Analytical Thinking Accessibility: Keyboard Navigation Blooms: Remember Difficulty: 1 Easy Learning Objective: 05-03 Apply the definitions and rules of probability. Topic: Rules of Probability |
| 33. | In a contingency table, the probability of the union of two events is found by taking the frequency of the intersection of the two events and dividing by the total.
FALSE Review the general law of addition P(A or B) = P(A) + P(B) – P(A∩B). The given statement is a joint probability P(A and B) rather than a union P(A or B). |
| AACSB: Analytical Thinking Accessibility: Keyboard Navigation Blooms: Understand Difficulty: 2 Medium Learning Objective: 05-06 Apply the concepts of probability to contingency tables. Topic: Contingency Tables |
| 34. | Bayes’ Theorem shows how to revise a prior probability to obtain a conditional or posterior probability when another event’s occurrence is known.
TRUE This is a rather general statement about Bayes’ contribution to statistics. |
| AACSB: Analytical Thinking Accessibility: Keyboard Navigation Blooms: Understand Difficulty: 2 Medium Learning Objective: 05-08 Use Bayes’ Theorem to calculate revised probabilities. Topic: Bayes’ Theorem |
| 35. | The union of two events is all outcomes in either or both, while the intersection is only those events in both.
TRUE Review definitions of union and intersection. |
| AACSB: Analytical Thinking Accessibility: Keyboard Navigation Blooms: Understand Difficulty: 2 Medium Learning Objective: 05-03 Apply the definitions and rules of probability. Topic: Rules of Probability |
| 36. | A contingency table is a cross-tabulation of frequencies for two categorical variables.
TRUE The categories define the rows and columns. |
| AACSB: Analytical Thinking Accessibility: Keyboard Navigation Blooms: Remember Difficulty: 1 Easy Learning Objective: 05-06 Apply the concepts of probability to contingency tables. Topic: Contingency Tables |
| 37. | The number of arrangements of sampled items drawn from a population is found with the formula for permutations (if order is important) or combinations (if order does not matter).
TRUE Review definitions of permutations and combinations. |
| AACSB: Analytical Thinking Accessibility: Keyboard Navigation Blooms: Remember Difficulty: 2 Medium Learning Objective: 05-09 Apply counting rules to calculate possible event arrangements. Topic: Counting Rules |
| 38. | If P(A) = .20 then the odds against event A‘s occurrence are 4 to 1.
TRUE Review definition of odds. |
| AACSB: Analytical Thinking Accessibility: Keyboard Navigation Blooms: Apply Difficulty: 2 Medium Learning Objective: 05-04 Calculate odds from given probabilities. Topic: Probability |
| 39. | The value of 7! is 5040.
TRUE Factorial n (written as n!) is 7! = 7 × 6 × 5 × 4 × 3 × 2 × 1 = 5040. |
| AACSB: Analytical Thinking Accessibility: Keyboard Navigation Blooms: Apply Difficulty: 1 Easy Learning Objective: 05-09 Apply counting rules to calculate possible event arrangements. Topic: Counting Rules |
Multiple Choice Questions
| 40. | Events A and B are mutually exclusive when:
Review definition of mutually exclusive. |
| AACSB: Analytical Thinking Accessibility: Keyboard Navigation Blooms: Understand Difficulty: 2 Medium Learning Objective: 05-03 Apply the definitions and rules of probability. Topic: Rules of Probability |
| 41. | If two events are complementary, then we know that:
Review definition of complementary events. |
| AACSB: Analytical Thinking Accessibility: Keyboard Navigation Blooms: Remember Difficulty: 2 Medium Learning Objective: 05-03 Apply the definitions and rules of probability. Topic: Rules of Probability |
| 42. | Regarding probability, which of the following is correct?
Review the rules of probability. |
| AACSB: Analytical Thinking Accessibility: Keyboard Navigation Blooms: Remember Difficulty: 2 Medium Learning Objective: 05-05 Determine when events are independent. Topic: Independent Events |
| 43. | Independent events A and B would be consistent with which of the following statements:
For independence, the product P(A)P(B) must equal P(A∩B). |
| AACSB: Analytical Thinking Accessibility: Keyboard Navigation Blooms: Apply Difficulty: 1 Easy Learning Objective: 05-05 Determine when events are independent. Topic: Independent Events |
| 44. | Find the probability that either event A or B occurs if the chance of A occurring is .5, the chance of B occurring is .3, and events A and B are independent.
Given that the events are independent, the product P(A)P(B) must equal P(A∩B). Thus, P(A or B) = P(A) + P(B) – P(A∩B) = .50 + .30 – (.50)(.30) = .80 – .15 = .65 using the General Law of Addition. |
| AACSB: Analytical Thinking Accessibility: Keyboard Navigation Blooms: Apply Difficulty: 2 Medium Learning Objective: 05-05 Determine when events are independent. Topic: Independent Events |
| 45. | Regarding the rules of probability, which of the following statements is correct?
Review the rules of probabilities. |
| AACSB: Analytical Thinking Accessibility: Keyboard Navigation Blooms: Remember Difficulty: 2 Medium Learning Objective: 05-03 Apply the definitions and rules of probability. Topic: Rules of Probability |
| 46. | Within a given population, 22 percent of the people are smokers, 57 percent of the people are males, and 12 percent are males who smoke. If a person is chosen at random from the population, what is the probability that the selected person is either a male or a smoker?
Use the General Law of Addition P(A or B) = P(A) + P(B) – P(A∩B). |
| AACSB: Analytical Thinking Accessibility: Keyboard Navigation Blooms: Apply Difficulty: 2 Medium Learning Objective: 05-03 Apply the definitions and rules of probability. Topic: Rules of Probability |
| 47. | Information was collected on those who attended the opening of a new movie. The analysis found that 56 percent of the moviegoers were female, 26 percent were under age 25, and 17 percent were females under the age of 25. Find the probability that a moviegoer is either female or under age 25.
Use the General Law of Addition P(A or B) = P(A) + P(B) – P(A∩B). |
| AACSB: Analytical Thinking Accessibility: Keyboard Navigation Blooms: Apply Difficulty: 3 Hard Learning Objective: 05-03 Apply the definitions and rules of probability. Topic: Rules of Probability |
| 48. | Given the contingency table shown here, find P(V).
This is a marginal probability P(V) = 40/200 = .20. |
| AACSB: Analytical Thinking Blooms: Apply Difficulty: 2 Medium Learning Objective: 05-06 Apply the concepts of probability to contingency tables. Topic: Contingency Tables |
| 49. | Given the contingency table shown here, find P(V | W).
This is a conditional probability P(V|W) = 19/80. |
| AACSB: Analytical Thinking Blooms: Apply Difficulty: 2 Medium Learning Objective: 05-06 Apply the concepts of probability to contingency tables. Topic: Contingency Tables |
| 50. | Given the contingency table shown here, find the probability P(V´), that is, the probability of the complement of V.
Calculate the probability of the complement of V by subtracting from its marginal probability P(V) = 40/200 to get P(V´) = 1 – P(V) = 1 – 40/200. |
| AACSB: Analytical Thinking Blooms: Apply Difficulty: 3 Hard Learning Objective: 05-06 Apply the concepts of probability to contingency tables. Topic: Contingency Tables |
| 51. | Given the contingency table shown here, find P(W∩S).
This is a joint probability P(W and S) = 24/200. |
| AACSB: Analytical Thinking Blooms: Apply Difficulty: 2 Medium Learning Objective: 05-06 Apply the concepts of probability to contingency tables. Topic: Contingency Tables |
| 52. | Given the contingency table shown here, find P(A or M).
Use the General Law of Addition P(A or M) = 100/200 + 50/200 – 25/200. |
| AACSB: Analytical Thinking Blooms: Apply Difficulty: 3 Hard Learning Objective: 05-06 Apply the concepts of probability to contingency tables. Topic: Contingency Tables |
| 53. | Given the contingency table shown here, find P(A2).
This is a marginal probability: P(A2) = 86/467. |
| AACSB: Analytical Thinking Blooms: Apply Difficulty: 2 Medium Learning Objective: 05-06 Apply the concepts of probability to contingency tables. Topic: Contingency Tables |
| 54. | Given the contingency table shown here, find P(A3∩B2).
This is a joint probability: P(A3 and B2) = 44/467. |
| AACSB: Analytical Thinking Blooms: Apply Difficulty: 2 Medium Learning Objective: 05-06 Apply the concepts of probability to contingency tables. Topic: Contingency Tables |
| 55. | Given the contingency table shown here, find P(A2 | B3).
This is a conditional probability: P(A2|B3) = 32/169. |
| AACSB: Analytical Thinking Blooms: Apply Difficulty: 2 Medium Learning Objective: 05-06 Apply the concepts of probability to contingency tables. Topic: Contingency Tables |
| 56. | Given the contingency table shown here, find P(A1 or B2).
Apply the General Law of Addition: P(A1 or B2) = 44/467 + 150/467 – 14/467. |
| AACSB: Analytical Thinking Blooms: Apply Difficulty: 3 Hard Learning Objective: 05-06 Apply the concepts of probability to contingency tables. Topic: Contingency Tables |
| 57. | Given the contingency table shown here, find P(A1∩A2).
This is a joint probability. The important thing here is that events A1 and A2 are mutually exclusive and so both events cannot occur. |
| AACSB: Analytical Thinking Blooms: Apply Difficulty: 2 Medium Learning Objective: 05-06 Apply the concepts of probability to contingency tables. Topic: Contingency Tables |
| 58. | Given the contingency table shown here, find the probability that either event A2 or event B2 will occur.
Use the General Law of Addition: P(A2 or B2) = 86/467 + 150/467 – 28/467. |
| AACSB: Analytical Thinking Blooms: Apply Difficulty: 3 Hard Learning Objective: 05-06 Apply the concepts of probability to contingency tables. Topic: Contingency Tables |
| 59. | Given the contingency table shown here, find P(B).
This is a marginal probability: P(B) = 90/200. |
| AACSB: Analytical Thinking Blooms: Apply Difficulty: 1 Easy Learning Objective: 05-06 Apply the concepts of probability to contingency tables. Topic: Contingency Tables |
| 60. | Given the contingency table shown here, find P(A or B).
Use the General Law of Addition: P(A or B) = 80/200 + 90/200 – 50/200. |
| AACSB: Analytical Thinking Blooms: Apply Difficulty: 2 Medium Learning Objective: 05-06 Apply the concepts of probability to contingency tables. Topic: Contingency Tables |
| 61. | Given the contingency table shown here, find P(B | A).
This is a conditional probability: P(B|A) = 50/80. |
| AACSB: Analytical Thinking Blooms: Apply Difficulty: 2 Medium Learning Objective: 05-06 Apply the concepts of probability to contingency tables. Topic: Contingency Tables |
| 62. | Given the contingency table shown here, what is the probability that a randomly chosen employee who is under age 25 would be absent 2 or more days?
This is a conditional probability: P(B‘|A) = 30/80. |
| AACSB: Analytical Thinking Blooms: Apply Difficulty: 2 Medium Learning Objective: 05-06 Apply the concepts of probability to contingency tables. Topic: Contingency Tables |
| 63. | Oxnard Casualty wants to ensure that their e-mail server has 99.98 percent reliability. They will use several independent servers in parallel, each of which is 95 percent reliable. What is the smallest number of independent file servers that will accomplish the goal?
1 – P(F1∩F2∩F3) = 1 – (.05) (.05) (.05) = 1 – .000125 = .999875, so 3 servers will do. |
| AACSB: Analytical Thinking Accessibility: Keyboard Navigation Blooms: Apply Difficulty: 3 Hard Learning Objective: 05-05 Determine when events are independent. Topic: Independent Events |
| 64. | Given the contingency table shown here, does the decision to retire appear independent of the employee type?
Survey question: Do you plan on retiring or keep working when you turn 65?
Does the product of the marginal probabilities equal their joint probability? This can be checked by asking whether P(M and R) = P(M) P(R). In this example, because (31/124)(52/124) = 13/124, we can see that M and R are independent events. |
| AACSB: Analytical Thinking Blooms: Apply Difficulty: 2 Medium Learning Objective: 05-06 Apply the concepts of probability to contingency tables. Topic: Contingency Tables |
| 65. | Given the contingency table shown here, find the probability that a randomly chosen employee is a line worker who plans to retire at age 65.
Survey question: Do you plan on retiring or keep working when you turn 65?
This is a joint probability: P(L and R) = 39/124. |
| AACSB: Analytical Thinking Blooms: Apply Difficulty: 2 Medium Learning Objective: 05-06 Apply the concepts of probability to contingency tables. Topic: Contingency Tables |
| 66. | Given the contingency table shown here, find P(R∩L).
Survey question: Do you plan on retiring or keep working when you turn 65?
This is a joint probability: P(R∩L) = 39/124. |
| AACSB: Analytical Thinking Blooms: Apply Difficulty: 2 Medium Learning Objective: 05-06 Apply the concepts of probability to contingency tables. Topic: Contingency Tables |
| 67. | Given the contingency table shown here, find P(W | M).
Survey question: Do you plan on retiring or keep working when you turn 65?
This is a conditional probability: P(W | M) = 18/31. |
| AACSB: Analytical Thinking Blooms: Apply Difficulty: 2 Medium Learning Objective: 05-06 Apply the concepts of probability to contingency tables. Topic: Contingency Tables |
| 68. | Given the contingency table shown here, find P(L or W).
Survey question: Do you plan on retiring or keep working when you turn 65?
Use the General Law of Addition: P(L or W) = 93/124 + 72/124 – 54/124. |
| AACSB: Analytical Thinking Blooms: Apply Difficulty: 2 Medium Learning Objective: 05-06 Apply the concepts of probability to contingency tables. Topic: Contingency Tables |
| 69. | Ramjac Company wants to set up k independent file servers, each capable of running the company’s intranet. Each server has average “uptime” of 98 percent. What must k be to achieve 99.999 percent probability that the intranet will be “up”?
1 – P(F1∩F2∩F3) = 1 – (.02)(.02)(.02) = 1 – .000008 = .999992, so 3 servers will do. |
| AACSB: Analytical Thinking Accessibility: Keyboard Navigation Blooms: Apply Difficulty: 3 Hard Learning Objective: 05-05 Determine when events are independent. Topic: Independent Events |
| 70. | Given the contingency table shown here, what is the probability that a mother in the study smoked during pregnancy?
This is a marginal probability: P(smoked) = 1122/4331. |
| AACSB: Analytical Thinking Blooms: Apply Difficulty: 1 Easy Learning Objective: 05-06 Apply the concepts of probability to contingency tables. Topic: Contingency Tables |
| 71. | Given the contingency table shown here, what is the probability that a mother smoked during pregnancy if her education level was below high school?
This is a conditional probability: P(smoked | below high school) = 393/1033. |
| AACSB: Analytical Thinking Blooms: Apply Difficulty: 2 Medium Learning Objective: 05-06 Apply the concepts of probability to contingency tables. Topic: Contingency Tables |
| 72. | Given the contingency table shown here, what is the probability that a mother smoked during pregnancy and had a college degree?
This is a joint probability: P(smoked and college) = 48/4331. |
| AACSB: Analytical Thinking Blooms: Apply Difficulty: 2 Medium Learning Objective: 05-06 Apply the concepts of probability to contingency tables. Topic: Contingency Tables |
| 73. | Given the contingency table shown here, what is the probability that a mother smoked during pregnancy or that she graduated from college?
Use the General Law of Addition: 1122/4331 + 598/4331 – 48/4331. |
| AACSB: Analytical Thinking Blooms: Apply Difficulty: 3 Hard Learning Objective: 05-06 Apply the concepts of probability to contingency tables. Topic: Contingency Tables |
| 74. | Given the contingency table shown here, if a mother attended some college but did not have a degree, what is the probability that she did not smoke during her pregnancy?
This is a conditional probability: 635/756. |
| AACSB: Analytical Thinking Blooms: Apply Difficulty: 3 Hard Learning Objective: 05-06 Apply the concepts of probability to contingency tables. Topic: Contingency Tables |
| 75. | Given the contingency table shown here, find the probability that a mother with some college smoked during pregnancy.
This is a conditional probability: 121/7561. |
| AACSB: Analytical Thinking Blooms: Apply Difficulty: 2 Medium Learning Objective: 05-06 Apply the concepts of probability to contingency tables. Topic: Contingency Tables |
| 76. | Given the contingency table shown here, if a survey participant is selected at random, what is the probability he/she is an undergrad who favors the change to a quarter system?
This is a joint probability: P(U and S) = 27/200. |
| AACSB: Analytical Thinking Blooms: Apply Difficulty: 2 Medium Learning Objective: 05-06 Apply the concepts of probability to contingency tables. Topic: Contingency Tables |
| 77. | Given the contingency table shown here, if a faculty member is chosen at random, what is the probability he/she opposes the change to a quarter system?
This is a marginal probability: P(N | F) = 20/50 = .40. |
| AACSB: Analytical Thinking Blooms: Apply Difficulty: 2 Medium Learning Objective: 05-06 Apply the concepts of probability to contingency tables. Topic: Contingency Tables |
| 78. | Given the contingency table shown here, what is the probability that a participant selected at random is a graduate student who opposes the change to a quarter system?
This is a joint probability: P(G and N) = 27/200. |
| AACSB: Analytical Thinking Blooms: Apply Difficulty: 2 Medium Learning Objective: 05-06 Apply the concepts of probability to contingency tables. Topic: Contingency Tables |
| 79. | Given the contingency table shown here, what is the probability that a student attends a public school in a rural area?
This is a marginal probability. |
| AACSB: Analytical Thinking Blooms: Apply Difficulty: 1 Easy Learning Objective: 05-06 Apply the concepts of probability to contingency tables. Topic: Contingency Tables |
| 80. | Given the contingency table shown here, if a randomly chosen student attends a religious school, what is the probability the location is rural?
This is a conditional probability: P(R | L) = 5/30 = .167. |
| AACSB: Analytical Thinking Blooms: Apply Difficulty: 2 Medium Learning Objective: 05-06 Apply the concepts of probability to contingency tables. Topic: Contingency Tables |
| 81. | Given the contingency table shown here, if a randomly chosen student attends school in an inner-city location, what is the probability that it is a public school?
This is a conditional probability: P(P | I) = 35/70 = .500. |
| AACSB: Analytical Thinking Blooms: Apply Difficulty: 2 Medium Learning Objective: 05-06 Apply the concepts of probability to contingency tables. Topic: Contingency Tables |
| 82. | Given the contingency table shown here, find P(E).
This is a marginal probability: P(E) = 300/1000 = .300. |
| AACSB: Analytical Thinking Blooms: Apply Difficulty: 1 Easy Learning Objective: 05-06 Apply the concepts of probability to contingency tables. Topic: Contingency Tables |
| 83. | Given the contingency table shown here, find P(E | F).
This is a conditional probability: P(E | F) = 160/470 = .340. |
| AACSB: Analytical Thinking Blooms: Apply Difficulty: 2 Medium Learning Objective: 05-06 Apply the concepts of probability to contingency tables. Topic: Contingency Tables |
| 84. | Given the contingency table shown here, find P(A∩M).
This is a joint probability: P(A∩M) = 210/1000 = .210. |
| AACSB: Analytical Thinking Blooms: Apply Difficulty: 1 Easy Learning Objective: 05-06 Apply the concepts of probability to contingency tables. Topic: Contingency Tables |
| 85. | Given the contingency table shown here, find P(F or G).
Use the General Law of Addition: P(F or G) = 470/1000 + 340/1000 – 160/1000. |
| AACSB: Analytical Thinking Blooms: Apply Difficulty: 3 Hard Learning Objective: 05-06 Apply the concepts of probability to contingency tables. Topic: Contingency Tables |
| 86. | Given the contingency table shown here, find the probability that a randomly chosen individual is a female economics major.
This is a joint probability: P(F and E) = 160/1000 = .16. |
| AACSB: Analytical Thinking Blooms: Apply Difficulty: 2 Medium Learning Objective: 05-06 Apply the concepts of probability to contingency tables. Topic: Contingency Tables |
| 87. | Debbie has two stocks, X and Y. Consider the following events:
X = the event that the price of stock X has increased
This is a joint probability that also entails the notation for an event’s complement. |
| AACSB: Analytical Thinking Accessibility: Keyboard Navigation Blooms: Understand Difficulty: 1 Easy Learning Objective: 05-03 Apply the definitions and rules of probability. Topic: Rules of Probability |
| 88. | If P(A | B) = 0.40 and P(B) = 0.30, find P(A∩B).
Use the definition for conditional probability. |
| AACSB: Analytical Thinking Accessibility: Keyboard Navigation Blooms: Apply Difficulty: 3 Hard Learning Objective: 05-03 Apply the definitions and rules of probability. Topic: Rules of Probability |
| 89. | A company is producing two types of ski goggles. Thirty percent of the production is of type A, and the rest is of type B. Five percent of all type A goggles are returned within 10 days after the sale, whereas only two percent of type B are returned. If a pair of goggles is returned within the first 10 days after the sale, the probability that the goggles returned are of type B is:
Review Bayes’ Theorem, and perhaps make a table or tree. |
| AACSB: Analytical Thinking Accessibility: Keyboard Navigation Blooms: Apply Difficulty: 3 Hard Learning Objective: 05-08 Use Bayes’ Theorem to calculate revised probabilities. Topic: Bayes’ Theorem |
| 90. | Given the contingency table shown here, find the joint probability that a call sampled at random out of this population is local and 2-5 minutes long.
This is a joint probability. You must add the column frequencies. |
| AACSB: Analytical Thinking Blooms: Apply Difficulty: 2 Medium Learning Objective: 05-06 Apply the concepts of probability to contingency tables. Topic: Contingency Tables |
| 91. | Given the contingency table shown here, if a call is sampled at random, find the marginal probability that the call is long distance.
You must first add the column frequencies. |
| AACSB: Analytical Thinking Blooms: Apply Difficulty: 2 Medium Learning Objective: 05-06 Apply the concepts of probability to contingency tables. Topic: Contingency Tables |
| 92. | If a call is sampled at random, the conditional probability that the call is not “6+” minutes long given that it is a long distance call is:
Calculate the conditional probability 1 – 10/300 = .9667. |
| AACSB: Analytical Thinking Blooms: Apply Difficulty: 3 Hard Learning Objective: 05-06 Apply the concepts of probability to contingency tables. Topic: Contingency Tables |
| 93. | The following table gives a classification of the 10,000 shareholders of Oxnard Xylophone Distributors, Inc. A few numbers are missing from the table. Given that a shareholder holding 500-999 shares is picked, there is a 0.625 probability that the shareholder will be a woman. Consequently, what is the number of men holding 1000 or more shares?
Multiply by the column total and subtract to fill in the remaining frequencies. |
| AACSB: Analytical Thinking Blooms: Apply Difficulty: 3 Hard Learning Objective: 05-06 Apply the concepts of probability to contingency tables. Topic: Contingency Tables |
| 94. | In any sample space P(A | B) and P(B | A):
Use the definition of conditional probability. |
| AACSB: Analytical Thinking Accessibility: Keyboard Navigation Blooms: Remember Difficulty: 3 Hard Learning Objective: 05-03 Apply the definitions and rules of probability. Topic: Rules of Probability |
| 95. | If P(A∩B) = 0.50, can P(A) = 0.20?
The given information contains a contradiction, because P(A∩B) cannot exceed P(A). |
| AACSB: Analytical Thinking Accessibility: Keyboard Navigation Blooms: Apply Difficulty: 2 Medium Learning Objective: 05-03 Apply the definitions and rules of probability. Topic: Rules of Probability |
| 96. | The following relationship always holds true for events A and B in a sample space.
Use the definition of conditional probability: P(A | B) = P(A∩B)/P(B). |
| AACSB: Analytical Thinking Accessibility: Keyboard Navigation Blooms: Understand Difficulty: 3 Hard Learning Objective: 05-03 Apply the definitions and rules of probability. Topic: Rules of Probability |
| 97. | The following probabilities are given about events A and B in a sample space: P(A) = 0.30, P(B) = 0.40, P(A or B) = 0.60. We can say that:
Apply the General Rule of Addition: P(A or B) = P(A) + P(B) – P(A∩B). |
| AACSB: Analytical Thinking Accessibility: Keyboard Navigation Blooms: Apply Difficulty: 3 Hard Learning Objective: 05-03 Apply the definitions and rules of probability. Topic: Rules of Probability |
| 98. | If P(A) = 0.35, P(B) = 0.60, and P(A or B) = 0.70, then:
Apply the General Rule of Addition: P(A or B) = P(A) + P(B) – P(A∩B). |
| AACSB: Analytical Thinking Accessibility: Keyboard Navigation Blooms: Apply Difficulty: 3 Hard Learning Objective: 05-03 Apply the definitions and rules of probability. Topic: Rules of Probability |
| 99. | The following table shows the survival experience of 1,000 males who retire at age 65:
Based on these data, the probability that a 75-year-old male will survive to age 80 is:
Given that 775 have survived to 75, the probability is 596 divided by 775. |
| AACSB: Analytical Thinking Blooms: Apply Difficulty: 3 Hard Learning Objective: 05-03 Apply the definitions and rules of probability. Topic: Rules of Probability |
| 100. | Given the contingency table shown here, find P(G | M).
This is a conditional probability: P(G | M) = 18/54. |
| AACSB: Analytical Thinking Blooms: Apply Difficulty: 2 Medium Learning Objective: 05-06 Apply the concepts of probability to contingency tables. Topic: Contingency Tables |
| 101. | Given the contingency table shown here, find P(V or S).
Use the General Rule of Addition: P(V or S) = 72/400 + 100/400 – 19/400. |
| AACSB: Analytical Thinking Blooms: Apply Difficulty: 3 Hard Learning Objective: 05-06 Apply the concepts of probability to contingency tables. Topic: Contingency Tables |
| 102. | Given the contingency table shown here, find P(V | S).
This is a conditional probability: P(V | S) = 19/100. |
| AACSB: Analytical Thinking Blooms: Apply Difficulty: 2 Medium Learning Objective: 05-06 Apply the concepts of probability to contingency tables. Topic: Contingency Tables |
| 103. | The manager of Ardmore Pharmacy knows that 25 percent of the customers entering the store buy prescription drugs, 65 percent buy over-the-counter drugs, and 18 percent buy both types of drugs. What is the probability that a randomly selected customer will buy at least one of these two types of drugs?
Use the General Rule of Addition: P(A or B) = P(A) + P(B) – P(A∩B) = .25 + .65 – .18. |
| AACSB: Analytical Thinking Accessibility: Keyboard Navigation Blooms: Apply Difficulty: 2 Medium Learning Objective: 05-03 Apply the definitions and rules of probability. Topic: Rules of Probability |
| 104. | Two events are complementary (i.e., they are complements) if:
Review rules of probability. |
| AACSB: Analytical Thinking Accessibility: Keyboard Navigation Blooms: Remember Difficulty: 2 Medium Learning Objective: 05-03 Apply the definitions and rules of probability. Topic: Rules of Probability |
| 105. | Which statement is false?
Review rules of probability and counting rules. |
| AACSB: Analytical Thinking Accessibility: Keyboard Navigation Blooms: Apply Difficulty: 3 Hard Learning Objective: 05-03 Apply the definitions and rules of probability. Topic: Rules of Probability |
| 106. | The number of unique orders in which five items (A, B, C, D, E) can be arranged is:
Apply rules of counting: 5 × 4 × 3 × 2 × 1 = 120. |
| AACSB: Analytical Thinking Accessibility: Keyboard Navigation Blooms: Apply Difficulty: 3 Hard Learning Objective: 05-09 Apply counting rules to calculate possible event arrangements. Topic: Counting Rules |
| 107. | If four items are chosen at random without replacement from seven items, in how many ways can the four items be arranged, treating each arrangement as a different event (i.e., if order is important)?
This is 7P4. |
| AACSB: Analytical Thinking Accessibility: Keyboard Navigation Blooms: Apply Difficulty: 3 Hard Learning Objective: 05-09 Apply counting rules to calculate possible event arrangements. Topic: Counting Rules |
| 108. | How many ways can we choose three items at random without replacement from five items (A, B, C, D, E) if the order of the selected items is not important?
This is 5C3. |
| AACSB: Analytical Thinking Accessibility: Keyboard Navigation Blooms: Apply Difficulty: 3 Hard Learning Objective: 05-09 Apply counting rules to calculate possible event arrangements. Topic: Counting Rules |
| 109. | The value of 6C2 is:
Apply the formula for combinations. |
| AACSB: Analytical Thinking Accessibility: Keyboard Navigation Blooms: Remember Difficulty: 2 Medium Learning Objective: 05-09 Apply counting rules to calculate possible event arrangements. Topic: Counting Rules |
| 110. | The value of 4P2 is:
Apply the formula for permutations. |
| AACSB: Analytical Thinking Accessibility: Keyboard Navigation Blooms: Remember Difficulty: 2 Medium Learning Objective: 05-09 Apply counting rules to calculate possible event arrangements. Topic: Counting Rules |
| 111. | The probability that event A occurs, given that event B has occurred, is an example of:
Review definition of conditional probability. |
| AACSB: Analytical Thinking Accessibility: Keyboard Navigation Blooms: Remember Difficulty: 1 Easy Learning Objective: 05-03 Apply the definitions and rules of probability. Topic: Rules of Probability |
| 112. | If each of two independent file servers has a reliability of 93 percent and either alone can run the website, then the overall website availability is:
Follow the textbook example of reliability for independent events. |
| AACSB: Analytical Thinking Accessibility: Keyboard Navigation Blooms: Remember Difficulty: 2 Medium Learning Objective: 05-05 Determine when events are independent. Topic: Independent Events |
| 113. | In a certain city, 5 percent of all drivers have expired licenses, 10 percent have an unpaid parking ticket, and 1 percent have both an expired license and an unpaid parking ticket. Are these events independent?
For independence we would require P(A)P(B) = P(A∩B). |
| AACSB: Analytical Thinking Accessibility: Keyboard Navigation Blooms: Apply Difficulty: 2 Medium Learning Objective: 05-05 Determine when events are independent. Topic: Independent Events |
| 114. | In a certain city, 5 percent of all drivers have expired licenses and 10 percent have an unpaid parking ticket. If these events are independent, what is the probability that a driver has both an expired license and an unpaid parking ticket?
By independence P(A∩B) = P(A)P(B) = (.05)(.10). |
| AACSB: Analytical Thinking Accessibility: Keyboard Navigation Blooms: Apply Difficulty: 2 Medium Learning Objective: 05-05 Determine when events are independent. Topic: Independent Events |
| 115. | If two events are collectively exhaustive, what is the probability that one or the other will occur?
Review definition of probabilities (collectively exhaustive covers all the possibilities). |
| AACSB: Analytical Thinking Accessibility: Keyboard Navigation Blooms: Apply Difficulty: 2 Medium Learning Objective: 05-03 Apply the definitions and rules of probability. Topic: Rules of Probability |
| 116. | Which best exemplifies a subjective probability?
Subjective probabilities are not based on empirical frequencies. |
| AACSB: Analytical Thinking Accessibility: Keyboard Navigation Blooms: Analyze Difficulty: 2 Medium Learning Objective: 05-02 Distinguish among the three views of probability. Topic: Probability |
| 117. | Which best exemplifies the classical definition of probability?
Classical probability is determined a priori by the nature of the experiment. |
| AACSB: Analytical Thinking Accessibility: Keyboard Navigation Blooms: Analyze Difficulty: 2 Medium Learning Objective: 05-02 Distinguish among the three views of probability. Topic: Probability |
| 118. | Which best exemplifies the empirical definition of probability?
Empirical probabilities are based on observed frequencies. |
| AACSB: Analytical Thinking Accessibility: Keyboard Navigation Blooms: Analyze Difficulty: 2 Medium Learning Objective: 05-02 Distinguish among the three views of probability. Topic: Probability |
| 119. | From the following tree, find the probability that a randomly chosen person will get the flu vaccine and will also get the flu.
Multiply down the branch: .70 × .10 = .07. |
| AACSB: Analytical Thinking Blooms: Apply Difficulty: 2 Medium Learning Objective: 05-07 Interpret a tree diagram. Topic: Tree Diagrams |
| 120. | From the following tree, find the probability that a randomly chosen person will not get a vaccination and will not get the flu.
Multiply down the branch: .30 × .60 = .18. |
| AACSB: Analytical Thinking Blooms: Apply Difficulty: 2 Medium Learning Objective: 05-07 Interpret a tree diagram. Topic: Tree Diagrams |
| 121. | From the following tree, find the probability that a randomly chosen person will get the flu.
Multiply down two branches and add .07 to .12. That is (.70)(.10) + (.30)(.40). |
| AACSB: Analytical Thinking Blooms: Apply Difficulty: 2 Medium Learning Objective: 05-07 Interpret a tree diagram. Topic: Tree Diagrams |
| 122. | At Joe’s Restaurant, 80 percent of the diners are new customers (N), while 20 percent are returning customers (R). Fifty percent of the new customers pay by credit card, compared with 70 percent of the regular customers. If a customer pays by credit card, what is the probability that the customer is a new customer?
Review Bayes’ Theorem, and perhaps make a table or tree. |
| AACSB: Analytical Thinking Accessibility: Keyboard Navigation Blooms: Remember Difficulty: 3 Hard Learning Objective: 05-08 Use Bayes’ Theorem to calculate revised probabilities. Topic: Bayes’ Theorem |
| 123. | At Dolon General Hospital, 30 percent of the patients have Medicare insurance (M) while 70 percent do not have Medicare insurance (M´). Twenty percent of the Medicare patients arrive by ambulance, compared with 10 percent of the non-Medicare patients. If a patient arrives by ambulance, what is the probability that the patient has Medicare insurance?
Review Bayes’ Theorem, and perhaps make a table or tree. |
| AACSB: Analytical Thinking Accessibility: Keyboard Navigation Blooms: Remember Difficulty: 3 Hard Learning Objective: 05-08 Use Bayes’ Theorem to calculate revised probabilities. Topic: Bayes’ Theorem |
Short Answer Questions
| 124. | Axolotl, Inc. decided to double the number of independent network file servers, each server being capable of running its entire internal corporate network. Bob suggested that “doubling the number of file servers in our corporate network will just double the chances of failure.” Is Bob right? Has the company made an error?
No, as long as the file servers are independent, doubling the number of file servers will reduce the chance of failure. For example, if they had two file servers each with .95 reliability (i.e., a .05 chance of being “down”), then their chance of total failure would be P(F1∩F2) = P(F1)P(F2) = (.05)(.05) = .0025. Feedback: No, as long as the file servers are independent, doubling the number of file servers will reduce the chance of failure. For example, if they had two file servers each with .95 reliability (i.e., a .05 chance of being “down”), then their chance of total failure would be P(F1∩F2) = P(F1)P(F2) = (.05)(.05) = .0025. But if they go to four independent file servers, the chance of being “down” is P(F1∩F2∩F3∩F4) = P(F1)P(F2)P(F3)P(F4) = (.05)(.05)(.05)(.05) = .00000625. |
| AACSB: Reflective Thinking Blooms: Apply Difficulty: 3 Hard Learning Objective: 05-05 Determine when events are independent. Topic: Independent Events |
| 125. | A baseball player said, “Yes, I’m in a batting slump. But the law of averages has got to catch up sooner or later. I’m bound to improve.” Discuss the logic of this statement.
If the player’s probability of getting a hit each time at the plate is constant, then the player’s chances of getting a hit do not change. Feedback: If the player’s probability of getting a hit each time at the plate is constant (i.e., if his at bats are independent events), then the player’s chances of getting a hit do not change, just as one coin flip cannot affect the next flip. But ballplayers are not coins. If the player’s mental outlook deteriorates because of his slump, he may continue hitting poorly (i.e., his not getting a hit today may lower the probability of getting a hit the next time). Conversely, if the player believes that his “luck” is due to change, he may become more effective (i.e., a mental turnaround could increase his chances of getting a hit). Unlike a coin experiment, a ballplayer’s at bats may not be independent events, because human psychology is involved. |
| AACSB: Reflective Thinking Blooms: Apply Difficulty: 2 Medium Learning Objective: 05-02 Distinguish among the three views of probability. Topic: Probability |
| 126. | Recently, a few companies have been accused of issuing stock options to their top executives at the lowest stock price of the year. Executives can later sell these stock shares at a higher price and make a large profit. In some companies, this happened several years in a row. An executive argued, “It’s just a coincidence. There are about 250 trading days in a year, so there is a reasonable chance that my option price just happened to be lowest of the year.” Discuss.
It’s possible, but not very likely. For example, the probability that options would be issued at the yearly low two years in a row would be (1/250)(1/250) = .000016. Feedback: It’s possible, but not very likely. For example, the probability that options would be issued at the yearly low two years in a row would be (1/250)(1/250) = .000016. For three years, it would be (1/250)3 = .000000064. In this case, the probabilities are a strong indication of the need for a deeper investigation. |
| AACSB: Reflective Thinking Blooms: Apply Difficulty: 3 Hard Learning Objective: 05-05 Determine when events are independent. Topic: Independent Events |
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.
Original price was: $30.00.$20.00Current price is: $20.00.
