Applied Statistics in Business and Economics David Doane 6e - Test Bank

Applied Statistics in Business and Economics David Doane 6e – Test Bank

 

Instant Download – Complete Test Bank With Answers

 

 

Sample Questions Are Posted Below

 

Applied Statistics in Business and Economics, 6e (Doane)

Chapter 5   Probability

 

1) A sample space is the set of all possible outcomes in an experiment.

 

Answer:  TRUE

Explanation:  Review the definition of sample space.

Difficulty: 1 Easy

Topic:  05.01 Random Experiments

Learning Objective:  05-01 Describe the sample space of a random experiment.

Bloom’s:  Remember

AACSB:  Analytical Thinking

Accessibility:  Keyboard Navigation

 

2) The sum of all the probabilities of simple events in a sample space equals one.

 

Answer:  TRUE

Explanation:  Simple events are non-overlapping, so they sum to unity.

Difficulty: 1 Easy

Topic:  05.03 Rules of Probability

Learning Objective:  05-03 Apply the definitions and rules of probability.

Bloom’s:  Understand

AACSB:  Analytical Thinking

Accessibility:  Keyboard Navigation

 

3) The sum of the probabilities of all compound events in a sample space equals one.

 

Answer:  FALSE

Explanation:  Compound events may overlap, so you cannot simply add their probabilities.

Difficulty: 1 Easy

Topic:  05.02 Probability

Learning Objective:  05-03 Apply the definitions and rules of probability.

Bloom’s:  Understand

AACSB:  Analytical Thinking

Accessibility:  Keyboard Navigation

 

4) Probability is the measure of the relative likelihood that an event will occur.

 

Answer:  TRUE

Explanation:  This is one of three ways to view probability.

Difficulty: 1 Easy

Topic:  05.02 Probability

Learning Objective:  05-03 Apply the definitions and rules of probability.

Bloom’s:  Remember

AACSB:  Analytical Thinking

Accessibility:  Keyboard Navigation

 

5) The probability of the union of two events P(A or B) can exceed one.

 

Answer:  FALSE

Explanation:  Review the General Rule of Addition. No event probability can exceed one.

Difficulty: 2 Medium

Topic:  05.03 Rules of Probability

Learning Objective:  05-03 Apply the definitions and rules of probability.

Bloom’s:  Understand

AACSB:  Analytical Thinking

Accessibility:  Keyboard Navigation

 

6) The empirical view of probability is based on relative frequencies.

 

Answer:  TRUE

Explanation:  This is one view of probability, calculated from experience.

Difficulty: 1 Easy

Topic:  05.02 Probability

Learning Objective:  05-02 Distinguish among the three views of probability.

Bloom’s:  Remember

AACSB:  Analytical Thinking

Accessibility:  Keyboard Navigation

 

7) Grandma’s predicting rain based on how much her arthritis is acting up is an example of the classical view of probability.

 

Answer:  FALSE

Explanation:  This would be a subjective probability, where no data are available.

Difficulty: 2 Medium

Topic:  05.02 Probability

Learning Objective:  05-02 Distinguish among the three views of probability.

Bloom’s:  Apply

AACSB:  Analytical Thinking

Accessibility:  Keyboard Navigation

 

8) The odds of an event can be calculated by dividing the event’s probability by the probability of its complement.

 

Answer:  TRUE

Explanation:  Review the definition of odds.

Difficulty: 2 Medium

Topic:  05.02 Probability

Learning Objective:  05-04 Calculate odds from given probabilities.

Bloom’s:  Remember

AACSB:  Analytical Thinking

Accessibility:  Keyboard Navigation

 

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.

 

Answer:  FALSE

Explanation:  The union is all outcomes in either A or B.

Difficulty: 2 Medium

Topic:  05.03 Rules of Probability

Learning Objective:  05-03 Apply the definitions and rules of probability.

Bloom’s:  Remember

AACSB:  Analytical Thinking

Accessibility:  Keyboard Navigation

 

10) The general law of addition for probabilities says P(A or B) = P(A) + P(B) − P(A ∩ B).

 

Answer:  TRUE

Explanation:  We must subtract the probability of the event intersection from the sum.

Difficulty: 1 Easy

Topic:  05.03 Rules of Probability

Learning Objective:  05-03 Apply the definitions and rules of probability.

Bloom’s:  Remember

AACSB:  Analytical Thinking

Accessibility:  Keyboard Navigation

 

11) Events A and B are mutually exclusive if P(A ∩ B) = 0.

 

Answer:  TRUE

Explanation:  Nonoverlapping events cannot both occur (intersection is the empty set).

Difficulty: 2 Medium

Topic:  05.03 Rules of Probability

Learning Objective:  05-03 Apply the definitions and rules of probability.

Bloom’s:  Remember

AACSB:  Analytical Thinking

Accessibility:  Keyboard Navigation

 

12) Independent events are mutually exclusive.

 

Answer:  FALSE

Explanation:  Only if P(A ∩ B) = 0 would they be mutually exclusive.

Difficulty: 2 Medium

Topic:  05.04 Independent Events

Learning Objective:  05-05 Determine when events are independent.

Bloom’s:  Remember

AACSB:  Analytical Thinking

Accessibility:  Keyboard Navigation

 

 

13) If events A and B are mutually exclusive, then P(A) + P(B) = 0.

 

Answer:  FALSE

Explanation:  Their intersection would have zero probability, but not their sum.

Difficulty: 2 Medium

Topic:  05.03 Rules of Probability

Learning Objective:  05-03 Apply the definitions and rules of probability.

Bloom’s:  Remember

AACSB:  Analytical Thinking

Accessibility:  Keyboard Navigation

 

14) P(A | B) is the joint probability of events A and B divided by the probability of A.

 

Answer:  FALSE

Explanation:  You would divide the joint probability by P(B) rather than by P(A).

Difficulty: 2 Medium

Topic:  05.03 Rules of Probability

Learning Objective:  05-03 Apply the definitions and rules of probability.

Bloom’s:  Remember

AACSB:  Analytical Thinking

Accessibility:  Keyboard Navigation

 

15) Two events A and B are independent if P(A | B) is the same as P(A).

 

Answer:  TRUE

Explanation:  This is the definition of independence. Event B does not affect event A.

Difficulty: 1 Easy

Topic:  05.04 Independent Events

Learning Objective:  05-05 Determine when events are independent.

Bloom’s:  Remember

AACSB:  Analytical Thinking

Accessibility:  Keyboard Navigation

 

16) If events A and B are dependent, it can be concluded that one event causes the other.

 

Answer:  FALSE

Explanation:  Maybe there is causation, but both events could be affected by something else.

Difficulty: 2 Medium

Topic:  05.04 Independent Events

Learning Objective:  05-05 Determine when events are independent.

Bloom’s:  Understand

AACSB:  Analytical Thinking

Accessibility:  Keyboard Navigation

 

 

 

17) For any event A, the probability of A is always 0 ≤ P(A) ≤ 1.

 

Answer:  TRUE

Explanation:  Any probability must be between 0 and 1.

Difficulty: 1 Easy

Topic:  05.02 Probability

Learning Objective:  05-03 Apply the definitions and rules of probability.

Bloom’s:  Remember

AACSB:  Analytical Thinking

Accessibility:  Keyboard Navigation

18) If events A and B are mutually exclusive, the joint probability of the events is zero.

 

Answer:  TRUE

Explanation:  The intersection of nonoverlapping events is the empty set.

Difficulty: 2 Medium

Topic:  05.03 Rules of Probability

Learning Objective:  05-03 Apply the definitions and rules of probability.

Bloom’s:  Remember

AACSB:  Analytical Thinking

Accessibility:  Keyboard Navigation

 

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.

 

Answer:  TRUE

Explanation:  For example, the speeds of vehicles on a freeway are not listable (nonintegers).

Difficulty: 1 Easy

Topic:  05.01 Random Experiments

Learning Objective:  05-01 Describe the sample space of a random experiment.

Bloom’s:  Understand

AACSB:  Analytical Thinking

Accessibility:  Keyboard Navigation

 

20) If A and B are independent events, then P(A or B) = P(A) P(B).

 

Answer:  FALSE

Explanation:  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).

Difficulty: 2 Medium

Topic:  05.04 Independent Events

Learning Objective:  05-05 Determine when events are independent.

Bloom’s:  Remember

AACSB:  Analytical Thinking

Accessibility:  Keyboard Navigation

 

21) If A and B are mutually exclusive events, then P(A ∩ B) = P(A) + P(B).

 

Answer:  FALSE

Explanation:  Their intersection is empty so it has zero probability.

Difficulty: 2 Medium

Topic:  05.03 Rules of Probability

Learning Objective:  05-03 Apply the definitions and rules of probability.

Bloom’s:  Remember

AACSB:  Analytical Thinking

Accessibility:  Keyboard Navigation

22) The probability of A and its complement (A´) will always sum to one.

 

Answer:  TRUE

Explanation:  An event and its complement comprise the sample space.

Difficulty: 2 Medium

Topic:  05.03 Rules of Probability

Learning Objective:  05-03 Apply the definitions and rules of probability.

Bloom’s:  Remember

AACSB:  Analytical Thinking

Accessibility:  Keyboard Navigation

 

23) If event A occurs, then its complement (A´) will also occur.

 

Answer:  FALSE

Explanation:  An event and its complement are nonoverlapping.

Difficulty: 1 Easy

Topic:  05.03 Rules of Probability

Learning Objective:  05-03 Apply the definitions and rules of probability.

Bloom’s:  Remember

AACSB:  Analytical Thinking

Accessibility:  Keyboard Navigation

 

24) The sum of the probabilities of two mutually exclusive events is one.

 

Answer:  FALSE

Explanation:  Review definitions for event probabilities. Their probabilities would sum to one only if these two events comprise the entire sample space.

Difficulty: 2 Medium

Topic:  05.03 Rules of Probability

Learning Objective:  05-03 Apply the definitions and rules of probability.

Bloom’s:  Remember

AACSB:  Analytical Thinking

Accessibility:  Keyboard Navigation

 

 

 

25) P(A ∩ B) = .50 is an example of a joint probability.

 

Answer:  TRUE

Explanation:  Intersection of two events yields a joint probability.

Difficulty: 1 Easy

Topic:  05.03 Rules of Probability

Learning Objective:  05-03 Apply the definitions and rules of probability.

Bloom’s:  Remember

AACSB:  Analytical Thinking

Accessibility:  Keyboard Navigation

26) The general law of addition for probabilities says P(A or B) = P(A) P(B).

 

Answer:  FALSE

Explanation:  Review the General Law of Addition: P(A or B) = P(A) + P(B) − P(A ∩ B).

Difficulty: 1 Easy

Topic:  05.03 Rules of Probability

Learning Objective:  05-03 Apply the definitions and rules of probability.

Bloom’s:  Remember

AACSB:  Analytical Thinking

Accessibility:  Keyboard Navigation

 

27) If P(A) = .50, P(B) = .30, and P(A ∩ B) = .15, then A and B are independent events.

 

Answer:  TRUE

Explanation:  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).

Difficulty: 2 Medium

Topic:  05.04 Independent Events

Learning Objective:  05-05 Determine when events are independent.

Bloom’s:  Apply

AACSB:  Analytical Thinking

Accessibility:  Keyboard Navigation

 

28) Insurance company life tables are an example of the classical (a priori) approach to probability.

 

Answer:  FALSE

Explanation:  Actuarial data are based on relative frequencies (empirical probability).

Difficulty: 2 Medium

Topic:  05.02 Probability

Learning Objective:  05-02 Distinguish among the three views of probability.

Bloom’s:  Understand

AACSB:  Analytical Thinking

Accessibility:  Keyboard Navigation

 

 

 

29) When two events cannot occur at the same time, they are said to be mutually exclusive.

 

Answer:  TRUE

Explanation:  This is the definition of mutually exclusive events.

Difficulty: 2 Medium

Topic:  05.03 Rules of Probability

Learning Objective:  05-03 Apply the definitions and rules of probability.

Bloom’s:  Remember

AACSB:  Analytical Thinking

Accessibility:  Keyboard Navigation

30) The probability of events A or B occurring can be found by summing their probabilities.

 

Answer:  FALSE

Explanation:  Event probabilities cannot be summed unless their intersection is the empty set.

Difficulty: 2 Medium

Topic:  05.03 Rules of Probability

Learning Objective:  05-03 Apply the definitions and rules of probability.

Bloom’s:  Understand

AACSB:  Analytical Thinking

Accessibility:  Keyboard Navigation

 

31) When two events A and B are independent, the probability of their intersection can be found by multiplying their probabilities.

 

Answer:  TRUE

Explanation:  This is true by the definition of independent events.

Difficulty: 2 Medium

Topic:  05.04 Independent Events

Learning Objective:  05-05 Determine when events are independent.

Bloom’s:  Remember

AACSB:  Analytical Thinking

Accessibility:  Keyboard Navigation

 

32) Two events are mutually exclusive when they contain no outcomes in common.

 

Answer:  TRUE

Explanation:  This is the definition of mutually exclusive events.

Difficulty: 1 Easy

Topic:  05.03 Rules of Probability

Learning Objective:  05-03 Apply the definitions and rules of probability.

Bloom’s:  Remember

AACSB:  Analytical Thinking

Accessibility:  Keyboard Navigation

 

 

 

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.

 

Answer:  FALSE

Explanation:  Review the general law of addition P(A or B) = P(A) + P(B) − P(A ∩ B).

Difficulty: 2 Medium

Topic:  05.05 Contingency Tables

Learning Objective:  05-06 Apply the concepts of probability to contingency tables.

Bloom’s:  Understand

AACSB:  Analytical Thinking

Accessibility:  Keyboard Navigation

34) Bayes’ Theorem shows how to revise a prior probability to obtain a conditional or posterior probability when another event’s occurrence is known.

 

Answer:  TRUE

Explanation:  This is a rather general statement about Bayes’ contribution to statistics.

Difficulty: 2 Medium

Topic:  05.07 Bayes’ Theorem

Learning Objective:  05-08 Use Bayes’ Theorem to calculate revised probabilities.

Bloom’s:  Understand

AACSB:  Analytical Thinking

Accessibility:  Keyboard Navigation

 

35) The union of two events is all outcomes in either or both, while the intersection is only those events in both.

 

Answer:  TRUE

Explanation:  Review the definitions of union and intersection.

Difficulty: 2 Medium

Topic:  05.03 Rules of Probability

Learning Objective:  05-03 Apply the definitions and rules of probability.

Bloom’s:  Understand

AACSB:  Analytical Thinking

Accessibility:  Keyboard Navigation

 

36) A contingency table is a cross-tabulation of frequencies for two categorical variables.

 

Answer:  TRUE

Explanation:  The categories define the rows and columns.

Difficulty: 1 Easy

Topic:  05.05 Contingency Tables

Learning Objective:  05-06 Apply the concepts of probability to contingency tables.

Bloom’s:  Remember

AACSB:  Analytical Thinking

Accessibility:  Keyboard Navigation

 

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).

 

Answer:  TRUE

Explanation:  Review the definitions of permutations and combinations.

Difficulty: 2 Medium

Topic:  05.08 Counting Rules

Learning Objective:  05-09 Apply counting rules to calculate possible event arrangements.

Bloom’s:  Remember

AACSB:  Analytical Thinking

Accessibility:  Keyboard Navigation

38) If P(A) = .20 then the odds against event A‘s occurrence are 4 to 1.

 

Answer:  TRUE

Explanation:  Review the definition of odds.

Difficulty: 2 Medium

Topic:  05.02 Probability

Learning Objective:  05-04 Calculate odds from given probabilities.

Bloom’s:  Apply

AACSB:  Analytical Thinking

Accessibility:  Keyboard Navigation

 

39) The value of 7! is 5040.

 

Answer:  TRUE

Explanation:  Factorial n (written as n!) is 7! = 7 × 6 × 5 × 4 × 3 × 2 × 1 = 5040.

Difficulty: 1 Easy

Topic:  05.08 Counting Rules

Learning Objective:  05-09 Apply counting rules to calculate possible event arrangements.

Bloom’s:  Apply

AACSB:  Analytical Thinking

Accessibility:  Keyboard Navigation

 

 

 

40) Events A and B are mutually exclusive when

1. A) their joint probability is zero.
2. B) they are independent events.
3. C) P(A)P(B) = 0
4. D) P(A)P(B) = P(A | B)

 

Answer:  A

Explanation:  By the definition of mutually exclusive P(A and B) = 0. That is, both events cannot happen.

Difficulty: 2 Medium

Topic:  05.03 Rules of Probability

Learning Objective:  05-03 Apply the definitions and rules of probability.

Bloom’s:  Understand

AACSB:  Analytical Thinking

Accessibility:  Keyboard Navigation

 

41) If two events are complementary, then we know that

1. A) the sum of their probabilities is one.
2. B) the joint probability of the two events is one.
3. C) their intersection has a nonzero probability.
4. D) they are independent events.

 

Answer:  A

Explanation:  Together, complementary events comprise the sample space.

Difficulty: 2 Medium

Topic:  05.03 Rules of Probability

Learning Objective:  05-03 Apply the definitions and rules of probability.

Bloom’s:  Remember

AACSB:  Analytical Thinking

Accessibility:  Keyboard Navigation

 

 

42) Regarding probability, which of the following is correct?

1. A) When events A and B are mutually exclusive, then P(A ∩ B) = P(A) + P(B).
2. B) The union of events A and B consists of all outcomes in the sample space that are contained in both event A and event B.
3. C) When two events A and B are independent, the joint probability of the events can be found by multiplying the probabilities of the individual events.
4. D) The probability of the union of two events can exceed one.

 

Answer:  C

Explanation:  Review the rules of probability. When events are independent, their joint probability is their product. However, this is not true in general.

Difficulty: 2 Medium

Topic:  05.04 Independent Events

Learning Objective:  05-05 Determine when events are independent.

Bloom’s:  Remember

AACSB:  Analytical Thinking

Accessibility:  Keyboard Navigation

 

43) Independent events A and B would be consistent with which of the following statements?

4. A) P(A) = .3, P(B) = .5, P(A ∩ B) = .4.
5. B) P(A) = .4, P(B) = .5, P(A ∩ B) = .2.
6. C) P(A) = .5, P(B) = .4, P(A ∩ B) = .3.
7. D) P(A) = .4, P(B) = .3, P(A ∩ B) = .5.

 

Answer:  B

Explanation:  For independence, the product P(A)P(B) must equal P(A ∩ B).

Difficulty: 1 Easy

Topic:  05.04 Independent Events

Learning Objective:  05-05 Determine when events are independent.

Bloom’s:  Apply

AACSB:  Analytical Thinking

Accessibility:  Keyboard Navigation

 

 

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.

1. A) .80
2. B) .15
3. C) .65
4. D) .85

 

Answer:  C

Explanation:  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.

Difficulty: 2 Medium

Topic:  05.04 Independent Events

Learning Objective:  05-05 Determine when events are independent.

Bloom’s:  Apply

AACSB:  Analytical Thinking

Accessibility:  Keyboard Navigation

 

45) Regarding the rules of probability, which of the following statements is correct?

1. A) If A and B are independent events, then P(B) = P(A)P(B).
2. B) The sum of two mutually exclusive events is one.
3. C) The probability of A and its complement will sum to one.
4. D) If event A occurs, then its complement will also occur.

 

Answer:  C

Explanation:  By definition P(A and A‘) = 1.

Difficulty: 2 Medium

Topic:  05.03 Rules of Probability

Learning Objective:  05-03 Apply the definitions and rules of probability.

Bloom’s:  Remember

AACSB:  Analytical Thinking

Accessibility:  Keyboard Navigation

 

 

 

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?

1. A) .67
2. B) .79
3. C) .22
4. D) .43

 

Answer:  A

Explanation:  Use the General Law of Addition P(A or B) = P(A) + P(B) − P(A ∩ B).

Difficulty: 2 Medium

Topic:  05.03 Rules of Probability

Learning Objective:  05-03 Apply the definitions and rules of probability.

Bloom’s:  Apply

AACSB:  Analytical Thinking

Accessibility:  Keyboard Navigation

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.

1. A) .79
2. B) .82
3. C) .65
4. D) .50

 

Answer:  C

Explanation:  Use the General Law of Addition P(A or B) = P(A) + P(B) − P(A ∩ B).

Difficulty: 3 Hard

Topic:  05.03 Rules of Probability

Learning Objective:  05-03 Apply the definitions and rules of probability.

Bloom’s:  Apply

AACSB:  Analytical Thinking

Accessibility:  Keyboard Navigation

 

 

 

48) Given the contingency table shown here, find P(V).

 

Cell Phone Service Provider
County	Sprint (S)			AT&T (A)			Verizon (V)			Row Total
Macomb (M)		17			25			8			50
Oakland (O)		19			38			13			70
Wayne (W)		24			37			19			80
Col Total		60			100			40			200

 

1. A) .20
2. B) .40
3. C) .50
4. D) .80

 

Answer:  A

Explanation:  This is a marginal probability P(V) = 40/200 = .20.

Difficulty: 2 Medium

Topic:  05.05 Contingency Tables

Learning Objective:  05-06 Apply the concepts of probability to contingency tables.

Bloom’s:  Apply

AACSB:  Analytical Thinking

Accessibility:  Keyboard Navigation

49) Given the contingency table shown here, find P(V | W).

 

Cell Phone Service Provider
County	Sprint (S)			AT&T (A)			Verizon (V)			Row Total
Macomb (M)		17			25			8			50
Oakland (O)		19			38			13			70
Wayne (W)		24			37			19			80
Col Total		60			100			40			200

 

1. A) .4000
2. B) .0950
3. C) .2375
4. D) .5875

 

Answer:  C

Explanation:  This is a conditional probability P(V | W) = 19/80.

Difficulty: 2 Medium

Topic:  05.05 Contingency Tables

Learning Objective:  05-06 Apply the concepts of probability to contingency tables.

Bloom’s:  Apply

AACSB:  Analytical Thinking

Accessibility:  Keyboard Navigation

 

50) Given the contingency table shown here, find the probability P(V’), that is, the probability of the complement of V.

 

Cell Phone Service Provider
County	Sprint (S)			AT&T (A)			Verizon (V)			Row Total
Macomb (M)		17			25			8			50
Oakland (O)		19			38			13			70
Wayne (W)		24			37			19			80
Col Total		60			100			40			200

 

1. A) .30
2. B) .50
3. C) .80
4. D) .15

 

Answer:  C

Explanation:  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.

Difficulty: 3 Hard

Topic:  05.05 Contingency Tables

Learning Objective:  05-06 Apply the concepts of probability to contingency tables.

Bloom’s:  Apply

AACSB:  Analytical Thinking

Accessibility:  Keyboard Navigation

 

 

51) Given the contingency table shown here, find P(W ∩ S).

 

Cell Phone Service Provider
County	Sprint (S)			AT&T (A)			Verizon (V)			Row Total
Macomb (M)		17			25			8			50
Oakland (O)		19			38			13			70
Wayne (W)		24			37			19			80
Col Total		60			100			40			200

 

1. A) .12
2. B) .30
3. C) .40
4. D) .58

 

Answer:  A

Explanation:  This is a joint probability P(W and S) = 24/200.

Difficulty: 2 Medium

Topic:  05.05 Contingency Tables

Learning Objective:  05-06 Apply the concepts of probability to contingency tables.

Bloom’s:  Apply

AACSB:  Analytical Thinking

Accessibility:  Keyboard Navigation

 

52) Given the contingency table shown here, find P(A or M).

 

Cell Phone Service Provider
County	Sprint (S)			AT&T (A)			Verizon (V)			Row Total
Macomb (M)		17			25			8			50
Oakland (O)		19			38			13			70
Wayne (W)		24			37			19			80
Col Total		60			100			40			200

 

1. A) .2500
2. B) .7500
3. C) .6250
4. D) .1250

 

Answer:  C

Explanation:  Use the General Law of Addition P(A or M) = 100/200 + 50/200 − 25/200.

Difficulty: 3 Hard

Topic:  05.05 Contingency Tables

Learning Objective:  05-06 Apply the concepts of probability to contingency tables.

Bloom’s:  Apply

AACSB:  Analytical Thinking

Accessibility:  Keyboard Navigation

 

53) Given the contingency table shown here, find P(A2).

 

	A1			A2			A3			A4			Row Total
B1		12			26			42			68			148
B2		14			28			44			64			150
B3		18			32			47			72			169
Col Total		44			86			133			204			467

 

1. A) .1842
2. B) .1766
3. C) .8163
4. D) .0578

 

Answer:  A

Explanation:  This is a marginal probability: P(A2) = 86/467.

Difficulty: 2 Medium

Topic:  05.05 Contingency Tables

Learning Objective:  05-06 Apply the concepts of probability to contingency tables.

Bloom’s:  Apply

AACSB:  Analytical Thinking

Accessibility:  Keyboard Navigation

 

54) Given the contingency table shown here, find P(A3 ∩ B2).

 

	A1			A2			A3			A4			Row Total
B1		12			26			42			68			148
B2		14			28			44			64			150
B3		18			32			47			72			169
Col Total		44			86			133			204			467

 

1. A) .3212
2. B) .2933
3. C) .0942
4. D) .1006

 

Answer:  C

Explanation:  This is a joint probability: P(A3 and B2) = 44/467.

Difficulty: 2 Medium

Topic:  05.05 Contingency Tables

Learning Objective:  05-06 Apply the concepts of probability to contingency tables.

Bloom’s:  Apply

AACSB:  Analytical Thinking

Accessibility:  Keyboard Navigation

 

55) Given the contingency table shown here, find P(A2 | B3).

 

	A1			A2			A3			A4			Row Total
B1		12			26			42			68			148
B2		14			28			44			64			150
B3		18			32			47			72			169
Col Total		44			86			133			204			467

 

1. A) .0685
2. B) .1893
3. C) .3721
4. D) .1842

 

Answer:  B

Explanation:  This is a conditional probability: P(A2 | B3) = 32/169.

Difficulty: 2 Medium

Topic:  05.05 Contingency Tables

Learning Objective:  05-06 Apply the concepts of probability to contingency tables.

Bloom’s:  Apply

AACSB:  Analytical Thinking

Accessibility:  Keyboard Navigation

 

 

 

56) Given the contingency table shown here, find P(A1 or B2).

 

	A1			A2			A3			A4			Row Total
B1		12			26			42			68			148
B2		14			28			44			64			150
B3		18			32			47			72			169
Col Total		44			86			133			204			467

 

1. A) .0933
2. B) .3182
3. C) .0300
4. D) .3854

 

Answer:  D

Explanation:  Apply the General Law of Addition: P(A1 or B2) = 44/467 + 150/467 − 14/467.

Difficulty: 3 Hard

Topic:  05.05 Contingency Tables

Learning Objective:  05-06 Apply the concepts of probability to contingency tables.

Bloom’s:  Apply

AACSB:  Analytical Thinking

Accessibility:  Keyboard Navigation

57) Given the contingency table shown here, find P(A1 ∩ A2).

 

	A1			A2			A3			A4			Row Total
B1		12			26			42			68			148
B2		14			28			44			64			150
B3		18			32			47			72			169
Col Total		44			86			133			204			467

 

1. A) .00
2. B) .09
3. C) .28
4. D) .38

 

Answer:  A

Explanation:  This is a joint probability. The important thing here is that events A1 and A2 are mutually exclusive and so both events cannot occur.

Difficulty: 2 Medium

Topic:  05.05 Contingency Tables

Learning Objective:  05-06 Apply the concepts of probability to contingency tables.

Bloom’s:  Apply

AACSB:  Analytical Thinking

Accessibility:  Keyboard Navigation

 

58) Given the contingency table shown here, find the probability that either event A2 or event B2 will occur.

 

	A1			A2			A3			A4			Row Total
B1		12			26			42			68			148
B2		14			28			44			64			150
B3		18			32			47			72			169
Col Total		44			86			133			204			467

 

1. A) .4454
2. B) .5054
3. C) .0600

 

Answer:  A

Explanation:  Use the General Law of Addition: P(A2 or B2) =86/467 + 150/467 − 28/467.

Difficulty: 3 Hard

Topic:  05.05 Contingency Tables

Learning Objective:  05-06 Apply the concepts of probability to contingency tables.

Bloom’s:  Apply

AACSB:  Analytical Thinking

Accessibility:  Keyboard Navigation

59) Given the contingency table shown here, find P(B).

 

Age
Absences	Under 25 (A)			25 or More (A’)			Row Total
Under 2 days (B)		50			40				90
2 or more days (B’)		30			80				110
Column Total		80			120				200

 

1. A) .85
2. B) .25
3. C) .45
4. D) .22

 

Answer:  C

Explanation:  This is a marginal probability: P(B) = 90/200.

Difficulty: 1 Easy

Topic:  05.05 Contingency Tables

Learning Objective:  05-06 Apply the concepts of probability to contingency tables.

Bloom’s:  Apply

AACSB:  Analytical Thinking

Accessibility:  Keyboard Navigation

 

60) Given the contingency table shown here, find P(A or B).

 

Age
Absences	Under 25 (A)			25 or More (A’)				Row Total
Under 2 days (B)		50				40				90
2 or more days (B’)		30				80				110
Column Total		80				120				200

 

1. A) .25
2. B) .85
3. C) .60
4. D) .42

 

Answer:  C

Explanation:  Use the General Law of Addition: P(A or B) = 80/200 + 90/200 − 50/200.

Difficulty: 2 Medium

Topic:  05.05 Contingency Tables

Learning Objective:  05-06 Apply the concepts of probability to contingency tables.

Bloom’s:  Apply

AACSB:  Analytical Thinking

Accessibility:  Keyboard Navigation

61) Given the contingency table shown here, find P(B | A).

 

Age
Absences	Under 25 (A)			25 or More (A’)				Row Total
Under 2 days (B)		50				40				90
2 or more days (B’)		30				80				110
Column Total		80				120				200

 

1. A) .250
2. B) .555
3. C) .855
4. D) .625

 

Answer:  D

Explanation:  This is a conditional probability: P(B | A) = 50/80.

Difficulty: 2 Medium

Topic:  05.05 Contingency Tables

Learning Objective:  05-06 Apply the concepts of probability to contingency tables.

Bloom’s:  Apply

AACSB:  Analytical Thinking

Accessibility:  Keyboard Navigation

 

 

 

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?

 

Age
Absences	Under 25 (A)			25 or More (A’)				Row Total
Under 2 days (B)		50				40				90
2 or more days (B’)		30				80				110
Column Total		80				120				200

 

1. A) .625
2. B) .375
3. C) .150
4. D) .273

 

Answer:  B

Explanation:  This is a conditional probability: P(B ‘ | A) = 30/80.

Difficulty: 2 Medium

Topic:  05.05 Contingency Tables

Learning Objective:  05-06 Apply the concepts of probability to contingency tables.

Bloom’s:  Apply

AACSB:  Analytical Thinking

Accessibility:  Keyboard Navigation

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. A) 1
2. B) 2
3. C) 3
4. D) 4

 

Answer:  C

Explanation:  1 − P(F1 ∩ F2 ∩ F3) = 1 − (.05)(.05)(.05) = 1 − .000125 = .999875, so 3 servers will do.

Difficulty: 3 Hard

Topic:  05.04 Independent Events

Learning Objective:  05-05 Determine when events are independent.

Bloom’s:  Apply

AACSB:  Analytical Thinking

Accessibility:  Keyboard Navigation

 

 

 

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?

 

Employee	Retire (R)			Work (W)			Total
Management (M)		13			18				31
Line worker (L)		39			54				93
Total		52			72				124

 

1. A) Yes
2. B) No

 

Answer:  A

Explanation:  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.

Difficulty: 2 Medium

Topic:  05.05 Contingency Tables

Learning Objective:  05-06 Apply the concepts of probability to contingency tables.

Bloom’s:  Apply

AACSB:  Analytical Thinking

Accessibility:  Keyboard Navigation

 

 

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?

 

Employee	Retire (R)			Work (W)			Total
Management (M)		13			18				31
Line worker (L)		39			54				93
Total		52			72				124

 

1. A) .227
2. B) .419
3. C) .750
4. D) .315

 

Answer:  D

Explanation:  This is a joint probability: P(L and R) = 39/124.

Difficulty: 2 Medium

Topic:  05.05 Contingency Tables

Learning Objective:  05-06 Apply the concepts of probability to contingency tables.

Bloom’s:  Apply

AACSB:  Analytical Thinking

Accessibility:  Keyboard Navigation

 

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?

 

Employee	Retire (R)			Work (W)			Total
Management (M)		13			18				31
Line worker (L)		39			54				93
Total		52			72				124

 

1. A) .250
2. B) .315
3. C) .425
4. D) .850

 

Answer:  B

Explanation:  This is a joint probability: P(R ∩ L) = 39/124.

Difficulty: 2 Medium

Topic:  05.05 Contingency Tables

Learning Objective:  05-06 Apply the concepts of probability to contingency tables.

Bloom’s:  Apply

AACSB:  Analytical Thinking

Accessibility:  Keyboard Navigation

 

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?

 

Employee	Retire (R)			Work (W)			Total
Management (M)		13			18				31
Line worker (L)		39			54				93
Total		52			72				124

 

1. A) .145
2. B) .250
3. C) .581
4. D) .687

 

Answer:  C

Explanation:  This is a conditional probability: P(W | M) = 18/31.

Difficulty: 2 Medium

Topic:  05.05 Contingency Tables

Learning Objective:  05-06 Apply the concepts of probability to contingency tables.

Bloom’s:  Apply

AACSB:  Analytical Thinking

Accessibility:  Keyboard Navigation

 

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?

 

Employee	Retire (R)			Work (W)			Total
Management (M)		13			18				31
Line worker (L)		39			54				93
Total		52			72				124

 

1. A) .750
2. B) .588
3. C) .435
4. D) .895

 

Answer:  D

Explanation:  Use the General Law of Addition: P(L or W) = 93/124 + 72/124 − 54/124.

Difficulty: 2 Medium

Topic:  05.05 Contingency Tables

Learning Objective:  05-06 Apply the concepts of probability to contingency tables.

Bloom’s:  Apply

AACSB:  Analytical Thinking

Accessibility:  Keyboard Navigation

 

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. A) 1
2. B) 2
3. C) 3
4. D) 4

 

Answer:  C

Explanation:  1 − P(F1∩F2∩F3) = 1 − (.02)(.02)(.02) = 1 − .000008 = .999992, so 3 servers will do.

Difficulty: 3 Hard

Topic:  05.04 Independent Events

Learning Objective:  05-05 Determine when events are independent.

Bloom’s:  Apply

AACSB:  Analytical Thinking

Accessibility:  Keyboard Navigation

 

70) Given the contingency table shown here, what is the probability that a mother in the study smoked during pregnancy?

 

Mother’s Education	Smoked during Pregnancy			Didn’t Smoke during Pregnancy			Row Total
Below High School		393			640			1,033
High School		560			1,370			1,930
Some College		121			635			756
College Degree		48			550			598
Col Total		1,122			3,195			4,317

 

1. A) .2599
2. B) .3174
3. C) .5000
4. D) .7401

 

Answer:  A

Explanation:  This is a marginal probability: P(smoked) = 1122/4317.

Difficulty: 1 Easy

Topic:  05.05 Contingency Tables

Learning Objective:  05-06 Apply the concepts of probability to contingency tables.

Bloom’s:  Apply

AACSB:  Analytical Thinking

Accessibility:  Keyboard Navigation

 

 

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?

 

Mother’s Education	Smoked during Pregnancy			Didn’t Smoke during Pregnancy			Row Total
Below High School		393			640			1,033
High School		560			1,370			1,930
Some College		121			635			756
College Degree		48			550			598
Col Total		1,122			3,195			4,317

 

1. A) .2385
2. B) .0907
3. C) .3503
4. D) .3804

 

Answer:  D

Explanation:  This is a conditional probability: P(smoked | below high school) = 393/1033.

Difficulty: 2 Medium

Topic:  05.05 Contingency Tables

Learning Objective:  05-06 Apply the concepts of probability to contingency tables.

Bloom’s:  Apply

AACSB:  Analytical Thinking

Accessibility:  Keyboard Navigation

 

 

72) Given the contingency table shown here, what is the probability that a mother smoked during pregnancy and had a college degree?

 

Mother’s Education	Smoked during Pregnancy			Didn’t Smoke during Pregnancy			Row Total
Below High School		393			640			1,033
High School		560			1,370			1,930
Some College		121			635			756
College Degree		48			550			598
Col Total		1,122			3,195			4,317

 

1. A) .0111
2. B) .0428
3. C) .0803
4. D) .2385

 

Answer:  A

Explanation:  This is a joint probability: P(smoked and college) = 48/4317.

Difficulty: 2 Medium

Topic:  05.05 Contingency Tables

Learning Objective:  05-06 Apply the concepts of probability to contingency tables.

Bloom’s:  Apply

AACSB:  Analytical Thinking

Accessibility:  Keyboard Navigation

 

 

73) Given the contingency table shown here, what is the probability that a mother smoked during pregnancy or that she graduated from college?

 

Mother’s Education	Smoked during Pregnancy			Didn’t Smoke during Pregnancy			Row Total
Below High School		393			640			1,033
High School		560			1,370			1,930
Some College		121			635			756
College Degree		48			550			598
Col Total		1,122			3,195			4,317

 

1. A) .0111
2. B) .2591
3. C) .3873
4. D) .7850

 

Answer:  C

Explanation:  Use the General Law of Addition: 1122/4317 + 598/4317 − 48/4317.

Difficulty: 3 Hard

Topic:  05.05 Contingency Tables

Learning Objective:  05-06 Apply the concepts of probability to contingency tables.

Bloom’s:  Apply

AACSB:  Analytical Thinking

Accessibility:  Keyboard Navigation

 

 

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?

 

Mother’s Education	Smoked during Pregnancy			Didn’t Smoke during Pregnancy			Row Total
Below High School		393			640			1,033
High School		560			1,370			1,930
Some College		121			635			756
College Degree		48			550			598
Col Total		1,122			3,195			4,317

 

1. A) .2736
2. B) .8399
3. C) .8752
4. D) .9197

 

Answer:  B

Explanation:  This is a conditional probability: 635/756.

Difficulty: 3 Hard

Topic:  05.05 Contingency Tables

Learning Objective:  05-06 Apply the concepts of probability to contingency tables.

Bloom’s:  Apply

AACSB:  Analytical Thinking

Accessibility:  Keyboard Navigation

 

 

75) Given the contingency table shown here, find the probability that a mother with some college smoked during pregnancy.

 

Mother’s Education	Smoked during Pregnancy			Didn’t Smoke during Pregnancy			Row Total
Below High School		393			640			1,033
High School		560			1,370			1,930
Some College		121			635			756
College Degree		48			550			598
Col Total		1,122			3,195			4,317

 

1. A) .1078
2. B) .1746
3. C) .1601
4. D) .1117

 

Answer:  C

Explanation:  This is a conditional probability: 121/756.

Difficulty: 2 Medium

Topic:  05.05 Contingency Tables

Learning Objective:  05-06 Apply the concepts of probability to contingency tables.

Bloom’s:  Apply

AACSB:  Analytical Thinking

Accessibility:  Keyboard Navigation

 

 

 

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?

 

Group Surveyed
Opinion:	Undergrads (U)			Graduates (G)			Faculty (F)				Row Total
Oppose Change (N)		73			27				20				120
Favor Change (S)		27			23				30				80
Col Total		100			50				50				200

 

1. A) .270
2. B) .135
3. C) .338
4. D) .756

 

Answer:  B

Explanation:  This is a joint probability: P(U and S) =27/200.

Difficulty: 2 Medium

Topic:  05.05 Contingency Tables

Learning Objective:  05-06 Apply the concepts of probability to contingency tables.

Bloom’s:  Apply

AACSB:  Analytical Thinking

Accessibility:  Keyboard Navigation

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?

 

Group Surveyed
Opinion:	Undergrads (U)			Graduates (G)			Faculty (F)				Row Total
Oppose Change (N)		73			27				20				120
Favor Change (S)		27			23				30				80
Col Total		100			50				50				200

 

1. A) .10
2. B) .25
3. C) .40
4. D) .60

 

Answer:  C

Explanation:  This is a marginal probability: P(N | F) = 20/50 = .40.

Difficulty: 2 Medium

Topic:  05.05 Contingency Tables

Learning Objective:  05-06 Apply the concepts of probability to contingency tables.

Bloom’s:  Apply

AACSB:  Analytical Thinking

Accessibility:  Keyboard Navigation

 

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?

 

Group Surveyed
Opinion:	Undergrads (U)			Graduates (G)			Faculty (F)				Row Total
Oppose Change (N)		73			27				20				120
Favor Change (S)		27			23				30				80
Col Total		100			50				50				200

 

1. A) .135
2. B) .250
3. C) .375
4. D) .540

 

Answer:  A

Explanation:  This is a joint probability: P(G and N) = 27/200.

Difficulty: 2 Medium

Topic:  05.05 Contingency Tables

Learning Objective:  05-06 Apply the concepts of probability to contingency tables.

Bloom’s:  Apply

AACSB:  Analytical Thinking

Accessibility:  Keyboard Navigation

 

 

79) Given the contingency table shown here, what is the probability that a student attends a public school in a rural area?

 

What type of school do you attend?

 

Location	Public (P)			Religious (R)				Other Private (O)				Row Total
Inner City (I)		35				15				20				70
Suburban (S)		45				10				25				80
Rural (R)		25				5				5				35
Col Total		105				30				50				185

 

1. A) .238
2. B) .714
3. C) .135
4. D) .567

 

Answer:  C

Explanation:  This is a joint probability: P(P and R) = 25/185 = .135.

Difficulty: 1 Easy

Topic:  05.05 Contingency Tables

Learning Objective:  05-06 Apply the concepts of probability to contingency tables.

Bloom’s:  Apply

AACSB:  Analytical Thinking

Accessibility:  Keyboard Navigation

 

 

80) Given the contingency table shown here, if a randomly chosen student attends a religious school, what is the probability the location is rural?

 

What type of school do you attend?

 

Location	Public (P)			Religious (R)				Other Private (O)				Row Total
Inner City (I)		35				15				20				70
Suburban (S)		45				10				25				80
Rural (R)		25				5				5				35
Col Total		105				30				50				185

 

1. A) .142
2. B) .162
3. C) .167
4. D) .333

 

Answer:  C

Explanation:  This is a conditional probability: P(R | L) = 5/30 = .167.

Difficulty: 2 Medium

Topic:  05.05 Contingency Tables

Learning Objective:  05-06 Apply the concepts of probability to contingency tables.

Bloom’s:  Apply

AACSB:  Analytical Thinking

Accessibility:  Keyboard Navigation

 

 

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?

 

What type of school do you attend?

 

Location	Public (P)			Religious (R)				Other Private (O)				Row Total
Inner City (I)		35				15				20				70
Suburban (S)		45				10				25				80
Rural (R)		25				5				5				35
Col Total		105				30				50				185

 

1. A) .189
2. B) .333
3. C) .500
4. D) .567

 

Answer:  C

Explanation:  This is a conditional probability: P(P | I) = 35/70 = .500.

Difficulty: 2 Medium

Topic:  05.05 Contingency Tables

Learning Objective:  05-06 Apply the concepts of probability to contingency tables.

Bloom’s:  Apply

AACSB:  Analytical Thinking

Accessibility:  Keyboard Navigation

 

82) Given the contingency table shown here, find P(E).

 

Major
Gender	Accounting (A)			Gen. Mgmt. (G)			Economics (E)			Row Total
Male (M)		210			180			140			530
Female (F)		150			160			160			470
Col Total		360			340			300			1000

 

1. A) .180
2. B) .300
3. C) .529
4. D) .641

 

Answer:  B

Explanation:  This is a marginal probability: P(E) = 300/1000 = .300.

Difficulty: 1 Easy

Topic:  05.05 Contingency Tables

Learning Objective:  05-06 Apply the concepts of probability to contingency tables.

Bloom’s:  Apply

AACSB:  Analytical Thinking

Accessibility:  Keyboard Navigation

 

83) Given the contingency table shown here, find P(E | F).

 

Major
Gender	Accounting (A)			Gen. Mgmt. (G)			Economics (E)			Row Total
Male (M)		210			180			140			530
Female (F)		150			160			160			470
Col Total		360			340			300			1000

 

1. A) .160
2. B) .300
3. C) .340
4. D) .533

 

Answer:  C

Explanation:  This is a conditional probability: P(E | F) = 160/470 = .340.

Difficulty: 2 Medium

Topic:  05.05 Contingency Tables

Learning Objective:  05-06 Apply the concepts of probability to contingency tables.

Bloom’s:  Apply

AACSB:  Analytical Thinking

Accessibility:  Keyboard Navigation

 

84) Given the contingency table shown here, find P(A ∩ M).

 

Major
Gender	Accounting (A)			Gen. Mgmt. (G)			Economics (E)			Row Total
Male (M)		210			180			140			530
Female (F)		150			160			160			470
Col Total		360			340			300			1000

 

1. A) .210
2. B) .360
3. C) .396
4. D) .583

 

Answer:  A

Explanation:  This is a joint probability: P(A ∩ M) = 210/1000 = .210.

Difficulty: 1 Easy

Topic:  05.05 Contingency Tables

Learning Objective:  05-06 Apply the concepts of probability to contingency tables.

Bloom’s:  Apply

AACSB:  Analytical Thinking

Accessibility:  Keyboard Navigation

 

 

85) Given the contingency table shown here, find P(F or G).

 

Major
Gender	Accounting (A)			Gen. Mgmt. (G)			Economics (E)			Row Total
Male (M)		210			180			140			530
Female (F)		150			160			160			470
Col Total		360			340			300			1000

 

1. A) .160
2. B) .470
3. C) .650
4. D) .810

 

Answer:  C

Explanation:  Use the General Law of Addition: P(F or G) = 470/1000 + 340/1000 − 160/1000.

Difficulty: 3 Hard

Topic:  05.05 Contingency Tables

Learning Objective:  05-06 Apply the concepts of probability to contingency tables.

Bloom’s:  Apply

AACSB:  Analytical Thinking

Accessibility:  Keyboard Navigation

 

86) Given the contingency table shown here, find the probability that a randomly chosen individual is a female economics major.

 

Major
Gender	Accounting (A)			Gen. Mgmt. (G)			Economics (E)			Row Total
Male (M)		210			180			140			530
Female (F)		150			160			160			470
Col Total		360			340			300			1000

 

1. A) .3404
2. B) .4700
3. C) .1600
4. D) .5333

 

Answer:  C

Explanation:  This is a joint probability: P(F and E) = 160/1000 = .16.

Difficulty: 2 Medium

Topic:  05.05 Contingency Tables

Learning Objective:  05-06 Apply the concepts of probability to contingency tables.

Bloom’s:  Apply

AACSB:  Analytical Thinking

Accessibility:  Keyboard Navigation

 

 

87) Debbie has two stocks, X and Y. Consider the following events:

X = the event that the price of stock X has increased

Y = the event that the price of stock Y has increased

The event “the price of stock X has increased and the price of stock Y has not increased” may be written as

1. A) X ‘ ∩ Y
2. B) X or Y ′
3. C) X ∩ Y ′
4. D) X or Y

 

Answer:  C

Explanation:  This is a joint probability that also entails the notation for an event’s complement.

Difficulty: 1 Easy

Topic:  05.03 Rules of Probability

Learning Objective:  05-03 Apply the definitions and rules of probability.

Bloom’s:  Understand

AACSB:  Analytical Thinking

Accessibility:  Keyboard Navigation

 

88) If P(A | B) = .40 and P(B) = .30, find P(A ∩ B).

1. A) .171
2. B) .525
3. C) .571
4. D) .120

 

Answer:  D

Explanation:  Use the definition for conditional probability.

Difficulty: 3 Hard

Topic:  05.03 Rules of Probability

Learning Objective:  05-03 Apply the definitions and rules of probability.

Bloom’s:  Apply

AACSB:  Analytical Thinking

Accessibility:  Keyboard Navigation

 

 

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

1. A) .014
2. B) .140
3. C) .070
4. D) .483

 

Answer:  D

Explanation:  Review Bayes’ Theorem, and perhaps make a table or tree.

Difficulty: 3 Hard

Topic:  05.07 Bayes’ Theorem

Learning Objective:  05-08 Use Bayes’ Theorem to calculate revised probabilities.

Bloom’s:  Apply

AACSB:  Analytical Thinking

Accessibility:  Keyboard Navigation

 

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.

 

	Call Length (minutes)
Type of Phone Call	0−1				2− 5				6+					Total
Local		150				250				100					500
Long Distance		170				120				10					300
Total		320				370				110					800

 

1. A) .5000
2. B) .3125
3. C) .4000
4. D) .4625

 

Answer:  B

Explanation:  This is a joint probability: 250/800 = .3125

Difficulty: 2 Medium

Topic:  05.05 Contingency Tables

Learning Objective:  05-06 Apply the concepts of probability to contingency tables.

Bloom’s:  Apply

AACSB:  Analytical Thinking

Accessibility:  Keyboard Navigation

 

 

91) Given the contingency table shown here, if a call is sampled at random, find the marginal probability that the call is long distance.

 

	Call Length (minutes)
Type of Phone Call	0−1				2− 5				6+					Total
Local		150				250				100					500
Long Distance		170				120				10					300
Total		320				370				110					800

 

1. A) .3750
2. B) .6250
3. C) .4000
4. D) 300/500

 

Answer:  A

Explanation:  500/800 = .6250.

Difficulty: 2 Medium

Topic:  05.05 Contingency Tables

Learning Objective:  05-06 Apply the concepts of probability to contingency tables.

Bloom’s:  Apply

AACSB:  Analytical Thinking

Accessibility:  Keyboard Navigation

 

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

 

	Call Length (minutes)
Type of Phone Call	0−1				2− 5				6+					Total
Local		150				250				100					500
Long Distance		170				120				10					300
Total		320				370				110					800

 

1. A) 120/300
2. B) 10/300
3. C) .9667
4. D) .6667

 

Answer:  C

Explanation:  Calculate the conditional probability 1 − 10/300 = .9667.

Difficulty: 3 Hard

Topic:  05.05 Contingency Tables

Learning Objective:  05-06 Apply the concepts of probability to contingency tables.

Bloom’s:  Apply

AACSB:  Analytical Thinking

Accessibility:  Keyboard Navigation

 

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 .625 probability that the shareholder will be a woman. Consequently, what is the number of men holding 1000 or more shares?

 

	Number of Shares Held
Shareholders		0−499					500−999						1,000+						Total
Men			3,000																		4,000
Women			2,000																		4,800
Joint Accounts			0						1,000					200							1,200
Total			5,000						4,000					1,000							10,000

 

1. A) 1,000
2. B) 250
3. C) 7,500
4. D) 500

 

Answer:  D

Explanation:  Multiply by the column total and subtract to fill in the remaining frequencies.

Difficulty: 3 Hard

Topic:  05.05 Contingency Tables

Learning Objective:  05-06 Apply the concepts of probability to contingency tables.

Bloom’s:  Apply

AACSB:  Analytical Thinking

Accessibility:  Keyboard Navigation

 

94) In any sample space P(A | B) and P(B | A)

1. A) are always equal to one another.
2. B) are never equal to one another.
3. C) are reciprocals of one another.
4. D) are equal only if P(A) = P(B).

 

Answer:  D

Explanation:  Use the definition of conditional probability.

Difficulty: 3 Hard

Topic:  05.03 Rules of Probability

Learning Objective:  05-03 Apply the definitions and rules of probability.

Bloom’s:  Remember

AACSB:  Analytical Thinking

Accessibility:  Keyboard Navigation

 

 

95) If P(A ∩ B) = .50, can P(A) = .20?

1. A) Only if P(A | B) = .10
2. B) Not unless P(B) = .30
3. C) Only if P(B ∩ A) = .60
4. D) If P(A) = .20, then P(A ∩ B) cannot equal .50.

 

Answer:  D

Explanation:  The given information contains a contradiction, because P(A ∩ B) cannot exceed P(A).

Difficulty: 2 Medium

Topic:  05.03 Rules of Probability

Learning Objective:  05-03 Apply the definitions and rules of probability.

Bloom’s:  Apply

AACSB:  Analytical Thinking

Accessibility:  Keyboard Navigation

 

96) The following relationship always holds true for events A and B in a sample space.

1. A) P(A | B) = P(B | A)
2. B) P(A ∩ B) = P(A | B) P(B)
3. C) P(A | B) = P(B | A) P(A)

 

Answer:  B

Explanation:  Use the definition of conditional probability: P(A | B) = P(A ∩ B) / P(B).

Difficulty: 3 Hard

Topic:  05.03 Rules of Probability

Learning Objective:  05-03 Apply the definitions and rules of probability.

Bloom’s:  Understand

AACSB:  Analytical Thinking

Accessibility:  Keyboard Navigation

 

97) The following probabilities are given about events A and B in a sample space: P(A) = .30, P(B) = .40, P(A or B) = .60. We can say that

70. A) P(A ∩ B) = .70.
71. B) P(A) = P(A ∩ B).
72. C) P(A ∩ B) = .10.
73. D) A and B are independent events.

 

Answer:  C

Explanation:  Apply the General Rule of Addition: P(A or B) = P(A) + P(B) − P(A ∩ B).

Difficulty: 3 Hard

Topic:  05.03 Rules of Probability

Learning Objective:  05-03 Apply the definitions and rules of probability.

Bloom’s:  Apply

AACSB:  Analytical Thinking

Accessibility:  Keyboard Navigation

 

 

98) If P(A) = .35, P(B) = .60, and P(A or B) = .70, then

1. A) A and B are mutually exclusive.
2. B) P(A ∩ B) = .15.
3. C) P(A ∩ B) = .25.
4. D) P(A ∩ B) = .35.

 

Answer:  C

Explanation:  Apply the General Rule of Addition: P(A or B) = P(A) + P(B) − P(A ∩ B).

Difficulty: 3 Hard

Topic:  05.03 Rules of Probability

Learning Objective:  05-03 Apply the definitions and rules of probability.

Bloom’s:  Apply

AACSB:  Analytical Thinking

Accessibility:  Keyboard Navigation

 

99) The following table shows the survival experience of 1,000 males who retire at age 65:

 

Age	Number of Males Surviving
65		1,000
70		907
75		775
80		596
85		383

 

Based on these data, the probability that a 75-year-old male will survive to age 80 is

1. A) .596
2. B) 1 − .596 = .404
3. C) 1 − .775 = .225
4. D) .769

 

Answer:  D

Explanation:  Given that 775 have survived to 75, the probability is 596 divided by 775.

Difficulty: 3 Hard

Topic:  05.03 Rules of Probability

Learning Objective:  05-03 Apply the definitions and rules of probability.

Bloom’s:  Apply

AACSB:  Analytical Thinking

Accessibility:  Keyboard Navigation

 

 

100) Given the contingency table shown here, find P(G | M).

 

Vehicle Type		Somerset (S)				Oakland (O)					Great Lakes (G)					Row Total
Car (C)			44				49					36						129
Minivan (M)			21				15					18						54
Full-Size Van (F)			2				3					3						8
SUV (V)			19				27					26						72
Truck (T)			14				6					17						37
Col Total			100				100					100						300

 

1. A) .1800
2. B) .0450
3. C) .3333
4. D) .1350

 

Answer:  C

Explanation:  This is a conditional probability: P(G | M) = 18/54.

Difficulty: 2 Medium

Topic:  05.05 Contingency Tables

Learning Objective:  05-06 Apply the concepts of probability to contingency tables.

Bloom’s:  Apply

AACSB:  Analytical Thinking

Accessibility:  Keyboard Navigation

 

 

101) Given the contingency table shown here, find P(V or S).

 

Vehicle Type		Somerset (S)				Oakland (O)					Great Lakes (G)					Row Total
Car (C)			44				49					36						129
Minivan (M)			21				15					18						54
Full-Size Van (F)			2				3					3						8
SUV (V)			19				27					26						72
Truck (T)			14				6					17						37
Col Total			100				100					100						300

 

1. A) .5100
2. B) .4300
3. C) .0475
4. D) .4775

 

Answer:  A

Explanation:  Use the General Rule of Addition: P(V or S) = 72/300 + 100/300 − 19/300.

Difficulty: 3 Hard

Topic:  05.05 Contingency Tables

Learning Objective:  05-06 Apply the concepts of probability to contingency tables.

Bloom’s:  Apply

AACSB:  Analytical Thinking

Accessibility:  Keyboard Navigation

 

 

102) Given the contingency table shown here, find P(V | S).

 

Vehicle Type		Somerset (S)				Oakland (O)					Great Lakes (G)					Row Total
Car (C)			44				49					36						129
Minivan (M)			21				15					18						54
Full-Size Van (F)			2				3					3						8
SUV (V)			19				27					26						72
Truck (T)			14				6					17						37
Col Total			100				100					100						300

 

1. A) .2639
2. B) .1900
3. C) .0475
4. D) .4144

 

Answer:  B

Explanation:  This is a conditional probability: P(V | S) = 19/100.

Difficulty: 2 Medium

Topic:  05.05 Contingency Tables

Learning Objective:  05-06 Apply the concepts of probability to contingency tables.

Bloom’s:  Apply

AACSB:  Analytical Thinking

Accessibility:  Keyboard Navigation

 

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?

1. A) .90
2. B) .85
3. C) .72
4. D) .65

 

Answer:  C

Explanation:  Use the General Rule of Addition: P(A or B) = P(A) + P(B) − P(A ∩ B) = .25 + .65 − .18.

Difficulty: 2 Medium

Topic:  05.03 Rules of Probability

Learning Objective:  05-03 Apply the definitions and rules of probability.

Bloom’s:  Apply

AACSB:  Analytical Thinking

Accessibility:  Keyboard Navigation

 

 

104) Two events are complementary (i.e., they are complements) if

1. A) the sum of their probabilities equals one.
2. B) they are disjoint and their probabilities sum to one.
3. C) the joint probability of the two events equals one.
4. D) they are independent events with equal probabilities.

 

Answer:  B

Explanation:  Review the rules of probability.

Difficulty: 2 Medium

Topic:  05.03 Rules of Probability

Learning Objective:  05-03 Apply the definitions and rules of probability.

Bloom’s:  Remember

AACSB:  Analytical Thinking

Accessibility:  Keyboard Navigation

 

105) Which statement is false?

1. A) If P(A) = .05, then the odds against event A’s occurrence are 19 to 1.
2. B) If A and B are mutually exclusive events, then P(A or B) = 0.
3. C) The number of permutations of five things taken two at a time is 20.

 

Answer:  B

Explanation:  Review rules of probability and counting rules.

Difficulty: 3 Hard

Topic:  05.03 Rules of Probability

Learning Objective:  05-03 Apply the definitions and rules of probability.

Bloom’s:  Apply

AACSB:  Analytical Thinking

Accessibility:  Keyboard Navigation

 

106) The number of unique orders in which five items (A, B, C, D, E) can be arranged is

1. A) 5
2. B) 840
3. C) 120
4. D) 24

 

Answer:  C

Explanation:  Apply rules of counting: 5 × 4 × 3 × 2 × 1 = 120.

Difficulty: 3 Hard

Topic:  05.08 Counting Rules

Learning Objective:  05-09 Apply counting rules to calculate possible event arrangements.

Bloom’s:  Apply

AACSB:  Analytical Thinking

Accessibility:  Keyboard Navigation

 

 

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)?

1. A) 35
2. B) 840
3. C) 5040
4. D) 24

 

Answer:  B

Explanation:  This is 7P4.

Difficulty: 3 Hard

Topic:  05.08 Counting Rules

Learning Objective:  05-09 Apply counting rules to calculate possible event arrangements.

Bloom’s:  Apply

AACSB:  Analytical Thinking

Accessibility:  Keyboard Navigation

 

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?

1. A) 60
2. B) 120
3. C) 10
4. D) 24

 

Answer:  C

Explanation:  This is 5C3.

Difficulty: 3 Hard

Topic:  05.08 Counting Rules

Learning Objective:  05-09 Apply counting rules to calculate possible event arrangements.

Bloom’s:  Apply

AACSB:  Analytical Thinking

Accessibility:  Keyboard Navigation

 

 

 

109) The value of 6C2 is

1. A) 15
2. B) 30
3. C) 720
4. D) 12

 

Answer:  A

Explanation:  Apply the formula for combinations.

Difficulty: 2 Medium

Topic:  05.08 Counting Rules

Learning Objective:  05-09 Apply counting rules to calculate possible event arrangements.

Bloom’s:  Remember

AACSB:  Analytical Thinking

Accessibility:  Keyboard Navigation

 

110) The value of 4P2 is

1. A) 8
2. B) 6
3. C) 24
4. D) 12

 

Answer:  D

Explanation:  Apply the formula for permutations.

Difficulty: 2 Medium

Topic:  05.08 Counting Rules

Learning Objective:  05-09 Apply counting rules to calculate possible event arrangements.

Bloom’s:  Remember

AACSB:  Analytical Thinking

Accessibility:  Keyboard Navigation

 

111) The probability that event A occurs, given that event B has occurred, is an example of:

1. A) a marginal probability.
2. B) a conditional probability.
3. C) a joint probability.
4. D) more than one of the above.

 

Answer:  B

Explanation:  Review the definition of conditional probability P(A | B).

Difficulty: 1 Easy

Topic:  05.03 Rules of Probability

Learning Objective:  05-03 Apply the definitions and rules of probability.

Bloom’s:  Remember

AACSB:  Analytical Thinking

Accessibility:  Keyboard Navigation

 

 

 

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

1. A) .9951
2. B) .8649
3. C) .9300
4. D) .9522

 

Answer:  A

Explanation:  Failure probability for either server is 1 − .93 = .07. We are told that failures are independent events, so we can multiply their probabilities. The probability that both will fail is P(F1 and F2) = P(F1) P(F2) = (.07)(.07) = .0049, so the probability that one or the other will operate is 1 − .0049 = .9951

Difficulty: 2 Medium

Topic:  05.04 Independent Events

Learning Objective:  05-05 Determine when events are independent.

Bloom’s:  Remember

AACSB:  Analytical Thinking

Accessibility:  Keyboard Navigation

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?

1. A) No
2. B) Yes
3. C) Can’t tell from given information

 

Answer:  A

Explanation:  For independence we would require P(A)P(B) = P(A ∩ B).

Difficulty: 2 Medium

Topic:  05.04 Independent Events

Learning Objective:  05-05 Determine when events are independent.

Bloom’s:  Apply

AACSB:  Analytical Thinking

Accessibility:  Keyboard Navigation

 

 

 

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?

1. A) .010
2. B) .005
3. C) .001
4. D) Cannot be determined

 

Answer:  B

Explanation:  Because they are independent events then P(A ∩ B) = P(A)P(B) = (.05)(.10).

Difficulty: 2 Medium

Topic:  05.04 Independent Events

Learning Objective:  05-05 Determine when events are independent.

Bloom’s:  Apply

AACSB:  Analytical Thinking

Accessibility:  Keyboard Navigation

 

115) If two events are collectively exhaustive, what is the probability that one or the other will occur?

1. A) 1.00
2. B) .00
3. C) .50
4. D) Cannot tell from given information

 

Answer:  A

Explanation:  Review definition of probabilities (collectively exhaustive covers all the possibilities).

Difficulty: 2 Medium

Topic:  05.03 Rules of Probability

Learning Objective:  05-03 Apply the definitions and rules of probability.

Bloom’s:  Apply

AACSB:  Analytical Thinking

Accessibility:  Keyboard Navigation

 

 

116) Which best exemplifies a subjective probability?

1. A) The probability that a female age 30 will have an accident in a week’s car rental at Hertz
2. B) The probability that a pair of dice will come up 7 in a given throw
3. C) The probability that the summer Olympic games will be held in Chicago in 2020
4. D) The probability that a checked bag on Flight 1872 will weigh more than 40 pounds

 

Answer:  C

Explanation:  Subjective probabilities are not based on empirical frequencies.

Difficulty: 2 Medium

Topic:  05.02 Probability

Learning Objective:  05-02 Distinguish among the three views of probability.

Bloom’s:  Analyze

AACSB:  Analytical Thinking

Accessibility:  Keyboard Navigation

 

117) Which best exemplifies the classical definition of probability?

1. A) The probability that a male age 50 will have an accident in a week’s car rental at Alamo
2. B) The probability that a pair of dice will come up 7 when they are rolled
3. C) The probability that the winter Olympic games will be held in Europe in 2022
4. D) The probability that a checked bag on Flight 1872 will weigh more than 30 pounds

 

Answer:  B

Explanation:  Classical probability is determined a priori by the nature of the experiment.

Difficulty: 2 Medium

Topic:  05.02 Probability

Learning Objective:  05-02 Distinguish among the three views of probability.

Bloom’s:  Analyze

AACSB:  Analytical Thinking

Accessibility:  Keyboard Navigation

118) Which best exemplifies the empirical definition of probability?

1. A) The probability that a Chinese athlete will win the diving competition in the next Olympics
2. B) The probability that a fair coin will come up heads when it is flipped
3. C) The probability that your own bank will become insolvent within 12 months
4. D) The probability that a checked bag on Flight 1872 will weigh less than 30 pounds

 

Answer:  D

Explanation:  Empirical probabilities are based on observed frequencies.

Difficulty: 2 Medium

Topic:  05.02 Probability

Learning Objective:  05-02 Distinguish among the three views of probability.

Bloom’s:  Analyze

AACSB:  Analytical Thinking

Accessibility:  Keyboard Navigation

 

 

 

119) From the following tree, find the probability that a randomly chosen person will get the flu vaccine and will also get the flu.

 

1. A) .10
2. B) .07
3. C) .19
4. D) .70

 

Answer:  B

Explanation:  Multiply down the branch: .70 × .10 = .07.

Difficulty: 2 Medium

Topic:  05.06 Tree Diagrams

Learning Objective:  05-07 Interpret a tree diagram.

Bloom’s:  Apply

AACSB:  Analytical Thinking

Accessibility:  Keyboard Navigation

 

 

120) From the following tree, find the probability that a randomly chosen person will not get a vaccination and will not get the flu.

 

 

1. A) .18
2. B) .60
3. C) .19
4. D) .70

 

Answer:  A

Explanation:  Multiply down the branch: .30 × .60 = .18.

Difficulty: 2 Medium

Topic:  05.06 Tree Diagrams

Learning Objective:  05-07 Interpret a tree diagram.

Bloom’s:  Apply

AACSB:  Analytical Thinking

Accessibility:  Keyboard Navigation

 

 

121) From the following tree, find the probability that a randomly chosen person will get the flu.

 

1. A) .19
2. B) .07
3. C) .81
4. D) .70

 

Answer:  A

Explanation:  Multiply down two branches and add .07 to .12. That is (.70)(.10) + (.30)(.40).

Difficulty: 2 Medium

Topic:  05.06 Tree Diagrams

Learning Objective:  05-07 Interpret a tree diagram.

Bloom’s:  Apply

AACSB:  Analytical Thinking

Accessibility:  Keyboard Navigation

 

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?

1. A) .7407
2. B) .8000
3. C) .5400
4. D) .5000

 

Answer:  A

Explanation:  Review Bayes’ Theorem, and perhaps make a table or tree.

Difficulty: 3 Hard

Topic:  05.07 Bayes’ Theorem

Learning Objective:  05-08 Use Bayes’ Theorem to calculate revised probabilities.

Bloom’s:  Remember

AACSB:  Analytical Thinking

Accessibility:  Keyboard Navigation

 

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?

1. A) .7000
2. B) .5000
3. C) .4615
4. D) .1300

 

Answer:  C

Explanation:  Review Bayes’ Theorem, and perhaps make a table or tree.

Difficulty: 3 Hard

Topic:  05.07 Bayes’ Theorem

Learning Objective:  05-08 Use Bayes’ Theorem to calculate revised probabilities.

Bloom’s:  Remember

AACSB:  Analytical Thinking

Accessibility:  Keyboard Navigation