Statistics for Management and Economics Abbreviated 10e Gerald Keller - Test Bank

Statistics for Management and Economics Abbreviated 10e Gerald Keller – Test Bank

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Sample Questions Are Posted Below

CHAPTER 6: PROBABILITY

TRUE/FALSE

The relative frequency approach to probability uses long term relative frequencies, often based on past data.

ANS: T PTS: 1 DIF: Easy OBJ: SFME.KELL.15.06.01

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.03

KEY: Bloom’s: Comprehension

Predicting the outcome of a football game is using the subjective approach to probability.

ANS: T PTS: 1 DIF: Easy OBJ: SFME.KELL.15.06.01

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.03

KEY: Bloom’s: Comprehension

You think you have a 90% chance of passing your next advanced financial accounting exam. This is an example of subjective approach to probability.

ANS: T PTS: 1 DIF: Easy OBJ: SFME.KELL.15.06.01

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.03

KEY: Bloom’s: Application

P(A) + P(B) = 1 for any events A and B that are mutually exclusive.

ANS: F PTS: 1 DIF: Easy OBJ: SFME.KELL.15.06.01

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.03

KEY: Bloom’s: Comprehension

The collection of all the possible outcomes of a random experiment is called a sample space.

ANS: T PTS: 1 DIF: Easy OBJ: SFME.KELL.15.06.01

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.03

KEY: Bloom’s: Comprehension

If events A and B cannot occur at the same time, they are called mutually exclusive.

ANS: T PTS: 1 DIF: Easy OBJ: SFME.KELL.15.06.01

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.03

KEY: Bloom’s: Comprehension

If either event A or event B must occur, they are called mutually exclusive.

ANS: F PTS: 1 DIF: Easy OBJ: SFME.KELL.15.06.01

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.03

KEY: Bloom’s: Comprehension

If either event A or event B must occur, then A and B are mutually exclusive and collectively exhaustive events.

ANS: T PTS: 1 DIF: Easy OBJ: SFME.KELL.15.06.01

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.03

KEY: Bloom’s: Comprehension

If P(A) = 0.4 and P(B) = 0.6, then A and B must be collectively exhaustive.

ANS: F PTS: 1 DIF: Easy OBJ: SFME.KELL.15.06.01

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.03

KEY: Bloom’s: Comprehension

If P(A) = 0.4 and P(B) = 0.6, then A and B must be mutually exclusive.

ANS: F PTS: 1 DIF: Easy OBJ: SFME.KELL.15.06.01

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.03

KEY: Bloom’s: Comprehension

MULTIPLE CHOICE

Of the last 500 customers entering a supermarket, 50 have purchased a wireless phone. If the relative frequency approach for assigning probabilities is used, the probability that the next customer will purchase a wireless phone is

a.	0.10	c.	0.50
b.	0.90	d.	None of these choices.

ANS: A PTS: 1 DIF: Easy OBJ: SFME.KELL.15.06.01

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.03

KEY: Bloom’s: Application

If A and B are mutually exclusive events with P(A) = 0.75, then P(B):

a.	can be any value between 0 and 1.	c.	cannot be larger than 0.25.
b.	can be any value between 0 and 0.75.	d.	equals 0.25.

ANS: C PTS: 1 DIF: Easy OBJ: SFME.KELL.15.06.01

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.03

KEY: Bloom’s: Application

If you roll a balanced die 50 times, you should expect an even number to appear:

a.	on every other roll.	c.	25 times on average, over the long term.
b.	exactly 50 times out of 100 rolls.	d.	All of these choices are true.

ANS: C PTS: 1 DIF: Easy OBJ: SFME.KELL.15.06.01

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.03

KEY: Bloom’s: Application

An approach of assigning probabilities which assumes that all outcomes of the experiment are equally likely is referred to as the:

a.	subjective approach	c.	classical approach
b.	objective approach	d.	relative frequency approach

ANS: C PTS: 1 DIF: Easy OBJ: SFME.KELL.15.06.01

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.03

KEY: Bloom’s: Comprehension

The collection of all possible outcomes of an experiment is called:

a.	a simple event	c.	a sample
b.	a sample space	d.	a population

ANS: B PTS: 1 DIF: Easy OBJ: SFME.KELL.15.06.01

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.03

KEY: Bloom’s: Comprehension

Which of the following is an approach to assigning probabilities?

a.	Classical approach	c.	Subjective approach
b.	Relative frequency approach	d.	All of these choices are true.

ANS: B PTS: 1 DIF: Easy OBJ: SFME.KELL.15.06.01

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.03

KEY: Bloom’s: Comprehension

A sample space of an experiment consists of the following outcomes: 1, 2, 3, 4, and 5. Which of the following is a simple event?

a.	At least 3	c.	3
b.	At most 2	d.	15

ANS: C PTS: 1 DIF: Easy OBJ: SFME.KELL.15.06.01

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.03

KEY: Bloom’s: Comprehension

Which of the following is a requirement of the probabilities assigned to outcome O_i?

a.	P(O_i) £ 0 for each i	c.	0 £ P(O_i) £ 1 for each i
b.	P(O_i) ³ 1 for each i	d.	P(O_i) = 1 for each i

ANS: C PTS: 1 DIF: Easy OBJ: SFME.KELL.15.06.01

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.03

KEY: Bloom’s: Knowledge

If an experiment consists of five outcomes with P(O₁) = 0.10, P(O₂) = 0.20, P(O₃) = 0.30, P(O₄) = 0.25, then P(O₅) is

a.	0.75
b.	0.15
c.	0.50
d.	Cannot be determined from the information given.

ANS: B PTS: 1 DIF: Easy OBJ: SFME.KELL.15.06.01

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.03

KEY: Bloom’s: Application

If two events are collectively exhaustive, what is the probability that one or the other occurs?

a.	0.00
b.	0.50
c.	1.00
d.	Cannot be determined from the information given.

ANS: D PTS: 1 DIF: Easy OBJ: SFME.KELL.15.06.01

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.03

KEY: Bloom’s: Application

If two events are collectively exhaustive, what is the probability that both occur at the same time?

a.	0.00
b.	0.50
c.	1.00
d.	Cannot be determined from the information given.

ANS: D PTS: 1 DIF: Easy OBJ: SFME.KELL.15.06.01

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.03

KEY: Bloom’s: Application

If two events are mutually exclusive, what is the probability that one or the other occurs?

a.	0.00
b.	0.50
c.	1.00
d.	Cannot be determined from the information given.

ANS: D PTS: 1 DIF: Easy OBJ: SFME.KELL.15.06.01

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.03

KEY: Bloom’s: Application

If two events are mutually exclusive, what is the probability that both occur at the same time?

a.	0.00
b.	0.50
c.	1.00
d.	Cannot be determined from the information given.

ANS: A PTS: 1 DIF: Easy OBJ: SFME.KELL.15.06.01

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.03

KEY: Bloom’s: Application

If two events are mutually exclusive and collectively exhaustive, what is the probability that both occur?

a.	0.00
b.	0.50
c.	1.00
d.	Cannot be determined from the information given.

ANS: A PTS: 1 DIF: Easy OBJ: SFME.KELL.15.06.01

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.03

KEY: Bloom’s: Application

If the two events are mutually exclusive and collectively exhaustive, what is the probability that one or the other occurs?

a.	0.00
b.	0.50
c.	1.00
d.	Cannot be determined from the information given.

ANS: C PTS: 1 DIF: Easy OBJ: SFME.KELL.15.06.01

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.03

KEY: Bloom’s: Application

If events A and B are mutually exclusive and collectively exhaustive, what is the probability that event A occurs?

a.	0.25
b.	0.50
c.	1.00
d.	Cannot be determined from the information given.

ANS: D PTS: 1 DIF: Easy OBJ: SFME.KELL.15.06.01

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.03

KEY: Bloom’s: Application

If two equally likely events A and B are mutually exclusive and collectively exhaustive, what is the probability that event A occurs?

a.	0.00
b.	0.50
c.	1.00
d.	Cannot be determined from the information given.

ANS: B PTS: 1 DIF: Easy OBJ: SFME.KELL.15.06.01

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.03

KEY: Bloom’s: Application

If event A and event B cannot occur at the same time, then A and B are said to be

a.	mutually exclusive	c.	collectively exhaustive
b.	independent	d.	None of these choices.

ANS: A PTS: 1 DIF: Easy OBJ: SFME.KELL.15.06.01

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.03

KEY: Bloom’s: Comprehension

The collection of all possible events is called

a.	an outcome	c.	an event
b.	a sample space	d.	None of these choices.

ANS: B PTS: 1 DIF: Easy OBJ: SFME.KELL.15.06.01

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.03

KEY: Bloom’s: Comprehension

COMPLETION

A random experiment is an action or process that leads to one of several possible ____________________.

ANS: outcomes

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.06.01

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.03

KEY: Bloom’s: Knowledge

The outcomes of a sample space must be ____________________, which means that all possible outcomes must be included.

ANS: exhaustive

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.06.01

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.03

KEY: Bloom’s: Comprehension

The outcomes of a sample space must be ____________________, which means that no two outcomes can occur at the same time.

ANS: mutually exclusive

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.06.01

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.03

KEY: Bloom’s: Comprehension

A(n) ____________________ of a random experiment is a list of all possible outcomes of the experiment.

ANS: sample space

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.06.01

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.03

KEY: Bloom’s: Knowledge

The outcomes of a sample space must be ____________________ and ____________________.

ANS:

exhaustive; mutually exclusive

mutually exclusive; exhaustive

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.06.01

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.03

KEY: Bloom’s: Comprehension

There are ____________________ requirements of probabilities for the outcomes of a sample space.

ANS:

two

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.06.01

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.03

KEY: Bloom’s: Comprehension

An individual outcome of a sample space is called a(n) ____________________ event.

ANS: simple

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.06.01

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.03

KEY: Bloom’s: Knowledge

A(n) ____________________ is a collection or set of one or more simple events in a sample space.

ANS: event

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.06.01

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.03

KEY: Bloom’s: Knowledge

The probability of an event is the ____________________ of the probabilities of the simple events that constitute the event.

ANS: sum

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.06.01

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.03

KEY: Bloom’s: Knowledge

No matter which approach was used to assign probability (classical, relative frequency, or subjective) the one that is always used to interpret a probability is the ____________________ approach.

ANS: relative frequency

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.06.01

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.03

KEY: Bloom’s: Comprehension

SHORT ANSWER

Alana, Eva, and Stephanie, three candidates for the presidency of a college’s student body, are to address a student forum. The forum’s organizer is to select the order in which the candidates will give their speeches, and must do so in such a way that each possible order is equally likely to be selected.

a.	What is the random experiment?
b.	List the outcomes in the sample space.
c.	Assign probabilities to the outcomes.
d.	What is the probability that Stephanie will speak first?
e.	What is the probability that Alana will speak before Stephanie does?

ANS:

a.	The random experiment is to observe the order in which the three candidates give their speeches.
b.	S = {ABC, ACB, BAC, BCA, CAB, CBA}, where A = Alana, B = Eva, and C = Stephanie.
c.	The probability assigned to each outcome is 1/6.
d.	P(CAB, CBA) = 1/3
e.	P(ABC, ACB, BAC) = 1/2

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.06.01

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.03

KEY: Bloom’s: Application

There are three approaches to determining the probability that an outcome will occur: classical, relative frequency, and subjective. For each situation that follows, determine which approach is most appropriate.

a.	A Russian will win the French Open Tennis Tournament next year.
b.	The probability of getting any single number on a balanced die is 1/6.
c.	Based on the past, it’s reasonable to assume the average book sales for a certain textbook is 6,500 copies per month.

ANS:

a.	subjective
b.	classical
c.	relative frequency

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.06.01

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.03

KEY: Bloom’s: Comprehension

Hobby Shop Sales

Sales records of a hobby shop showed the following number of radio controlled trucks sold weekly for each of the last 50 weeks.

Number of Trucks Sold	Number of Weeks
0	20
1	15
2	10
3	4
4	1

{Hobby Shop Sales Narrative} Define the random experiment of interest to the store.

ANS:

The random experiment consists of observing the number of trucks sold in any given week.

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.06.01

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.03

KEY: Bloom’s: Application

{Hobby Shop Sales Narrative} List the outcomes in the sample space.

ANS:

S = {0, 1, 2, 3, 4}

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.06.01

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.03

KEY: Bloom’s: Application

{Hobby Shop Sales Narrative} What approach would you use in determining the probabilities for next week’s sales? Assign probabilities to the outcomes.

ANS:

The relative frequency approach was used.

Number of Trucks	Prob.
0	0.40
1	0.30
2	0.20
3	0.08
4	0.02

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.06.01

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.03

KEY: Bloom’s: Application

{Hobby Shop Sales Narrative} What is the probability of selling at least two trucks in any given week?

ANS:

P{2, 3, 4} = 0.30

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.06.01

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.03

KEY: Bloom’s: Application

{Hobby Shop Sales Narrative} What is the probability of selling between 1 and 3 (inclusive) trucks in any given week?

ANS:

P{1,2,3} = 0.58

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.06.01

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.03

KEY: Bloom’s: Application

Mutual Fund Price

An investor estimates that there is a 75% chance that a particular mutual fund’s price will increase to $100 per share over the next three weeks, based on past data.

{Mutual Fund Price Narrative} Which approach was used to produce this figure?

ANS:

The relative frequency approach

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.06.01

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.03

KEY: Bloom’s: Application

{Mutual Fund Price Narrative} Interpret the 75% probability.

ANS:

We interpret the 75% figure to mean that if we had an infinite number of funds with exactly the same economic and market characteristics as the one the investor will buy, 75% of them will increase in price to $100 over the next three weeks.

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.06.01

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.03

KEY: Bloom’s: Application

The sample space of the toss of a balanced die is S = {1, 2, 3, 4, 5, 6}. If the die is balanced, each simple event (outcome) has the same probability. Find the probability of the following events:

a.	Rolling an odd number
b.	Rolling a number less than or equal to 3
c.	Rolling a number greater than or equal to 5
d.	Rolling a number between 2 and 5, inclusive.

ANS:

a.	3/6
b.	3/6
c.	2/6
d.	4/6

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.06.01

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.03

KEY: Bloom’s: Application

Equity Loan Rates

A survey of banks estimated the following probabilities for the interest rate being charged on a equity loan based on a 30-year loan, based on past records.

Interest Rate	6.0%	6.5%	7.0%	7.5%	>7.5%
Probability	0.20	0.23	0.25	0.28	.04

{Equity Loan Rates Narrative} If a bank is selected at random from this distribution, what is the probability that the interest rate charged on a home loan exceeds 7.0%?

ANS:

0.32

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.06.01

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.03

KEY: Bloom’s: Application

{Equity Loan Rates Narrative} What is the most common interest rate?

ANS:

7.5%, since it occurred 28% of the time.

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.06.01

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.03

KEY: Bloom’s: Application

{Equity Loan Rates Narrative} What approach was used in estimating the probabilities for the interest rates?

ANS:

relative frequency approach

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.06.01

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.03

KEY: Bloom’s: Application

The probability of the intersection is called a joint probability.

ANS:

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.06.02

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.03

KEY: Bloom’s: Knowledge

Two or more events are said to be independent when the occurrence of one event has no effect on the probability that another will occur.

ANS:

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.06.02

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.03

KEY: Bloom’s: Comprehension

The union of events A and B is the event that occurs when either A or B or both occur. It is denoted as ‘A or B’.

ANS:

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.06.02

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.03

KEY: Bloom’s: Knowledge

If A and B are independent events with P(A) = 0.35 and P(B) = 0.55, then P(A|B) is 0.35/0.55 = .64.

ANS:

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.06.02

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.03

KEY: Bloom’s: Application

Two events A and B are said to be independent if P(A|B) = P(B).

ANS:

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.06.02

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.03

KEY: Bloom’s: Knowledge

The conditional probability of event B given event A is denoted by P(A|B).

ANS:

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.06.02

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.03

KEY: Bloom’s: Knowledge

If A and B are independent events with P(A) = .40 and P(B) = .50, then P(A and B) = .20.

ANS:

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.06.02

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.03

KEY: Bloom’s: Application

The intersection of two events A and B is the event that occurs when both A and B occur.

ANS:

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.06.02

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.03

KEY: Bloom’s: Knowledge

Two events A and B are independent if P(A and B) = 0.

ANS:

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.06.02

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.03

KEY: Bloom’s: Knowledge

The union of events A and B is the event that occurs when either A or B occurs but not both.

ANS:

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.06.02

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.03

KEY: Bloom’s: Knowledge

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

ANS:

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.06.02

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.03

KEY: Bloom’s: Knowledge

If P(A) = .30, P(B) = .60, and P(A and B) = .20, then P(A|B) = .40.

ANS:

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.06.02

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.03

KEY: Bloom’s: Application

Suppose the probability that a person owns both a cat and a dog is 0.10. Also suppose the probability that a person owns a cat but not a dog is 0.20. The marginal probability that someone owns a cat is 0.30.

ANS:

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.06.02

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.03

KEY: Bloom’s: Application

The probability of the intersection of two events A and B is denoted by P(A and B) and is called the:

a.	marginal probability
b.	joint probability
c.	conditional probability of A given B
d.	conditional probability of B given A

ANS:

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.06.02

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.03

KEY: Bloom’s: Comprehension

The intersection of events A and B is the event that occurs when:

a.	either A or B occurs but not both
b.	neither A nor B occur
c.	both A and B occur
d.	All of these choices are true.

ANS:

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.06.02

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.03

KEY: Bloom’s: Knowledge

The probability of event A given event B is denoted by

a.	P(A and B)
b.	P(A or B)
c.	P(A\|B)
d.	P(B\|A)

ANS:

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.06.02

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.03

KEY: Bloom’s: Knowledge

Which of the following is equivalent to P(A|B)?

a.	P(A and B)
b.	P(B\|A)
c.	P(A)/P(B)
d.	None of these choices.

ANS:

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.06.02

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.03

KEY: Bloom’s: Comprehension

Which of the following best describes the concept of marginal probability?

a.	It is a measure of the likelihood that a particular event will occur, regardless of whether another event occurs.
b.	It is a measure of the likelihood that a particular event will occur, if another event has already occurred.
c.	It is a measure of the likelihood of the simultaneous occurrence of two or more events.
d.	None of these choices.

ANS:

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.06.02

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.03

KEY: Bloom’s: Comprehension

If two events are independent, what is the probability that they both occur?

a.	0
b.	0.50
c.	1.00
d.	Cannot be determined from the information given

ANS:

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.06.02

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.03

KEY: Bloom’s: Application

If the outcome of event A is not affected by event B, then events A and B are said to be

a.	mutually exclusive
b.	independent
c.	collectively exhaustive
d.	None of these choices.

ANS:

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.06.02

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.03

KEY: Bloom’s: Comprehension

If A and B are disjoint events with P(A) = 0.70, then P(B):

a.	can be any value between 0 and 1
b.	can be any value between 0 and 0.70
c.	cannot be larger than 0.30
d.	cannot be determined with the information given

ANS:

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.06.02

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.03

KEY: Bloom’s: Application

If P(A) = 0.65, P(B) = 0.58, and P(A and B) = 0.76, then P(A or B) is:

a.	1.23
b.	0.47
c.	0.18
d.	0.11

ANS:

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.06.02

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.03

KEY: Bloom’s: Application

Suppose P(A) = 0.60, P(B) = 0.85, and A and B are independent. The probability of the complement of the event (A and B) is:

a.	.4 ´ .15 = .060
b.	0.40 + .15 = .55
c.	1 – (.40 + .15) = .45
d.	1 – (.6 ´ .85) = .490

ANS:

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.06.02

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.03

KEY: Bloom’s: Application

Which of the following statements is correct if the events A and B have nonzero probabilities?

a.	A and B cannot be both independent and disjoint
b.	A and B can be both independent and disjoint
c.	A and B are always independent
d.	A and B are always disjoint

ANS:

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.06.02

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.03

KEY: Bloom’s: Comprehension

A and B are disjoint events, with P(A) = 0.20 and P(B) = 0.30. Then P(A and B) is:

a.	0.50
b.	0.10
c.	0.00
d.	0.06

ANS:

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.06.02

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.03

KEY: Bloom’s: Application

If P(A) = 0.35, P(B) = 0.45, and P(A and B) = 0.25, then P(A|B) is:

a.	1.4
b.	1.8
c.	0.714
d.	0.556

ANS:

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.06.02

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.03

KEY: Bloom’s: Application

If A and B are independent events with P(A) = 0.60 and P(A|B) = 0.60, then P(B) is:

a.	1.20
b.	0.60
c.	0.36
d.	cannot be determined with the information given

ANS:

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.06.02

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.03

KEY: Bloom’s: Application

If A and B are independent events with P(A) = 0.20 and P(B) = 0.60, then P(A|B) is:

a.	0.20
b.	0.60
c.	0.40
d.	0.80

ANS:

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.06.02

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.03

KEY: Bloom’s: Application

If P(A) = 0.25 and P(B) = 0.65, then P(A and B) is:

a.	0.25
b.	0.40
c.	0.90
d.	cannot be determined from the information given

ANS:

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.06.02

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.03

KEY: Bloom’s: Application

Cars

Suppose X = the number of cars owned by a family in the U.S. The probability distribution of X is shown in the table below.

X	0	1	2	3
Probability	0.56	0.23	0.12	0.09

{Car Narrative}What is the chance that a family owns more than one car?

a.	0.23
b.	0.21
c.	0.44
d.	None of these choices.

ANS:

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.06.02

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.03

KEY: Bloom’s: Application

{Cars Narrative} Suppose you choose two families at random. What is the chance that they each own one car? (That means family A owns a car and family B owns a car.)

a.	0.23
b.	0.23 + 0.23 = 0.46
c.	0.23 + 0.23 – (0.23)*(0.23) = .4071
d.	(0.23)*(0.23) = 0.0529

ANS:

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.06.02

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.03

KEY: Bloom’s: Application

The ____________________ of events A and B is the event that occurs when both A and B occur.

ANS:

intersection

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.06.02

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.03

KEY: Bloom’s: Knowledge

The probability of an intersection of two events is called a(n) ____________________ probability.

ANS:

joint

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.06.02

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.03

KEY: Bloom’s: Comprehension

Suppose two events A and B are related. The ____________________ probability of A is the probability that A occurs, regardless of whether event B occurred or not.

ANS:

marginal

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.06.02

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.03

KEY: Bloom’s: Comprehension

If two events are mutually exclusive, their joint probability is ____________________.

ANS:

zero

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.06.02

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.03

KEY: Bloom’s: Comprehension

A conditional probability of A given B is written in probability notation as ____________________.

ANS:

P(A|B)

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.06.02

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.03

KEY: Bloom’s: Knowledge

If A and B are independent, then P(A|B) = ____________________.

ANS:

P(A)

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.06.02

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.03

KEY: Bloom’s: Knowledge

The ____________________ of two events A and B is the event that occurs when either A or B or both occur.

ANS:

union

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.06.02

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.03

KEY: Bloom’s: Knowledge

If A and B are mutually exclusive, their joint probability is ____________________.

ANS:

zero

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.06.02

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.03

KEY: Bloom’s: Comprehension

P(A|B) is the conditional probability of ____________________ given ____________________.

ANS:

A; B

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.06.02

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.03

KEY: Bloom’s: Comprehension

If P(A|B) = P(A) then events A and B are ____________________.

ANS:

independent

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.06.02

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.03

KEY: Bloom’s: Comprehension

Tea and Seltzer

Suppose 55 percent of adults drink tea, 45 percent drink seltzer, and 10 percent drink both.

{Tea and Seltzer Narrative} What is the probability that a randomly chosen adult does not drink seltzer?

ANS:

.55

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.06.02

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.03

KEY: Bloom’s: Application

{Tea and Seltzer Narrative} What is the probability that a randomly chosen adult drinks seltzer or tea or both?

ANS:

.90

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.06.02

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.03

KEY: Bloom’s: Application

{Tea and Seltzer Narrative} What is the probability that a randomly chosen adult doesn’t drink tea or seltzer?

ANS:

.10

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.06.02

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.03

KEY: Bloom’s: Application

Club Members

A survey of a club’s members indicates that 50% own a home, 80% own a car, and 90% of the homeowners who subscribe also own a car.

{Club Members Narrative} What is the probability that a subscriber owns both a car and a house?

ANS:

.45

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.06.02

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.03

KEY: Bloom’s: Application

{Club Members Narrative} What is the probability that a club member owns a car or a house, or both?

ANS:

.85

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.06.02

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.03

KEY: Bloom’s: Application

{Club Members Narrative} What is the probability that a club member owns neither a car nor a house?

ANS:

.15

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.06.02

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.03

KEY: Bloom’s: Application

Business Majors

Suppose 30% of business majors major in accounting. You take a random sample of 3 business majors.

{Business Majors Narrative} What is the chance that they all major in accounting?

ANS:

.027

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.06.02

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.03

KEY: Bloom’s: Application

{Business Majors Narrative} What is the chance that at least one majors in accounting?

ANS:

.657

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.06.02

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.03

KEY: Bloom’s: Application

{Business Majors Narrative} What is the chance that exactly one majors in accounting?

ANS:

.441

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.06.02

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.03

KEY: Bloom’s: Application

{Business Majors Narrative} What is the chance that none of them major in accounting?

ANS:

.343

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.06.02

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.03

KEY: Bloom’s: Application

Drunk Drivers

Six hundred accidents that occurred on a Saturday night were analyzed. Two items noted were the number of vehicles involved and whether alcohol played a role in the accident. The numbers are shown below:

	Number of Vehicles Involved
Did alcohol play a role?	1	2	3	Totals
Yes	75	125	50	250
No	50	225	75	350
Totals	125	350	125	600

{Drunk Drivers Narrative} What proportion of accidents involved more than one vehicle?

ANS:

475/600 or .79

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.06.02

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.03

KEY: Bloom’s: Application

{Drunk Drivers Narrative} What proportion of accidents involved alcohol and single vehicle?

ANS:

75/600 or .125

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.06.02

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.03

KEY: Bloom’s: Application

{Drunk Drivers Narrative} What proportion of accidents involved alcohol or a single vehicle?

ANS:

300/600 or .50

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.06.02

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.03

KEY: Bloom’s: Application

{Drunk Drivers Narrative} Given alcohol was involved, what proportion of accidents involved a single vehicle?

ANS:

75/250 or .30

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.06.02

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.03

KEY: Bloom’s: Application

{Drunk Drivers Narrative} If multiple vehicles were involved, what proportion of accidents involved alcohol?

ANS:

175/475 or .37

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.06.02

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.03

KEY: Bloom’s: Application

{Drunk Drivers Narrative} If 3 vehicles were involved, what proportion of accidents involved alcohol?

ANS:

50/125 or .40

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.06.02

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.03

KEY: Bloom’s: Application

{Drunk Drivers Narrative} If alcohol was not involved, what proportion of the accidents were single vehicle?

ANS:

50/350 or .143

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.06.02

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.03

KEY: Bloom’s: Application

{Drunk Drivers Narrative} If alcohol was not involved, what proportion of the accidents were multiple vehicle?

ANS:

300/350 or .857

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.06.02

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.03

KEY: Bloom’s: Application

Suppose A and B are two independent events for which P(A) = 0.20 and P(B) = 0.60.

a.	Find P(A\|B).
b.	Find P(B\|A).

ANS:

a.	0.20
b.	0.60

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.06.02

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.03

KEY: Bloom’s: Application

GPA and Class

A college professor classifies his students according to their grade point average (GPA) and their class rank. GPA is on a 0.0-4.0 scale, and class rank is defined as the under class (freshmen and sophomores) and the upper class (juniors and seniors). One student is selected at random.

	GPA
Class	Under 2.0	2.0 – 3.0	Over 3.0
Under	0.05	0.25	0.10
Upper	0.10	0.30	0.20

{GPA and Class Narrative} If the student selected is in the upper class, what is the probability that her GPA is between 2.0 and 3.0?

ANS:

0.50

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.06.02

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.03

KEY: Bloom’s: Application

{GPA and Class Narrative} If the GPA of the student selected is over 3.0, what is the probability that the student is in the lower class?

ANS:

0.333

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.06.02

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.03

KEY: Bloom’s: Application

{GPA and Class Narrative} What is the probability that the student is in the upper class?

ANS:

0.60

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.06.02

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.03

KEY: Bloom’s: Application

{GPA and Class Narrative} What is the probability that the student has GPA over 3.0?

ANS:

0.30

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.06.02

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.03

KEY: Bloom’s: Application

{GPA and Class Narrative} What is the probability that the student is in the lower class?

ANS:

0.40

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.06.02

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.03

KEY: Bloom’s: Application

{GPA and Class Narrative} What is the probability that the student is in the lower class and has GPA over 3.0?

ANS:

0.10

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.06.02

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.03

KEY: Bloom’s: Application

{GPA and Class Narrative} What is the probability that the student is in the upper class and has GPA under 2.0?

ANS:

0.10

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.06.02

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.03

KEY: Bloom’s: Application

{GPA and Class Narrative} Are being in the upper class and having a GPA over 3.0 related? Explain.

ANS:

Yes, since the product of the probabilities of the two events is not equal to the joint probability.

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.06.02

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.03

KEY: Bloom’s: Application

Marital Status

An insurance company has collected the following data on the gender and marital status of 570 customers.

	Marital Status
Gender	Single	Married	Divorced
Male	50	250	30
Female	100	100	40

Suppose that a customer is selected at random.

{Marital Status Narrative} Develop the joint probability table.

ANS:

Gender	Single	Married	Divorced
Male	.088	.439	.053
Female	.175	.175	.070

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.06.02

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.03

KEY: Bloom’s: Application

{Marital Status Narrative} Find the probability that the customer selected is a married female.

ANS:

0.175

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.06.02

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.03

KEY: Bloom’s: Application

{Marital Status Narrative} Find the probability that the customer selected is

a.	female and single
b.	married if the customer is male.
c.	not single

ANS:

a.	0.175
b.	0.757
c.	0.737

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.06.02

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.03

KEY: Bloom’s: Application

Financial Consultants

A Financial Consultant has classified his clients according to their gender and the composition of their investment portfolio (primarily bonds, primarily stocks, or a balanced mix of bonds and stocks). The proportions of clients falling into the various categories are shown in the following table:

	Portfolio Composition
Gender	Bonds	Stocks	Balanced
Male	0.18	0.20	0.25
Female	0.12	0.10	0.15

One client is selected at random, and two events A and B are defined as follows:

A: The client selected is male.

B: The client selected has a balanced portfolio.

{Financial Consultants Narrative} Find the following probabilities:

a.	P(A)
b.	P(B)

ANS:

a.	0.63
b.	0.40

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.06.02

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.03

KEY: Bloom’s: Application

{Financial Consultants Narrative} Express each of the following events in words:

a.	A or B
b.	A and B

ANS:

a.	The client selected either is male or has a balanced portfolio or both.
b.	The client selected is male and has a balanced portfolio.

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.06.02

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.03

KEY: Bloom’s: Application

{Financial Consultants Narrative} Find P(A and B).

ANS:

0.25

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.06.02

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.03

KEY: Bloom’s: Application

{Financial Consultants Narrative} Express each of the following probabilities in words:

a.	P(A\|B)
b.	P(B\|A)

ANS:

a.	The probability that the client selected is male, if the client has a balanced portfolio.
b.	The probability that the client selected has a balanced portfolio, if the client is male.

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.06.02

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.03

KEY: Bloom’s: Application

{Financial Consultants Narrative} Find the following probabilities:

a.	P(A\|B)
b.	P(B\|A)

ANS:

a.	0.625
b.	0.3968

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.06.02

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.03

KEY: Bloom’s: Application

Julius and Gabe go to a show during their Spring break and toss a balanced coin to see who will pay for the tickets. The probability that Gabe will pay three days in a row is 0.125.

ANS:

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.06.03

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.03

KEY: Bloom’s: Application

If events A and B have nonzero probabilities, then they can be both independent and mutually exclusive.

ANS:

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.06.02

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.03

KEY: Bloom’s: Comprehension

If the event of interest is A, the probability that A will not occur is the complement of A.

ANS:

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.06.03

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.03

KEY: Bloom’s: Comprehension

Assume that A and B are independent events with P(A) = 0.30 and P(B) = 0.50. The probability that both events will occur simultaneously is 0.80.

ANS:

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.06.02

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.03

KEY: Bloom’s: Application

Two events A and B are said to be independent if P(A) = P(A|B).

ANS:

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.06.03

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.03

KEY: Bloom’s: Knowledge

When A and B are mutually exclusive, P(A or B) can be found by adding P(A) and P(B).

ANS:

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.06.03

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.03

KEY: Bloom’s: Knowledge

Two events A and B are said to be independent if P(A|B) = P(B).

ANS:

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.06.03

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.03

KEY: Bloom’s: Knowledge

If A and B are two independent events with P(A) = 0.9 and P(B|A) = 0.5, then P(A and B) = 0.45.

ANS:

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.06.03

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.03

KEY: Bloom’s: Application

Two events A and B are said to be independent if P(A|B) = P(B|A).

ANS:

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.06.03

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.03

KEY: Bloom’s: Knowledge

The probability of the union of two mutually exclusive events A and B is 0.

ANS:

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.06.03

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.03

KEY: Bloom’s: Knowledge

Two events A and B are said to be mutually exclusive if P(A and B) = 1.0.

ANS:

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.06.03

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.03

KEY: Bloom’s: Knowledge

If P(A and B) = 1, then A and B must be mutually exclusive.

ANS:

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.06.03

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.03

KEY: Bloom’s: Knowledge

Events A and B are either independent or mutually exclusive.

ANS:

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.06.03

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.03

KEY: Bloom’s: Knowledge

If P(B) = .7 and P(B|A) = .4, then P(A and B) must be .28.

ANS:

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.06.03

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.03

KEY: Bloom’s: Application

If P(B) = .7 and P(A|B) = .7, then P(A and B) = 0.

ANS:

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.06.03

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.03

KEY: Bloom’s: Application

If the events A and B are independent with P(A) = 0.35 and P(B) = 0.45, then the probability that both events will occur simultaneously is:

a.	0
b.	0.16
c.	0.80
d.	Not enough information to tell.

ANS:

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.06.03

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.03

KEY: Bloom’s: Application

Two events A and B are said to be mutually exclusive if:

a.	P(A\|B) = 1
b.	P(A\|B) = P(A)
c.	P(A and B) =1
d.	P(A and B) = 0

ANS:

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.06.03

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.03

KEY: Bloom’s: Knowledge

If P(A) = 0.84, P(B) = 0.76, and P(A or B) = 0.90, then P(A and B) is:

a.	0.06
b.	0.14
c.	0.70
d.	0.83

ANS:

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.06.03

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.03

KEY: Bloom’s: Application

Which of the following statements is always correct?

a.	P(A and B) = P(A) * P(B)
b.	P(A or B) = P(A) + P(B)
c.	P(A) = 1 – P(A^c)
d.	None of these choices.

ANS:

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.06.03

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.03

KEY: Bloom’s: Knowledge

If P(A) = 0.20, P(B) = 0.30, and P(A and B) = 0, then A and B are:

a.	dependent events
b.	independent events
c.	mutually exclusive events
d.	complementary events

ANS:

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.06.03

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.03

KEY: Bloom’s: Application

If P(A) = 0.65, P(B) = 0.58, and P(A and B) = 0.76, then P(A or B) is:

a.	1.23
b.	0.47
c.	0.24
d.	None of these choices.

ANS:

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.06.03

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.03

KEY: Bloom’s: Application

Suppose P(A) = 0.30. The probability of the complement of A is:

a.	0.30
b.	0.70
c.	-0.30
d.	None of these choices.

ANS:

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.06.03

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.03

KEY: Bloom’s: Application

If events A and B are independent then:

a.	P(A and B) = P(A) * P(B)
b.	P(A and B) = P(A) + P(B)
c.	P(B\|A) = P(A)
d.	None of these choices.

ANS:

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.06.03

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.03

KEY: Bloom’s: Knowledge

If A and B are mutually exclusive events, with P(A) = 0.20 and P(B) = 0.30, then the probability that both events will occur simultaneously is:

a.	0.50
b.	0.06
c.	0
d.	None of these choices.

ANS:

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.06.03

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.03

KEY: Bloom’s: Comprehension

If A and B are independent events with P(A) = 0.60 and P(B) = 0.70, then P(A or B) equals:

a.	1.30
b.	0.88
c.	0.42
d.	Cannot tell from the given information.

ANS:

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.06.03

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.03

KEY: Bloom’s: Application

If A and B are mutually exclusive events with P(A) = 0.30 and P(B) = 0.40, then P(A or B) is:

a.	0.10
b.	0.12
c.	0.70
d.	None of these choices

ANS:

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.06.03

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.03

KEY: Bloom’s: Application

If A and B are any two events with P(A) = .8 and P(B|A) = .4, then P(A and B) is:

a.	.40
b.	.32
c.	1.20
d.	None of these choices.

ANS:

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.06.03

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.03

KEY: Bloom’s: Application

If A and B are any two events with P(A) = .8 and P(B|A^c) = .7, then P(A^c and B) is

a.	0.56
b.	0.14
c.	1.50
d.	None of these choices.

ANS:

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.06.03

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.03

KEY: Bloom’s: Application

The ____________________ rule says that P(A^c) = 1 – P(A).

ANS:

complement

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.06.03

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.03

KEY: Bloom’s: Comprehension

The ____________________ rule is used to calculate the joint probability of two events.

ANS:

multiplication

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.06.03

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.03

KEY: Bloom’s: Knowledge

If A and B are ____________________ events, the joint probability of A and B is the product of the probabilities of those two events.

ANS:

independent

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.06.03

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.03

KEY: Bloom’s: Comprehension

The ____________________ rule is used to calculate the probability of the union of two events.

ANS:

addition

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.06.03

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.03

KEY: Bloom’s: Knowledge

If A and B are ____________________ then the probability of the union of A and B is the sum of their individual probabilities.

ANS:

mutually exclusive

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.06.03

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.03

KEY: Bloom’s: Comprehension

The first set of branches of a probability tree represent ____________________ probabilities.

ANS:

marginal

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.06.03

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.03

KEY: Bloom’s: Comprehension

The second set of branches of a probability tree represent ____________________ probabilities.

ANS:

conditional

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.06.03

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.03

KEY: Bloom’s: Knowledge

When you multiply a first level branch with a second level branch on a probability tree you get a(n) ____________________ probability.

ANS:

joint

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.06.03

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.03

KEY: Bloom’s: Knowledge

If two events are complements, their probabilities sum to ____________________.

ANS:

one

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.06.03

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.03

KEY: Bloom’s: Knowledge

If two events are mutually exclusive their joint probability is ____________________.

ANS:

zero

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.06.03

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.03

KEY: Bloom’s: Knowledge

Suppose A and B are two independent events for which P(A) = 0.20 and P(B) = 0.60.

a.	Find P(A and B).
b.	Find P(A or B).

ANS:

a.	0.12
b.	0.68

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.06.03

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.03

KEY: Bloom’s: Application

College Professorship

A Ph.D. graduate has applied for a job with two colleges: A and B. The graduate feels that she has a 60% chance of receiving an offer from college A and a 50% chance of receiving an offer from college B. If she receives an offer from college B, she believes that she has an 80% chance of receiving an offer from college A. Let A = receiving an offer from college A, and let B = receiving an offer from college B.

{College Professorship Narrative} What is the probability that both colleges will make her an offer?

ANS:

(.5)(.8) = 0.40

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.06.03

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.03

KEY: Bloom’s: Application

{College Professorship Narrative} What is the probability that at least one college will make her an offer?

ANS:

.6 + .5 – .4 = 0.7

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.06.03

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.03

KEY: Bloom’s: Application

{College Professorship Narrative} If she receives an offer from college B, what is the probability that she will not receive an offer from college A?

ANS:

1 – 0.8 = 0.2.

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.06.03

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.03

KEY: Bloom’s: Application

Suppose P(A) = 0.50, P(B) = 0.40, and P(B|A) = 0.30.

a.	Find P(A and B).
b.	Find P(A or B).
c.	Find P(A\|B).

ANS:

a.	0.15
b.	0.75
c.	0.375

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.06.03

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.03

KEY: Bloom’s: Application

A survey of a magazine’s subscribers indicates that 50% own a house, 80% own a car, and 90% of the homeowners also own a car. What proportion of subscribers:

a.	own both a car and a house?
b.	own a car or a house, or both?
c.	own neither a car nor a house?

ANS:

a.	0.45
b.	0.85
c.	0.15

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.06.03

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.03

KEY: Bloom’s: Application

Suppose A and B are two mutually exclusive events for which P(A) = 0.30 and P(B) = 0.40.

a.	Find P(A and B).
b.	Find P(A or B).
c.	Are A and B independent events? Explain using probabilities.

ANS:

a.	0
b.	0.70
c.	No. P(A and B) = 0 because they are mutually exclusive events. If they were independent events, you would have P(A and B) = P(A) * P(B) = 0.12.

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.06.03

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.03

KEY: Bloom’s: Application

Suppose P(A) = 0.30, P(B) = 0.50, and P(B|A) = 0.60.

a.	Find P(A and B).
b.	Find P(A or B).
c.	Find P(A\|B).

ANS:

a.	0.18
b.	0.62
c.	0.36

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.06.03

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.03

KEY: Bloom’s: Application

Is it possible to have two events for which P(A) = 0.40, P(B) = 0.50, and P(A or B) = 0.30? Explain.

ANS:

Yes. In this situation, if P(A and B) = 0.60 it works.

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.06.03

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.03

KEY: Bloom’s: Application

A pharmaceutical firm has discovered a new diagnostic test for a certain disease that has infected 1% of the population. The firm has announced that 95% of those infected will show a positive test result, while 98% of those not infected will show a negative test result.

a.	What proportion of people don’t have the disease?
b.	What proportion who have the disease test negative?
c.	What proportion of those who don’t have the disease test positive?
d.	What proportion of test results are incorrect?
e.	What proportion of test results are correct?

ANS:

a.	0.99
b.	0.05
c.	0.02
d.	0.0203
e.	0.9797

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.06.03

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.03

KEY: Bloom’s: Application

{Marital Status Narrative} Find the probability that the customer selected is female or divorced.

ANS:

0.474

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.06.02

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.03

KEY: Bloom’s: Application

{Marital Status Narrative} Are gender and marital status mutually exclusive? Explain using probabilities.

ANS:

No, since P(female and married) = 0.175 > 0. (Any other combination shows this also.)

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.06.02

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.03

KEY: Bloom’s: Application

{Marital Status Narrative} Is marital status independent of gender? Explain using probabilities.

ANS:

No, since P(married / male) = 0.757 ¹ P(married) = 0.614. (Any other combination shows this also.)

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.06.03

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.03

KEY: Bloom’s: Application

Construction Bids

A construction company has submitted bids on two separate state contracts, A and B. The company feels that it has a 60% chance of winning contract A, and a 50% chance of winning contract B. Furthermore, the company believes that it has an 80% chance of winning contract A if it wins contract B.

{Construction Bids Narrative} What is the probability that the company will win both contracts?

ANS:

P(B and A) = P(B) * P(A|B) = (.50)(.80) = .40

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.06.03

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.03

KEY: Bloom’s: Application

{Construction Bids Narrative} What is the probability that the company will win at least one of the two contracts?

ANS:

P(A or B) = P(A) + P(B) – P(A and B) = .60 + .50 – .40 = .70

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.06.03

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.03

KEY: Bloom’s: Application

{Construction Bids Narrative} If the company wins contract B, what is the probability that it will not win contract A?

ANS:

P(A^c|B) = 1 – P(A|B) = 1 – .80 = .20

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.06.03

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.03

KEY: Bloom’s: Application

{Construction Bids Narrative} What is the probability that the company will win at most one of the two contracts?

ANS:

1 – P(A and B) = 0.60

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.06.03

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.03

KEY: Bloom’s: Application

{Construction Bids Narrative} What is the probability that the company will win neither contract?

ANS:

1 – P(A or B) = 0.30

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.06.03

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.03

KEY: Bloom’s: Application

Condo Sales and Interest Rates

The probability that condo sales will increase in the next 6 months is estimated to be 0.30. The probability that the interest rates on condo loans will go up in the same period is estimated to be 0.75. The probability that condo sales or interest rates will go up during the next 6 months is estimated to be 0.90.

{Condo Sales and Interest Rates Narrative} What is the probability that both condo sales and interest rates will increase during the next six months?

ANS:

0.15

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.06.03

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.03

KEY: Bloom’s: Application

{Condo Sales and Interest Rates Narrative} What is the probability that neither condo sales nor interest rates will increase during the next six months?

ANS:

0.10

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.06.03

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.03

KEY: Bloom’s: Application

{Condo Sales and Interest Rates Narrative} What is the probability that condo sales will increase but interest rates will not during the next six months?

ANS:

0.15

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.06.03

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.03

KEY: Bloom’s: Application

Bayes’ Law is a formula for revising an initial subjective (prior) probability value on the basis of new results, thus obtaining a new (posterior) probability value.

ANS:

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.06.04

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.03

KEY: Bloom’s: Comprehension

Although there is a formula defining Bayes’ law, you can also use a probability tree to conduct calculations.

ANS:

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.06.04

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.03

KEY: Bloom’s: Comprehension

Bayes’ Law allows us to compute conditional probabilities from other forms of probability.

ANS:

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.06.04

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.03

KEY: Bloom’s: Knowledge

Bayes’ Law says that P(A|B) = P(B|A)P(A).

ANS:

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.06.04

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.03

KEY: Bloom’s: Comprehension

Conditional probabilities are also called likelihood probabilities.

ANS:

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.06.04

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.03

KEY: Bloom’s: Knowledge

In applying Bayes’ Law, as the prior probabilities increase, the posterior probabilities decrease.

ANS:

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.06.04

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.03

KEY: Bloom’s: Comprehension

Prior probability of an event is the probability of the event before any information affecting it is given.

ANS:

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.06.04

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.03

KEY: Bloom’s: Comprehension

Bayes’ Law can be used to calculate posterior probabilities, prior probabilities, as well as new conditional probabilities.

ANS:

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.06.04

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.03

KEY: Bloom’s: Comprehension

Posterior probability of an event is the revised probability of the event after new information is available.

ANS:

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.06.04

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.03

KEY: Bloom’s: Comprehension

Prior probability is also called likelihood probability.

ANS:

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.06.04

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.03

KEY: Bloom’s: Knowledge

In general, a posterior probability is calculated by adding the prior and likelihood probabilities.

ANS:

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.06.04

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.03

KEY: Bloom’s: Knowledge

We can use the joint and marginal probabilities to compute conditional probabilities, for which a formula is available.

ANS:

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.06.04

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.03

KEY: Bloom’s: Knowledge

In problems where the joint probabilities are given, we can compute marginal probabilities by adding across rows and down columns.

ANS:

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.06.04

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.03

KEY: Bloom’s: Knowledge

If joint, marginal, and conditional probabilities are available, only joint probabilities can be used to determine whether two events are dependent or independent.

ANS:

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.06.05

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.03

KEY: Bloom’s: Knowledge

Suppose we have two events A and B. We can apply the addition rule to compute the probability that at least one of these events occurs.

ANS:

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.06.05

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.03

KEY: Bloom’s: Comprehension

Posterior probabilities can be calculated using the addition rule for mutually exclusive events.

ANS:

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.06.04

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.03

KEY: Bloom’s: Comprehension

Prior probabilities can be calculated using the multiplication rule for mutually exclusive events.

ANS:

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.06.04

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.03

KEY: Bloom’s: Comprehension

We can apply the multiplication rule to compute the probability that two events occur at the same time.

ANS:

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.06.05

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.03

KEY: Bloom’s: Comprehension

Which of the following statements is false?

a.	Thomas Bayes first employed the calculation of conditional probability in the eighteenth century.
b.	There is no formula defining Bayes’ Law.
c.	We use a probability tree to conduct all necessary calculations for Bayes’ Law.
d.	None of these choices.

ANS:

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.06.04

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.03

KEY: Bloom’s: Knowledge

A posterior probability value is a prior probability value that has been:

a.	modified on the basis of new information.
b.	multiplied by a conditional probability value.
c.	divided by a conditional probability value.
d.	added to a conditional probability value.

ANS:

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.06.04

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.03

KEY: Bloom’s: Knowledge

Initial estimates of the probabilities of events are known as:

a.	joint probabilities
b.	posterior probabilities
c.	prior probabilities
d.	conditional probabilities

ANS:

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.06.04

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.03

KEY: Bloom’s: Comprehension

Which of the following statements is false regarding a scenario using Bayes’ Law?

a.	Prior probabilities are called likelihood probabilities.
b.	Conditional probabilities are called posterior probabilities.
c.	Posterior probabilities are calculated by using prior probabilities that have been modified based on new information.
d.	None of these choices.

ANS:

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.06.04

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.03

KEY: Bloom’s: Knowledge

Bayes’ Law is used to compute:

a.	prior probabilities.
b.	joint probabilities.
c.	union probabilities.
d.	posterior probabilities.

ANS:

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.06.04

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.03

KEY: Bloom’s: Comprehension

Thomas ____________________ first employed the calculation of conditional probability.

ANS:

Bayes

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.06.04

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.03

KEY: Bloom’s: Knowledge

Bayes’ Law involves three different types of probabilities: 1) prior probabilities; 2) likelihood probabilities; and 3) ____________________ probabilities.

ANS:

posterior

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.06.04

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.03

KEY: Bloom’s: Comprehension

Bayes’ Law involves three different types of probabilities: 1) ____________________ probabilities; 2) likelihood probabilities; and 3) posterior probabilities.

ANS:

prior

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.06.04

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.03

KEY: Bloom’s: Comprehension

Bayes’ Law involves three different types of probabilities: 1) prior probabilities; 2) ____________________ probabilities; and 3) posterior probabilities.

ANS:

likelihood

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.06.04

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.03

KEY: Bloom’s: Comprehension

There are situations where we witness a particular event and we need to compute the probability of one of its possible causes. ____________________ is the technique we use to do this.

ANS:

Bayes’ Law

Bayes Law

Baye’s Law

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.06.04

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.03

KEY: Bloom’s: Knowledge

In the scenario of Bayes’ Law, P(A|B) is a(n) ____________________ probability, while P(B|A) is a posterior probability.

ANS:

likelihood

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.06.04

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.03

KEY: Bloom’s: Comprehension

In the scenario of Bayes’ Law, P(A|B) is a posterior probability, while P(B|A) is a(n) ____________________ probability.

ANS:

likelihood

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.06.04

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.03

KEY: Bloom’s: Comprehension

____________________ can find the probability that someone with a disease tests positive by using (among other things) the probability that someone who actually has the disease tests positive for it.

ANS:

Bayes’ Law

Bayes Law

Baye’s Law

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.06.04

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.03

KEY: Bloom’s: Comprehension

Certification Test

A standard certification test was given at three locations. 1,000 candidates took the test at location A, 600 candidates at location B, and 400 candidates at location C. The percentages of candidates from locations A, B, and C who passed the test were 70%, 68%, and 77%, respectively. One candidate is selected at random from among those who took the test.

{Certification Test Narrative} What is the probability that the selected candidate passed the test?

ANS:

(.5)(.7) + (.3)(.68) + (.2)(.77) = 0.708

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.06.04

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.03

KEY: Bloom’s: Application

{Certification Test Narrative} If the selected candidate passed the test, what is the probability that the candidate took the test at location B?

ANS:

(.3)(.68) / .708 = 0.288

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.06.04

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.03

KEY: Bloom’s: Application

{Certification Test Narrative} What is the probability that the selected candidate took the test at location C and failed?

ANS:

(.2)(.23) = 0.046

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.06.04

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.03

KEY: Bloom’s: Application

Cysts

After researching cysts of a particular type, a doctor learns that out of 10,000 such cysts examined, 1,500 are malignant and 8,500 are benign. A diagnostic test is available which is accurate 80% of the time (whether the cyst is malignant or not). The doctor has discovered the same type of cyst in a patient.

{Cysts Narrative} In the absence of any test, what is the probability that the cyst is malignant?

ANS:

M = Malignant, P(M) = .15

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.06.04

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.03

KEY: Bloom’s: Application

{Cysts Narrative} In the absence of any test, what is the probability that the cyst is benign?

ANS:

B = Benign, P(B) = .85

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.06.04

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.03

KEY: Bloom’s: Application

{Cysts Narrative} What is the probability that the patient will test positive?

ANS:

P(+)	= P(+ and M) + P(+ and B) = P(+/M) · P(M) + P(+/B) · P(B)
	= (.80)(.15) + (.20)(.85) = .29

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.06.04

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.03

KEY: Bloom’s: Application

{Cysts Narrative} What is the probability that the patient will test negative?

ANS:

P(-)	= 1 – P(+) = 1 – .29 = .71 or
P(-)	= P(- and M) + P(- and B) = P(-/M) · P(M) + P(-/B) · P(B)
	= (.20)(.15) + (.80)(.85) = .71

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.06.04

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.03

KEY: Bloom’s: Application

{Cysts Narrative} What is the probability that the patient has a benign tumor if he or she tests positive?

ANS:

P(B/+) = P(+ and B) / P(+) = P(+/B) · P(B) / P(+) = (.20)(.85) / (.29) = .586

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.06.04

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.03

KEY: Bloom’s: Application

{Cysts Narrative} What is the probability that the patient has a malignant cyst if he or she tests negative?

ANS:

P(M/-) = P(- and M) / P(-) = P(-/M) · P(M) / P(-) = (.20)(.15) / (.71) = .042

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.06.04

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.03

KEY: Bloom’s: Application

Messenger Service

Three messenger services deliver to a small town in Oregon. Service A has 60% of all the scheduled deliveries, service B has 30%, and service C has the remaining 10%. Their on-time rates are 80%, 60%, and 40% respectively. Define event O as a service delivers a package on time.

{Messenger Service Narrative} Calculate P(A and O).

ANS:

P(A and O) = P(A)P(O|A) = (.60)(.80) = 0.48

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.06.04; SFME.KELL.15.06.05

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.03

KEY: Bloom’s: Application

{Messenger Service Narrative} Calculate P(B and O).

ANS:

P(B and O) = P(B) P(O|B) = (.30)(.60) = 0.18

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.06.04; SFME.KELL.15.06.05

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.03

KEY: Bloom’s: Application

{Messenger Service Narrative} Calculate P(C and O).

ANS:

P(C and O) = P(C)P(O |C) = (.10)(.40) = 0.04

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.06.04; SFME.KELL.15.06.05

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.03

KEY: Bloom’s: Application

{Messenger Service Narrative} Calculate the probability that a package was delivered on time.

ANS:

P(O) = P(A and O) + P(B and O) + P(C and O) = .48 + .18 + .04 = 0.70

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.06.04; SFME.KELL.15.06.05

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.03

KEY: Bloom’s: Application

{Messenger Service Narrative} If a package was delivered on time, what is the probability that it was service A?

ANS:

P(A|O) = P(A and O) / P(O) = 0.48 / 0.70 = 0.686

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.06.04; SFME.KELL.15.06.05

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.03

KEY: Bloom’s: Application

{Messenger Service Narrative} If a package was delivered on time, what is the probability that it was service B?

ANS:

P(B|O) = P(B and O) / P(O) = 0.18 / 0.70 = 0.257

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.06.04; SFME.KELL.15.06.05

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.03

KEY: Bloom’s: Application

{Messenger Service Narrative} If a package was delivered on time, what is the probability that it was service C?

ANS:

P(C|O) = P(C and O) / P(O) = 0.04 / 0.70 = 0.057

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.06.04; SFME.KELL.15.06.05

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.03

KEY: Bloom’s: Application

{Messenger Service Narrative} If a package was delivered 40 minutes late, what is the probability that it was service A?

ANS:

P(A|O^c) = P(A and O^c) / P(O^c) = (0.60)(0.20) / 0.30 = 0.40

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.06.04; SFME.KELL.15.06.05

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.03

KEY: Bloom’s: Application

{Messenger Service Narrative} If a package was delivered 40 minutes late, what is the probability that it was service B?

ANS:

P(B|O^c) = P(B and O^c) / P(O^c) = (0.30)(0.40) / 0.30 = 0.40

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.06.04; SFME.KELL.15.06.05

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.03

KEY: Bloom’s: Application

{Messenger Service Narrative} If a package was delivered 40 minutes late, what is the probability that it was service C?

ANS:

P(C|O^c) = P(C and O^c) / P(O^c) = (0.10)(0.60) / 0.30 = 0.20

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.06.04; SFME.KELL.15.06.05

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.03

KEY: Bloom’s: Application

Statistics for Management and Economics Abbreviated 10e Gerald Keller - Test Bank

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Statistics for Management and Economics Abbreviated 10e Gerald Keller - Test Bank

Description

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