Chapter 5 Probability Created by Kathy Fritz. Can ultrasound accurately predict the gender of a baby? The paper “The Use of Three-Dimensional Ultrasound

Chapter 5

Probability

Created by Kathy Fritz

Can ultrasound accurately predict the gender of a baby?

The paper “The Use of Three-Dimensional Ultrasound for Fetal Gender Determination in the First Trimester” (The British Journal of Radiology [2003]: 448-451) describes a study of ultrasound gender prediction. An experienced radiologist looked at 159 first trimester ultrasound images and made a gender prediction for each one.

When each baby was born, the ultrasound gender prediction was compared to the baby’s actual gender.

This table summarizes the resulting data:

The paper also included gender predictions by a second radiologist, who looked at 154 first trimester ultrasound inmages.

Radiologist 2

Predicted Male Predicted Female

Baby is Male 81 8

Baby is Female 7 58

Notice that the gender prediction based on the ultrasound image is NOT always correct.

Radiologist 1

Predicted Male Predicted Female

Baby is Male 74 12

Baby is Female 14 59

How likely is it that a predicted

gender is correct?Is a predicted gender more likely to be

correct when the baby is male than

when the baby is female?If the predicted gender is female, should

you paint the nursery pink?

If you do, how likely is it that you will need

to repaint?

Does the skill of the radiologist make a

difference?

All of these questions can be answered using the methods

introduced in this chapter.

Interpreting Probabilities

ProbabilityRelative Frequency

Law of Large NumbersBasic Properties

Probability

We often find ourselves in situations where the outcome is uncertain:

When a ticketed passenger shows up at the airport, she faces two possible outcomes: (1) she is able to take the flight, or (2) she is denied a seat as a result of overbooking by the airline and must take a later flight.

Based on her past experience, the passenger believes that the chance of being denied a seat is small or unlikely.

To quantify the likelihood of an occurrence, a number between 0 and 1 can be assigned to

an outcome.

A probability is a number between 0 and 1 that reflects the likelihood of occurrence of

some outcome.

Subjective Approach to ProbabilityThe subjective interpretation of probability is when a probability is interpreted as a personal measure of the strength of the belief that an outcome will occur.

A probability of 1 represents a belief that the outcome will certainly occur.

A probability of 0 represents a belief that the outcome will certainly NOT occur.

Because different people may have different subjective beliefs, they may assign different probabilities to the same outcome.

All other probabilities fall between these two extremes.

Relative Frequency Approach

In the relative frequency interpretation of probability, the probability of an outcome, denoted by P(outcome), is interpreted as the proportion of the time that the outcome occurs in the long run.

Relative frequency can be computed by:A probability of 1 corresponds to an outcome that occurs 100% of the time.

A probability of 0 corresponds to an outcome that occurs 0% of the time.

A package delivery service promises 2-day delivery between 2 cities in California but is often able to deliver the packages in just 1 day. The company reports that the probability of next-day delivery is 0.3.

Suppose that you track the delivery of packages shipped with this company. With each new package shipped, you could compute the relative frequency of packages shipped so far that have arrived in 1 day:

One way to interpret this probability would be to say that in the long run, about 30 out of every

100 packages shipped arrive in 1 day.

Here is a graph displaying the relative frequencies for each of the first 15 packages shipped.



As the number of packages in the sequence increases, the relative frequency does not continue to

fluctuate wildly, but instead settles down and approaches a specific value, which is the probability of

interest.

Law of Large Numbers

As the number of observations increases, the proportion of the time that an outcome occurs gets close to the probability of that outcome.

The Law of Large Numbers is the basis for the relative frequency interpretation

of probabilities.

Some Basic Properties of Probability

1. The probability of any outcome is a number between 0 and 1.

2. If outcomes can’t occur at the same time, then the probability that any one of them will occur is the sum of their individual probabilities.

A large auto center sells cars made by many different manufacturers. Two of these are Honda and Toyota.

Suppose: P(Honda) = 0.25 and P(Toyota) = 0.14

Consider the make of the next car sold.

What is the probability that the next car sold is either a Honda or a Toyota?

P(Honda or Toyota) = 0.25 + 0.14 = 0.39

Why don’t these two probabilities have a sum of 1?Can the outcomes Honda and Toyota

and happen at the same time?

An interpretation for this value is that about 25 out of every 100 cars sold would be

Hondas.

Some Basic Properties of Probability

3. The probability that an outcome will not occur is equal to 1 minus the probability that the outcome will occur.

Recall the car dealership (P(Honda) = 0.25):

What is the probability that the next car sold is not a Honda?

P(not Honda) = 1 - 0.25 = 0.75

Because a probability represents a long-run relative frequency, in situations where exact probabilities are not known, it is

common to estimate probabilities based on observation.

Computing Probabilities

Chance ExperimentSample Space

EventClassical Approach to Probability

A chance experiment is any activity or situation in which there is uncertainty about which of two or more possible outcomes will result.

Suppose two six-sided dice are rolled and they both land on sixes.

Or a coin is flipped and it lands on heads.

Or record the color of the next 20 cars to pass an intersection.

Chance Experiment

These are all examples of chance experiments.

two six-sided dice are rolled

a coin is flipped

cars to passan intersection.

These are the outcomes of chance experiments.

The collection of all possible outcomes of a chance experiment is the sample space for the experiment.

Consider a chance experiment to investigate whether men or women are more likely to choose a hybrid engine over a traditional internal combustion engine when purchasing a Honda Civic at a particular dealership. The type of vehicle purchased (hybrid or traditional) will be determined and the customer’s gender will be recorded.

This is an example of a sample space.

Sample Space

A list of all possible outcomes are:

Sample space = {MH, FH, MT, FT}

An event is any collection of outcomes from the sample space of a chance experiment.

Chance Experiment

Recall the situation in which a person purchases a Honda Civic: Sample space = {MH, FH, MT, FT}

A simple event is an event consisting of exactly on outcome.

Each of these 4 outcomes are simple events.

An event can be represented by a name, such as hybrid, or by an uppercase letter, such as A, B, or

C.

Identify the following events:

traditional =

female = {FH, FT}

{MT, FT}

When the outcomes in the sample space of a chance experiment are equally likely, the probability of an event E, denoted by P(E), is the ratio of the number of outcomes favorable to E to the total number of outcomes in the sample space:

Classical Approach to Probability

The classical approach to probability works well for chance experiments that have a finite set of outcomes

that are equally likely.

Four students (Adam (A), Bettina (B), Carlos (C), and Debra(D)) submitted correct solutions to a math contest that had two prizes. The contest rules specify that if more than two correct responses are submitted, the winners will be selected at random from those submitting correct responses.

What is the sample space for selecting the two winners from the four correct responses?

Sample space = {AB, AC, AD, BC, BD, CD}

Because the winners are selected at random, the six possible outcomes are equally likely.



Let E be the event that both selected winners are the same sex.

What is the probability of E?



Let F be the event that at least one of the selected winners is female.

What is the probability of F?

Relative Frequency Approach to Probability

The probability of an event E, denoted by P(E), is defined to be the value approached by the relatively frequency of occurrence of E in a very long series of observations from a chance experiment. If the number of observations is large,

When a chance experiment is performed, some events may be likely

to occur, whereas others may not be as likely to occur. In cases like these, the classical approach is not appropriate.

Suppose that you perform a chance experiment that consists of flipping a cap from a 20-ounce bottle of soda and noting whether the cap lands with the top up or down.

You carry out this chance experiment by flipping the cap 1000 times and record if it lands top up or top down. The cap lands top up 694 times.

Do you think that the event U, the cap landing top up, and event D, the cap landing top

down, are equally likely? Why or Why not?

Probabilities of More Complex Events

UnionIntersection

ComplementMutually Exclusive Events

Independents Events

Consider the chance experiment that consists of selecting a student at random from those enrolled at a particular college.

There are 9000 students enrolled at the college

Here are some possible events:

F = event that the selected student is femaleO = event that the selected student is older than 30A = event that the selected student favors the

expansion of the athletic programS = event that the selected student is majoring is

one of the lab sciences

If E is an event, the complement of E, denoted EC, is the event that E does not occur.

Complement

Suppose that 4300 of the 9000 students favor the expansion of the athletic program.𝑃 (𝐴𝐶 )=1−0.48=0.52

The probability of EC can be computed from the probability of E as follows:

What is the probability of event A not occurring?

If E and F are events, the intersection of E and F is denoted by and is the new event that both E and F occur.

This is the symbol for “intersection”.

Intersection

Consider the events:O = event that the selected student is older than 30S = event that the selected student is majoring is one

of the lab science

This table summaries the occurrence of these events:

If E and F are events, the intersection of E and F is denoted by and is the new event that both E and F occur.

Intersection

S(Majoring in Lab Science)

SC

(Not Majoring in Lab

Science)

Total

O (Over 30) 400 1700 2100

OC (Not over 30) 1100 5800 6900

Total 1500 7500 9000

𝑃 (𝑂∩𝑆 )= 4009000

=0.04

What is the probability of a randomly selected student is older than 30 AND is majoring in a lab science?

The numbers in red corresponds to the intersections of the events.

Majoring in lab science AND

Over 30

Majoring in lab science AND

Not over 30

Not majoring in lab science AND

Over 30

Not majoring in lab science ANDNot over 30

Consider the events:O = event that the selected student is older than 30A = event that the selected student favors the

expansion of the athletic program

This table summaries the occurrence of these events:

If E and F are events, the union is denoted by . The event is the new event that E or F occur.

This is the symbol for “union”.

Union

If E and F are events, the union is denoted by . The event is the new event that E or F occur.

Union

A(Favors

Expansion)

AC

(Does Not Favor Expansion)

Total

O (Over 30) 1600 500 2100

OC (Not over 30) 2700 4200 6900

Total 4300 4700 9000

𝑃 (𝑂∪𝐴 )=1600+500+27009000

=0.53

What is the probability of a randomly selected student is older than 30 OR favors the expansion of the athletic program?

A(Favors

Expansion)

AC


Total

O (Over 30) 1600 500 2100

OC (Not over 30) 2700 4200 6900

Total 4300 4700 9000

The event A, favors sale of alcohol

A(Favors

AC


Total

O (Over 30) 1600 500 2100

OC (Not over 30) 2700 4200 6900

Total 4300 4700 9000

The event

O, over 30

You can use tables to compute the probability of an intersection of two events and the probability of a union of two events.

In many situations, you may ONLY know the probabilities of some events. In this case, it is often possible to create a “hypothetical 1000” table and then use the table to compute probabilities.

Hypothetical 1000

In the previous examples, this was possible because a student was to be selected at random and because the number of students falling into each of the cells of the appropriate table were

given.

The report “TV Drama/Comedy Viewers and Health Information” (www.cdc.gov/healthmarketing) describes a large survey that was conducted by the Centers for Disease Control (CDC). The CDC believed that the sample was representative of adult Americans.

Let’s investigate these events (taken from questions on the survey):L = event that a randomly selected adult American

reports learning something new about a health issue or disease from a TV show in the previous 6 months.

F = event that a randomly selected adult American is female.

Data from the survey were used to estimate the following probabilities:

CDC study continued

𝑃 (𝐿)=0.58𝑃 (𝐹 )=0.5𝑃 (𝐿∩𝐹 )=0.31

F (female) Not F Total

L (learned from TV)

Not L

Total 1000

Begin by labeling rows and columns of the table. Put the “hypothetical 1000” in the bottom right

cell.

P(L) tells you that 58% of the 1000 people should be in the L row: (0.58)(1000) = 580.

The L row and the Not L row have a sum of 1000.

580

420

P(F) tells you that the F row is (0.50)(1000) = 500 and

that the Not F row is 1000 - 500 = 500

500 500

P(L F) tells you that 31% of the 1000 people are both female and learned health information from a

TV show.

The cell for L and F is (0.31)(1000) = 310.

310

Fill in the remaining cells to complete the table.

190

270

230

What is the probability that a randomly selected adult American has learned something new about a health issue or disease from a TV show in the previous 6 months or is female?

0

Let’s look at the hypothetical table once more.

Suppose: P (A) = 0.6, P (B C) = 0.7, and P (A B) = 0.2

A AC Total

B

BC

Total 1000

It does not matter which event goes on the side or

on the top.

400

400600

200 100

300 700

300

What is the probability of A or B happening?

𝑃 ( 𝐴∪𝐵 )=200+100+4001000

=700

1000=0.7

400

200 100

Sometimes people call the emergency 9-1-1 number to report situations that are not considered emergencies (such as to report a lost dog). Let two events be:

M = event that the next call to 9-1-1 is for a medical emergency

N= event that the next call to 9-1-1 is not considered an emergency

Suppose that you know P(M) = 0.30 and P(N) = 0.20. Events M and N are mutually exclusive because the next call can’t be both a medical emergency and a call that is not considered an emergency.

Two events E and F are mutually exclusive if they can NOT occur at the same time.

Mutually Exclusive Events

P(M) = 0.30 and P(N) = 0.20

Mutually Exclusive Events

N (Non-emergency)

Not N Total

M (Medical Emergency) 0 300 300

Not M 200 500 700

Total 200 800 1000

A “hypothetical 1000” table is shown below. The uppermost cell must be 0 when the two events

are mutually exclusive.

If E and F are mutually exclusive events, then

and

Addition Rule for Mutually Exclusive Events

𝑃 (𝐸∩𝐹 )=0

𝑃 (𝐸∪𝐹 )=𝑃 (𝐸 )+𝑃 (𝐹 )

Independent EventsTwo events are independent if the probability that one event occurs is not affected by knowledge of whether the other event has occurred.

Suppose that you purchase a desktop computer system with a separate monitor and keyboard. Two possible events are:

Event 1: The monitor needs service while under warranty.

Event 2: The keyboard needs service while under warranty.

Because the two components operate independently of each other, learning that the

monitor has needed warranty service would not effect your assessment of the likelihood that the

keyboard will need repair.

Dependent EventsTwo events are dependent if knowing that one event has occurred changes the probability that the other event occurs.

Consider a university’s course registration process, which divides students into 12 priority groups. Overall, only 10% of all students receive all requested classes, but 75% of those in the first priority group receive all requested classes.You would say that the probability that a randomly selected student at this university receives all requested class is 0.10.However, if you know that the selected student is in the first priority group, you revise the probability that the student receives all requested classes to 0.75.These two events are said to be dependent events.

If two events, E and F, independent, the probability that both events occur is the product of the individual event probabilities.

Multiplication Rule for Two Independent Events

More generally, if there are k independent events, the probability that all the events occur is the product of all individual event probabilities.

The Diablo Canyon nuclear power plant in California has a warning system that includes a network of sirens. When the system is tested, individual sirens sometimes fail. The sirens operate independently of one another.

Imagine that you live near Diablo Canyon and that there are two sirens that can be heard from your home. You might be concerned about the probability that both Siren 1 and Siren 2 fail. (When the siren system is activated, about 5% of the individual sirens fail.)

Using the multiplication rule for independent events:

Conditional Probability

Sometimes the knowledge that one event has occurred changes our assessment of the likelihood that another event occurs.

Consider a population in which 0.1% of all the individuals have a certain disease. The presence of the disease cannot be discerned from appearances, but there is a diagnostic test available. Unfortunately, the test is not always correct.

Suppose that 80% of those with positive test results actually have the disease and the other 20% of those with positive test results actually do NOT have the disease (false positive).

Disease example continued . . .

Consider the chance experiment in which an individual is randomly selected from the population.

Let:

E = event that the individual has the disease

F = event that the individual's diagnostic test is positive

P(E|F) denotes the probability that event E (has disease) GIVEN that event F (tested positive) occurs.

The vertical line is read “given”.This is an example of conditional probability.

Conditional Probability

Conditional probability is a probability that takes into account a given condition has occurred.

P(A|B)

is read as

the probability of event A occurring GIVEN event B has occurred.

Radiologist 1

Predicted Male Predicted Female Total

Baby is Male 74 12 86

Baby is Female

14 5973

Total 88 71 159

Recall the example in the Chapter Preview section about gender predictions based on ultrasounds performed during the first trimester of pregnancy. The table below summarizes the data for Radiologist 1.

How likely is it that a predicted gender is correct?

𝑃 (𝑝𝑟𝑒𝑑𝑖𝑐𝑡𝑒𝑑𝑔𝑒𝑛𝑑𝑒𝑟 𝑖𝑠 𝑐𝑜𝑟𝑟𝑒𝑐𝑡 )=74+59159

=0.836

This question is about ALL 159 ultrasound predictions.

Gender prediction example continued.

Is a predicted gender more likely to be correct when the baby is male than when the baby is female?

Radiologist 1 is slightly more likely to be correct when the baby is male than when the baby is female.

Radiologist 1



Baby is Female

14 5973

Total 88 71 159

This question is based on two conditions:the 86 male babies or the 73 female babies.

The appropriate row total or column total is used as the denominator in the probability

calculation.

If the predicted gender is female, should you paint the nursery pink?

𝑃 ( 𝑓𝑒𝑚𝑎𝑙𝑒∨𝑝𝑟𝑒𝑑𝑖𝑐𝑡𝑒𝑑 𝑓𝑒𝑚𝑎𝑙𝑒 )


Radiologist 1



Baby is Female

14 5973

Total 88 71 159

This is a condition. In the probability statement, the condition follows the

vertical line “|”.

3

For Radiologist 1, when the predicted gender is female, about 83% of the time the baby is actually female.

So, if you painted the room pink, then the probability that you would need to repaint is about 0.17 (1 – 0.83).

Let’s take the gender prediction example a little further.

Suppose that two radiologists both work in the same clinic; Radiologist 1 works part-time while Radiologist 2 (from the Chapter Preview section) works full-time.

Let’s answer these questions:

1. What is the probability that a gender prediction based on a first-trimester ultrasound at this clinic is correct?

2. If the first-trimester ultrasound gender prediction is incorrect, what is the probability that the prediction was made by Radiologist 2?


From the data we know:

Let’s create a “hypothetical 1000” table to answer the two questions.

Prediction Correct

Prediction Incorrect Total

Radiologist 1 300

Radiologist 2 700

Total 1000Since the probability that the prediction is correct given that the prediction was made by

Radiologist 1 is 0.836, then the value for this cell is:

(300)(0.836) = 250.8 ≈ 251

(Cell values MUST be whole numbers since we are counting how many are in each event.)

251

Similarly, the value for this cell is:(700)(0.903) = 632.1 ≈ 632

632

You can now fill in the values for the remaining cells.

833 117

68

49


From the data we know:

Prediction Correct

Prediction Incorrect Total

Radiologist 1 300

Radiologist 2 700

Total 1000

251

632

833 117

68

49

What is the probability that a gender prediction based on a first-trimester ultrasound at this clinic is correct?

𝑃 (𝑝𝑟𝑒𝑑𝑖𝑐𝑡𝑖𝑜𝑛𝑐𝑜𝑟𝑟𝑒𝑐𝑡 )= 8331000

=0.833

If the first-trimester ultrasound gender prediction is incorrect, what is the probability that the prediction was made by Radiologist 2?

𝑃 (𝑅𝑎𝑑𝑖𝑜𝑙𝑜𝑔𝑖𝑠𝑡 2∨𝑝𝑟𝑒𝑑𝑖𝑐𝑡𝑖𝑜𝑛𝑖𝑛𝑐𝑜𝑟𝑟𝑒𝑐𝑡 )= 68117

=0.581

About 58.1% of the incorrect gender predictions at this clinic are made by Radiologist 2.

This seems high – but remember that Radiologist 2 does more than twice as many predictions as

Radiologist 1.

Calculating Probabilities –A More Formal Approach

Probability Formulas

The Complement RuleFor any event E,

The Addition RuleFor any two events E and F,

For mutually exclusive events, this simplifies to

Probability Formulas Continued

The Multiplication RuleFor any two events E and F,

For independent events, this simplifies to

Conditional ProbabilitiesFor any two events E and F with P(F) ≠ 0,

Revisit CDC’s study . . .Recall:L = event that a randomly selected adult American reports

learning something new about a health issue or disease from a TV show in the previous 6 months.

F = event that a randomly selected adult American is female.

Data from the survey were used to estimate the following probabilities:

What is the probability that a randomly selected adult American reports learning something new about a health issue or disease from a TV show in the previous 6 months or that a randomly selected adult American is female?

The article “Chances Are You Know Someone with a Tattoo, and He’s Not a Sailor” (Associated Press, June 11, 2006) summarized data from a representative sample of adults ages 18 to 50. T = the event that a randomly selected person has a tattooA = the event that a randomly selected person is

between 18 and 29 years old The following probabilities were estimated based on data from the sample:

Notice that the probability of “A given T” and “T given A”

are NOT the same!

Another Approach to ProbabilityA large electronics store sells two different portable DVD players, Brand 1 and Brand 2. Based on past records, the store manager reports that 70% of the DVD players sold are Brand 1 and 30% are Brand 2.

The manager also reports that 20% of the people who buy Brand 1 also purchase an extended warranty, and 40% of the people who buy Brand 2 purchase an extended warranty.Consider selecting a person at random from those who purchased a DVD player from this store, what is the probability that the person purchased extended warranty?

One way to do this problem would be to set up a Hypothetical 1000 table.

DVD Players Continued

P(Brand 1) = 0.7 P(Brand 2) = 0.3

The manager also reports that 20% of the people who buy Brand 1 also purchase an extended warranty, and 40% of the people who buy Brand 2 purchase an extended warranty.

Consider selecting a person at random from those who purchased a DVD player from this store, what is the probability that the person purchased extended warranty?

Brand 1 Brand 2 Total

Bought Extended Warranty

Not Bought Extended Warranty

Total 700 300 1000

140 120 260

560 180 740

𝑃 (ExtendedWarranty )= 2601000

=0.26

DVD Players Continued

P(Brand 1) = 0.7 P(Brand 2) = 0.3 The manager also reports that 20% of the people who buy Brand 1 also purchase an extended warranty, and 40% of the people who buy Brand 2 purchase an extended warranty.Consider selecting a person at random from those who purchased a DVD player from this store, what is the probability that the person purchased extended warranty?

Another approach to this problem is to use a tree diagram.

B1 = 0.7

B2 = 0.3

E = 0.2

E = 0.4

EC = 0.8

EC = 0.6

and

(0.7)(0.2) = 0.14and

or

(0.3)(0.4) = 0.12

P(E) = 0.14 + 0.12 = 0.26

This is an example of the Law of Total Probability!

The Law of Total Probability

If B1 and B2 are disjoint events with P(B1) + P(B2) = 1, then for any event E

More generally, if B1, B2, , Bk are disjoint events with P(B1) + P(B2) + + P(Bk) = 1, then for any event E

Let’s consider another type of problem . . .

Suppose the conditional probability of “a positive test result given that the person has cancer” is known. However, you would like to know the converse probability. That is, you would like to know the probability of the person having cancer given a positive test result.

A converse probability is the reversal of a conditional

probability.This converse probability can be computed using

Bayes’ Rule.

This formula was discovered in the 1700’s by the Reverend Thomas Bayes, an English

Presbyterian minister.

Bayes’ RuleIf B1 and B2 are disjoint events with P(B1) + P(B2) = 1, then for any event E

More generally, if B1, B2, , Bk are disjoint events with P(B1) + P(B2) + + P(Bk) = 1, then for any event E

Let’s look at an example.

Internet addiction has been defined by researchers as a disorder characterized by excessive time spent on the Internet, impaired judgment and decision-making ability, social withdrawal, and depression. In a study of adolescents, each participant was assessed using the Chen Internet Addiction Scale to determine if he or she suffered from Internet addiction.

The following probabilities are based on survey results:P(F) = 0.518 P(M) = 0.482

P(I|F) = 0.131 P(I|M)= 0.248

What is the probability that a randomly selected adolescent from the survey is female given that she has Internet addiction?

Although Bayes’ Rule is not listed in the AP® Statistics course description, you are expected to be able to solve “Bayes’-like” problems. Besides using the formula, you can also solve using tables or tree

diagrams.

Probability as a Basis for Making Decisions

Probability plays an important role in drawing conclusions from data.A professor planning to give a quiz that consists of 20 true-false questions is interested in knowing how someone who answers by guessing would do on such a quiz.To investigate, he asks the 500 students in his introductory psychology course to write the numbers from 1 to 20 on a piece of paper and then to arbitrarily write T or F next to each number.The students are forced to guess at the answer to each question, because they are not even told what the questions are! These answers are then collected and graded using the key for the quiz.

This table summarizes the number of correct answers on the quiz.

Quiz example continued.Number of

Correct Responses

Number of Students

Proportion of Students

Number of Correct

Responses

Number of Students


0 0 0.000 11 79 0.158

1 0 0.000 12 61 0.122

2 1 0.002 13 39 0.078

3 1 0.002 14 18 0.036

4 2 0.004 15 7 0.014

5 8 0.016 16 1 0.002

6 18 0.036 17 1 0.002

7 37 0.074 18 0 0.000

8 58 0.116 19 0 0.000

9 81 0.162 20 0 0.000

10 88 0.176

Would you be surprised if someone guessing on a 20-question true-false quiz got only 3 correct?

Only about 2 in 1000 guessers would get exactly 3 correct. Since this is so unlikely, this outcome is surprising!


Correct Responses

Number of Students


Number of Correct

Responses

Number of Students


0 0 0.000 11 79 0.158

1 0 0.000 12 61 0.122

2 1 0.002 13 39 0.078

3 1 0.002 14 18 0.036

4 2 0.004 15 7 0.014

5 8 0.016 16 1 0.002

6 18 0.036 17 1 0.002

7 37 0.074 18 0 0.000

8 58 0.116 19 0 0.000

9 81 0.162 20 0 0.000

10 88 0.176

If a score of 15 or more correct is a passing grade on the quiz, is it likely that someone who is guessing will pass?

It would be unlikely that a student who is guessing would be able to pass.

P(passing quiz) ≈ 0.014 + 0.002 + 0.002 + 0 + 0 + 0 = 0.018


Correct Responses

Number of Students


Number of Correct

Responses

Number of Students


0 0 0.000 11 79 0.158

1 0 0.000 12 61 0.122

2 1 0.002 13 39 0.078

3 1 0.002 14 18 0.036

4 2 0.004 15 7 0.014

5 8 0.016 16 1 0.002

6 18 0.036 17 1 0.002

7 37 0.074 18 0 0.000

8 58 0.116 19 0 0.000

9 81 0.162 20 0 0.000

10 88 0.176

The professor actually gives the quiz, and a student scores 16 correct. Do you think that the student was just guessing?

P(scores 16 or higher) ≈ 0.002 + 0.002 + 0 + 0 + 0 = 0.004

Begin by assuming that the student was guessing and determine whether a score at least as high as 16

is a likely or an likely occurrence.

There are two possible explanations for a score of 16:

1) The student was guessing and was REALLY lucky2) The student was not just guessing

Since the first explanation is highly unlikely, you could conclude that a student with a score of 16 was

not just guessing.

Quiz example continued.

What score on the quiz would it take to convince you that a student was not just guessing?

Consider this table showing approximate probabilities for a certain score or higher.

Score Approximately Probability

20 0.000

19 or better 0.000 + 0.000 = 0.000

18 or better 0.000 + 0.000 + 0.000 = 0.000

17 or better 0.002 + 0.000 + 0.000 + 0.000 = 0.002

16 or better 0.002 + 0.002 + 0.000 + 0.000 + 0.000 = 0.004

15 or better 0.014 + 0.002 + 0.002 + 0.000 + 0.000 + 0.000 = 0.018

14 or better 0.036 + 0.014 + 0.002 + 0.002 + 0.000 + 0.000 + 0.000 = 0.054

13 or better 0.078 + 0.036 + 0.014 + 0.002 + 0.002 + 0.000 + 0.000 + 0.000 = 0.132You might say that a score of 14 or higher is

reasonable evidence that someone is not just guessing, because the approximate probability that a

guesser would score this high is only 0.054.

Estimating Probabilities Empirically and Using Simulation

Estimating Probabilities Empirically

It is fairly common practice to use observed long-run proportions to estimate probabilities.

The process used to estimate probabilities is simple:1. Observe a large number of chance

outcomes under controlled circumstances.2. Interpreting probability as a long-run

relative frequency, estimate the probability of an event by using the observed proportion of occurrence.

To recruit a new faculty member, a university biology department intends to advertise for someone with a Ph.D. in biology and at least 10 years of college-level teaching experience.

A member of the department express the belief that requiring at least 10 years of teaching experience will exclude most potential applicants and will exclude more female applicants than male applicants.

The biology department would like to determine the probability an applicant would be excluded because of the experience requirement.

A similar university just completed a search in which there was no requirement for prior teaching experience. However, prior teaching experience was recorded. The resulting data is summarized in the following table. Number of Applicants

Less than 10 years experience

10 or more years experience

Total

Male 178 112 290

Female 99 21 120

Total 277 138 410

𝑃 (𝑎𝑝𝑝𝑙𝑖𝑐𝑎𝑛𝑡 𝑖𝑠𝑒𝑥𝑐𝑙𝑢𝑑𝑒𝑑 )=277410

=0.675

The probability that an applicant would be excluded due to the requirement of at least 10 years

experience is 67.5%.

This is just a little more than two-thirds of the applicants.

New faculty member example continued.

Now let’s determine if more females than males are excluded due to the experience requirement.

Number of Applicants

Less than 10 years experience

10 or more years experience

Total

Male 178 112 290

Female 99 21 120

Total 277 138 410

It appears that female applicants are more likely to be excluded due to the experience requirement than

male applicants.

About 82.5% of the female applicants are excluded due to the experience requirement.

About 61.4% of the male applicants are excluded due to the experience requirement.

Estimating Probabilities by Using SimulationSimulation provides a way to estimate probabilities when:

• You are unable to determine probabilities analytically

• You do not have the time or resources to determine probabilities

• It is impractical to estimate probabilities empirically by observation

Simulations involves generating “observations” in a situation that is similar to the real situation of interest.

Using Simulation to Approximate a Probability1. Design a method that uses a random mechanism

(such as a random number generator or table, the selection of a ball from a box, or the toss a coin) to represent an observation. Be sure that the important characteristics of the actual process are preserved.

2. Generate an observation using the method in Step 1, and determine if the event of interest has occurred.

3. Repeat Step 2 a large number of times.

4. Calculate the estimated probability by dividing the number of observations for which the event of interest occurred by the total number of observations generated.

Suppose that couples who wanted children were to continue having children until a boy was born. Would this change the proportion of boys in the population?

We will use simulation to estimate the proportion of boys in the population if couples were to continue having children until a boy was born.

1. You can use a single random digit to represent a child, where odd digits represent a male birth and even digits represent a female birth.

2. An observation is constructed by selecting a sequence of random digits. If the first random number obtained is odd (a boy), the observation is complete. If the first random number obtained is even (a girl), another digit is chosen. You would continue in this way until an odd digit is obtained.

Baby Boy Simulation Continued . . .

Below are four rows from the random digit table.

Row

6 0 9 3 8 7 6 7 9 9 5 6 2 5 6 5 8 4 2 6 4

7 4 1 0 1 0 2 2 0 4 7 5 1 1 9 4 7 9 7 5 1

8 6 4 7 3 6 3 4 5 1 2 3 1 1 8 0 0 4 8 2 0

9 8 0 2 8 7 9 3 8 4 0 4 2 0 8 9 1 2 3 3 2

Trial 1: girl, boy

Trial 2: boy

Trial 3: girl, boy

Trial 4: girl, boy

Trial 5: boy

Trial 6: boy

Trial 7: boy

Trial 8: girl, girl, boy

Trial 9: girl, boy

Trial 10: girl, girl, girl, girl, girl, girl, boy

Notice that even with only 10 trials, the proportion of boys is 10/22, which is

close to 0.5!

Documents

Chapter 5 Probability Created by Kathy Fritz. Can ultrasound accurately predict the gender of a baby? The paper “The Use of Three-Dimensional Ultrasound