# learn by watching

Try Exercise Case Study #1.

Try Exercise Case Studies #2-4.

Complete the Reflection Activity on workplace drug testing.

Probability is a way to describe how likely something is to happen.  We often use phases like the “odds” or the “chance” of something happening to describe this.  And we usually report our probabilities as fractions or as percentages.

## Creating contingency tables

In situations where we want to calculate probabilities based on data that we have collected, we can use something called a “contingency table” to put our data into an easy-to-analyze format.

Here is an example of how to use a contingency to table describing the results of a small survey.

Example 1: A survey of recent cars looked at the type of transmission (manual vs. automatic) as well as the color of the vehicle (neutral vs. brightly-colored).  Out of 200 cars, there were 3 cars that had a manual transmission.  Of the manual-transmission cars, one was black and the rest were brightly colored.  There were 137 cars altogether that were neutral colors (black, grey, silver, white).  Create a contingency table to represent the results of your survey.

Start by creating a table showing the different category options.  Use one category for the rows, and the other category for the columns.  It really doesn’t matter which way you choose, just be sure things are clearly labeled so your numbers end up in the correct locations.  In this case, I decided to put transmission types across my top row, and I put the color options down my left column.  There were 200 cars in the survey, so that goes in the bottom right hand corner — we often call this the “grand total”.

 MANUAL TRANSMISSION (STICK SHIFT) AUTOMATIC TRANSMISSION TOTAL NEUTRAL COLORS BRIGHT COLORS TOTAL 200

Next, we want to put the specific values we were given in the problem.  We know that there were 3 manual transmissions altogether.  So the 3 is in the TOTAL box of the “manual transmission” column.  One was black, so a “1” will go in the “neutral colors” box of the manual transmission column.  We were also told that there were 137 cars altogether that were neutral — so this will go as the TOTAL box in the “neutral colors” row.  Now our table looks like this:

 MANUAL TRANSMISSION (STICK SHIFT) AUTOMATIC TRANSMISSION TOTAL NEUTRAL COLORS 1 137 BRIGHT COLORS TOTAL 3 197 200

This leaves a lot of blank boxes!  We need to fill them in before we can think about calculating probabilities.  Where do the numbers come from? Well, remember that if we add across the row, we get a total.  And if we add down a column we get a total.  So as long as we are only missing one piece of information in the row/column, we can use what we know to find it.

Right now, I don’t know enough information about “automatic transmissions” to work on that column or “bright colors” to work on that row.  So let’s start with some we know, and hopefully we will get enough information later!

In the manual transmission column, I know there are a total of 3 cars.  Since 1 of the cars is neutral, the rest will be brightly colored.  So 3-1 = 2 brightly colored, manual transmission cars.

In the neutral colors row, I know there are a total of 137 cars.  Since 1 of those cars has a manual transmission, the rest will have an automatic.  So 137-1 = 136 neutral colored, automatic transmission cars.

In the TOTAL column, I know that 137 out of the 200 cars had neutral colors.  So the rest would have to be brightly colored.  This gives us 200 – 137 = 63 total brightly colored cars.

 MANUAL TRANSMISSION (STICK SHIFT) AUTOMATIC TRANSMISSION TOTAL NEUTRAL COLORS 1 137-1 = 136 137 BRIGHT COLORS 3-1 = 2 200-137 = 63 TOTAL 3 197 200

Now I have enough information to figure out how many bright colored cars also had automatic transmissions, the only box remaining.  I can use either

• the bright colored row (2 out of 62 total bright colored cars are accounted for, so the rest would be automatic transmissions) 63-2 = 61 or
• the automatic transmission column (136 out of the 197 cars had neutral colors, so the rest would have bright colors) 197-136 = 61

TIP: Notice that you get the same answer either way! It’s a good idea to add “across the row” and “down the column” of where you filled in the last box.  If you don’t get the correct totals in both directions, you have a mistake somewhere, and should take the time to find it before you proceed with any calculations.

So my final contingency table representing this situation would look like this:

 MANUAL TRANSMISSION (STICK SHIFT) AUTOMATIC TRANSMISSION TOTAL NEUTRAL COLORS 1 136 137 BRIGHT COLORS 2 61 63 TOTAL 3 197 200

This completed table will make it easy to calculate any probabilities that you would like.

Now, it’s time to look at calculating some probabilities!

## calculating probabilities

To do this, we simply want to create a fraction (ratio) that will compare the number of “favorable” outcomes to the total number of possibilities.  You will want to report your probabilities as UNSIMPLIFIED fractions (so a reader can identify which numbers you used off of the contingency table) as well as the associated percentage form.

Definition of Probability

The probability of an outcome can be calculated with a ratio comparing the number of favorable outcomes to total possible outcomes.

$P\lgroup outcome\rgroup = \frac{number\:of\:favorable\:outcomes}{total\:number\:of\:possible\:outcomes}$

Probabilities are also commonly recorded as percentages. To convert this fraction to a percentage, you’ll want to divide the two values, then multiply the resulting decimal by 100.

TIP: Since 0% likely would indicate an outcome was impossible, and 100% would indicate that all outcomes apply, our probability results will always lie between 0% and 100%.  If we are looking at the decimal versions of these probabilities, we should be getting a number between 0 and 1 when we divide the numerator by the denominator.

Example 1A:

1. Find the probability of choosing a brightly colored car.
2. Find the probability of choosing a car with a manual transmission
First, make sure you have your contingency table handy.
 MANUAL TRANSMISSION (STICK SHIFT) AUTOMATIC TRANSMISSION TOTAL NEUTRAL COLORS 1 136 137 BRIGHT COLORS 2 61 63 TOTAL 3 197 200

When finding a probability, I always identify the bottom number first (total # of all possibilities), then I choose the number for the top of my fraction calculation.  In each of these cases, the total number of possibilities is the “grand total” from the survey, all 200 cars.  So the denominator of each probability will be 200.

1. In this case we want to find the probability of choosing a brightly-colored car.  Because mathematicians are lazy, we will often write this as P(brightly colored car), where the P stands for “find the probability of” and the inside of the parentheses represents the “favorable outcome” I am trying to predict.  I don’t care at all about the kind of transmission, so since there were a total of 137 bright colored cars, we can say that:
$P\lgroup brightly\:colored\:car\rgroup=\frac{137}{200}$.In the exercises, you will be asked to write your results both as unsimplified fractions as well as percentage form.  To find the percentage, simply divide the probability fraction, then multiply the resulting decimal value by 100.  $\frac{137}{200}=137\div200=0.685\times100=68.5\%$.
2. Again, we are choosing from all 200 cars surveyed.  There were a total of 3 cars with a manual transmission.  So we can write the probability of choosing a car with a manual transmission as: $P\lgroup manual\:transmission\rgroup = \frac{3}{200}=.015=1.5\%$.

## probabilities using “and”

Sometimes we need to find a probability where multiple conditions must apply.  The word “AND” is commonly used in situations where you need BOTH conditions to apply to be considered a “favorable outcome”.

Example 1B:

1. Find the probability of choosing a car that is brightly colored AND has a manual transmission.
2. Find the probability of choosing a neutral-colored colored car with an automatic transmission.
3. Find the probability of choosing a car with an automatic transmission AND  a manual transmission.
Again, we are choosing a car out of the entire group surveyed.  So total number of possibilities is the “grand total” from the survey, all 200 cars.  This means the denominator of each probability will be 200.
1. For “favorable number of outcomes”, we need to choose something that has BOTH conditions.  This time,  we want to look at the box that is in the brightly colored row that ALSO lies in the manual transmission column.
 MANUAL TRANSMISSION (STICK SHIFT) AUTOMATIC TRANSMISSION TOTAL NEUTRAL COLORS 1 136 137 BRIGHT COLORS 2 61 63 TOTAL 3 197 200

$P\lgroup bright\: AND\: manual\rgroup = \frac{2}{200}=.01=1\%$.

2. This time we don’t specifically use the word AND.  However, because they are written together, the wording implies that we need BOTH the neutral colors row AND the automatic transmission column.
 MANUAL TRANSMISSION (STICK SHIFT) AUTOMATIC TRANSMISSION TOTAL NEUTRAL COLORS 1 136 137 BRIGHT COLORS 2 61 63 TOTAL 3 197 200

$P\lgroup neutral\: AND\: automatic\rgroup = \frac{136}{200}=.68=68\%$.

3. Here, we again see the word “AND”, so BOTH conditions need to apply.  However, there are no boxes where BOTH automatic transmissions AND manual transmissions overlap.
 MANUAL TRANSMISSION (STICK SHIFT) AUTOMATIC TRANSMISSION TOTAL NEUTRAL COLORS 1 136 137 BRIGHT COLORS 2 61 63 TOTAL 3 197 200

So, we can conclude that $P\lgroup automatic\: AND\: manual\rgroup = \frac{0}{200}=0\%$.

## PROBABILITIES USING “OR”

Other times, we need to find probabilities where a “favorable outcome” could include either of the outcomes.  Since mathematics and logic are so intertwined, the word “OR” actually has a very specific meaning.  It means one OR the other OR both.  As long as at least one of the conditions apply, it needs to be counted.

Example 1C:

1. Find the probability of choosing a car that is brightly colored OR has a manual transmission.
2. Find the probability of choosing a neutral-colored OR an automatic transmission.
3. Find the probability of choosing a car with an automatic transmission OR  a manual transmission.
Again, we are choosing a car out of the entire group surveyed.  So total number of possibilities is the “grand total” from the survey, all 200 cars.  This means the denominator of each probability will be 200.
1. For “favorable number of outcomes”, we need to choose something that has one or the other of the conditions.  This time,  we want to all the boxes in the “brightly colored” row as well as all the boxes that lie in the “manual transmission” column.
 MANUAL TRANSMISSION (STICK SHIFT) AUTOMATIC TRANSMISSION TOTAL NEUTRAL COLORS 1 136 137 BRIGHT COLORS 2 61 63 TOTAL 3 197 200

$P\lgroup bright\: OR\: manual\rgroup = \frac{1+2+61}{200}=\frac{64}{200}=.032=32\%$.

2. Same idea this time.  We want to include anything that shows up in the “neutral colors” row as well as anything that shows up in the “automatic transmission” column.
 MANUAL TRANSMISSION (STICK SHIFT) AUTOMATIC TRANSMISSION TOTAL NEUTRAL COLORS 1 136 137 BRIGHT COLORS 2 61 63 TOTAL 3 197 200

Borrowing from the process in our last calculation, we can find:
$P\lgroup neutral\: OR\: automatic\rgroup = \frac{1+136+61}{200}=\frac{198}{200}=0.99=99\%$.

Or, as my husband would say, 99% of the cars on the market are boring. Haha!

One thing you need to be careful about when finding OR probabilities is just using the row/column totals.  In this example, if we just added the totals from automatic transmissions (197) and from neutral colors (137), we’d end up with 334 cars in the numerator.  But that is more cars than were in the survey!  Why did this happen?  Well, by using the totals, the 136 neutral/automatic cars were double-counted.  You can use the totals, but you need to remember to remove anything that would have gotten counted twice.

$P\lgroup neutral\: OR\: automatic\rgroup = \frac{137+197-136}{200}=\frac{198}{200}=0.99=99\%$.  The same answer! Either way is fine. I’ve just found most students make fewer mistakes by just adding up the boxes that apply, where it is more obvious that they have not included the same group more than one time.

3. Here, we again see the word “OR”, so any car that meets one or the other or both conditions is counted.
 MANUAL TRANSMISSION (STICK SHIFT) AUTOMATIC TRANSMISSION TOTAL NEUTRAL COLORS 1 136 137 BRIGHT COLORS 2 61 63 TOTAL 3 197 200

$P\lgroup manual\: OR\: automatic\rgroup = \frac{1+136+2+61}{200}=\frac{200}{200}=1=100\%$.  Notice that all the boxes are included this time.

If using the totals method, we find that $P\lgroup manual\: OR\: automatic\rgroup = \frac{3+197-0}{200}=\frac{200}{200}=1=100\%$.  Since there was no overlap, we hadn’t double-counted anything. We refer to groups like this as “mutually exclusive” — each group is made up of unique items with no overlap.

## Conditional probabilities

Sometimes we want to figure out probabilities that do NOT involve the grand total.  We want to limit our probability calculation to a more specialized group.  Being able to look at probabilities from different group perspectives can often point out things that might not have been evident if we limited our search to just the entire group as a whole.  In conditional probabilities, we will be changing the denominator of our probability.  No longer will we be looking at the grand total (everything in our survey). Instead we will focus on a subgroup (maybe just a single row or column) — in which case that row or column’s total represents the denominator.

Examples: Let’s continue looking at our car survey contingency table,

 MANUAL TRANSMISSION (STICK SHIFT) AUTOMATIC TRANSMISSION TOTAL NEUTRAL COLORS 1 136 137 BRIGHT COLORS 2 61 63 TOTAL 3 197 200
1. If we are looking at a neutral-colored car, what is the probability that it has an automatic transmission?
2. What is the probability that a car has a a manual transmission given that it is neutrally colored?
3. If we have an automatic transmission, what is the probability that it is brightly colored?
The words “IF” and “GIVEN THAT” are the most commonly used terms to indicate that we want to limit our search and calculations to a smaller group.  Whatever comes AFTER that key word becomes the new denominator, since we want to limit our search to just those listed outcomes.
1. If we are looking at a neutral-colored car, what is the probability that it has an automatic transmission?  Here the key word “IF” at the beginning of the question means that we are only caring about the numbers in the “neutral colors” row.
 MANUAL TRANSMISSION (STICK SHIFT) AUTOMATIC TRANSMISSION TOTAL NEUTRAL COLORS 1 136 137 BRIGHT COLORS 2 61 63 TOTAL 3 197 200

The total of neutral colored cars, 137, becomes our new denominator.  Out of those cars, 136 of them have an automatic transmission.
$Probability = \frac{136}{137}=0.9927=99.27\%$

2. What is the probability that a car has a a manual transmission given that it is neutrally colored? Here the key word “GIVEN THAT” at the end of the question means that again, we are only caring about the numbers in the “neutral colors” row, and that becomes the denominator of our fraction.  The numerator is how many manual transmissions there are in that row.
 MANUAL TRANSMISSION (STICK SHIFT) AUTOMATIC TRANSMISSION TOTAL NEUTRAL COLORS 1 136 137 BRIGHT COLORS 2 61 63 TOTAL 3 197 200

$Probability = \frac{1}{137}=0.0073=0.73\%$

3. If we have an automatic transmission, what is the probability that it is brightly colored?  Here the key word “IF” at the beginning of the question means that we are only caring about the what comes next, which is the automatic transmission column.  So that total (197) becomes our new denominator.
 MANUAL TRANSMISSION (STICK SHIFT) AUTOMATIC TRANSMISSION TOTAL NEUTRAL COLORS 1 136 137 BRIGHT COLORS 2 61 63 TOTAL 3 197 200

Out of those 197 cars with an automatic transmission, only 61 are brightly colored.  And so I can conclude that
$Probability = \frac{61}{197}=0.3096=30.96\%$

In these cases, by looking at particular subgroups, we can see that automatics are highly favored by people who enjoy neutral colors.  And a majority of people that drive manuals prefer brightly-colored cars.  We don’t necessarily find these connections without looking at the conditional probabilities that we can find when we disaggregate data.  In our DSJ investigation for this chapter, we will see an example of this.  It turns out we get a different view of police interactions when we take the time to look at stop, search, and arrest data separated out by race.

Another advantage of using contingency tables is that it can scale to larger applications with more available options (more than 2 rows and/or columns). All the key words, principles, and methods will continue to apply.

Exercises: Try These!

The statistics for the following homework problems were based on the “Basic Needs Security Among Washington College Students” Study published in January 2023.

## CASE STUDY #1

 In the last 30 days were you ever hungry but didn’t eat because there wasn’t enough money for food? 2-YEAR COLLEGE STUDENTS 4-YEAR COLLEGE STUDENTS Total YES 1071 1106 NO 3241 3440 Total

Complete the contingency table above. Then find the following probabilities. Report answers as the fraction using the numbers from the table, as well as the resulting percentages.

1. What is the probability that a student taking the survey was a 2-year college student?
2. What is the probability that a student was a 2-year college student and did not experience food insecurity?
3. What is the probability that a student experienced food insecurity?
4. What is the probability that a student experienced food insecurity given that they were a 2-year college student?

## CASE STUDY #2

 In the past 12 months have you had to temporarily stay with a relative, friend, or couch surfing until finding other housing? YES NO Total MALE GENDER 211 2031 FEMALE GENDER 437 4436 OTHER GENDER 61 441 Total

Complete the contingency table above. Then find the following probabilities. Report answers as the fraction using the numbers from the table, as well as the resulting percentages.

1. What is the probability that a surveyed student was female-gendered?
2. What is the probability that a student had to find temporary housing due to loss of housing, economic hardship, or similar reason?
3. What is the probability that a student had to find temporary housing given that they were male-gendered?
4. What is the probability that a student had to find temporary housing given that they were female-gendered?
5. What is the probability that a student had to find temporary housing given that they were other-gendered?
6. Looking at the students who required temporary housing, what is the the probability that a student was female-gendered?

## CASE STUDY #3

 “I can afford to pay for childcare” AMERICAN INDIAN/ ALASKA NATIVE ASIAN/ ASIAN AMERICAN BLACK/ AFRICAN-AMERICAN HISPANIC/ LATINX PACIFIC ISLANDER/ NATIVE HAWAIIAN TWO OR MORE ETHNICITIES WHITE Total STRONGLY DISAGREE 4 10 24 24 2 26 92 DISAGREE 1 4 12 18 1 11 51 UNDECIDED 2 9 9 7 0 0 25 AGREE 0 11 5 6 0 2 37 STRONGLY AGREE 0 2 2 5 0 4 11 Total

This question was asked to all students who were parents, primary caregivers, or guardians of any dependents.  Fill in the contingency table above.  Then find the following probabilities. Report answers as the fraction using the numbers from the table, as well as the resulting percentages.:

1. What is the probability that a surveyed student was Hispanic/LatinX?
2. What is the probability that a student was Black/African American and “Strongly Agreed” that they could afford childcare?
3. What is the probability that a student “Strongly Disagreed” or “Disagreed” that they could afford childcare?
4. What is the probability that a student “Strongly Disagreed or “Disagreed” that they could afford childcare given that they were Black/African-American?
5. What is the probability that a student “Strongly Disagreed or “Disagreed” that they could afford childcare given that they were White?
6. What is the probability that a student was Asian/Asian-American, given that they “Agreed” or “Strongly Agreed” that they could afford childcare?

## CASE STUDY #4

 “In the past 12 months, I was able to access the health care I needed.” STUDENTS REPORTING DISABILITY (physical, mental/emotional, or learning disability) STUDENTS NOT REPORTING DISABILITY Total ALWAYS TRUE 1228 4383 SOMETIMES TRUE 1273 2961 NEVER TRUE 170 489 DOES NOT APPLY 98 631 Total

Complete the contingency table above. Then find the following probabilities. Report answers as the fraction using the numbers from the table, as well as the resulting percentages.

1. What is the probability that a student reported a disability on the survey?
2. What is the probability that a student was always able to access needed healthcare?
3. What is the probability that a student was always able to access needed healthcare, given that they reported a disability?
4. If a student was never able to access needed healthcare, what was the probability that they had a disability?
5. What was the probability that a student replied “Does Not Apply?” to having access to needed healthcare?  Why might someone have given this answer?

## CASE STUDY #5

 “In the past 12 months, I was able to access the mental/ behavioral health care I needed.” STUDENT IS A MEMBER OF LGBTQI+ COMMUNITY STUDENT IS NOT PART OF LGBTQI+ COMMUNITY Total ALWAYS TRUE 751 2958 SOMETIMES TRUE 929 2191 NEVER TRUE 381 1041 DOES NOT APPLY 246 2010 Total

Complete the contingency table above. Then find the following probabilities. Report answers as the fraction using the numbers from the table, as well as the resulting percentages.

1. What is the probability that a student identifies as a member of the LGBTQI+ community?
2. What is the probability that a student did not try to access to mental/behavioral health services in the past year (i.e. answered “does not apply”)?
3. What is the probability that a student was never able to access mental/behavioral health care needed, given that they were a member of the LGBTQI+ community?
4. What is the probability that a student was never able to access mental/behavioral health care needed, given that they were not a member of the LGBTQI+ community?
5. What is the probability that a student always or sometimes able to access the mental/behavioral health care when needed? (hint: always agree, sometimes agree, or does not apply)
6. What is the probability that a student who identifies with the LGBTQI+ community is always or sometimes able to access the mental/behavioral health care when needed? (hint: always agree, sometimes agree, or does not apply)

## using contingency tables to better understand medical testing

Another useful application of contingency tables is reporting medical testing results.  Let’s consider some specialized vocabulary that gets used in these situations.

A medical test provides a POSITIVE result (the test shows evidence of a condition or disease) or a NEGATIVE result (the test shows no evidence of the condition or disease).

Tests, however, are not always accurate.  Sometimes someone does not have the condition/disease, but they get a positive result from the test.  This is called a FALSE POSITIVE.  Sometimes, someone has the condition/disease, but the the test provides a negative result.  We call this a FALSE NEGATIVE.

Example: A group of 500 women take a pregnancy test.  40% of the women tested are ACTUALLY pregnant.  The pregnancy test reports being 99% accurate.  What is the probability that if a woman gets a negative result, that they are actually pregnant?

To start, let’s set up a contingency table.  Our two sets of conditions are the ACTUAL status and the TEST RESULT status.  We know that the grand total in this study was 500 women.

 ACTUALLY PREGNANT ACTUALLY NOT PREGNANT TOTAL TEST IS POSITIVE TEST IS NEGATIVE TOTAL 500

We are told that 40% of the women taking the test were actually pregnant.  Since there were 500 women in this study, 40% of 500 = 0.40 * 500 = 200 women were actually pregnant.  So, with 500 people in the study, that left 500 – 200 = 300 women who were actually not-pregnant.  These numbers can go in the “totals” box.

 ACTUALLY PREGNANT ACTUALLY NOT PREGNANT TOTAL TEST IS POSITIVE TEST IS NEGATIVE TOTAL 200 300 500

Now, we can use the fact that the pregnancy test reports being 99% accurate.  What does that mean?  It means that 99% of women get a correct result.

99% of actually pregnant women get a positive result.

99% of actually not-pregnant women get a negative result.

 ACTUALLY PREGNANT ACTUALLY NOT-PREGNANT TOTAL TEST IS POSITIVE 0.99 * 200 = 198 TEST IS NEGATIVE 0.99 * 300 = 297 TOTAL 200 300 500

So, of the actually pregnant women 200-198 = 2 actually pregnant women tested negative.  We’d call these “false negatives”.

Of the actually not-pregnant women 300-297 = 3 actually not-pregnant women tested positive.  We’d call these “false positives”.

Let’s use this to finish filling in the contingency table — including finding the totals.

 ACTUALLY PREGNANT ACTUALLY NOT PREGNANT TOTAL TEST IS POSITIVE 198 3 201 TEST IS NEGATIVE 2 297 299 TOTAL 200 300 500

Now we can use the table to calculate the requested probability: What is the probability that if a woman gets a negative result, that they are actually pregnant?

Notice the “if” in the question.  This is one of the key words indicating a conditional probability. This means we are going to limit our group to the “test is negative” row.  So the bottom number of the probability fraction is the
test is negative” total, 299.  The top of the probability fraction is how many in that row are actually pregnant, 2.

So our probability is 2/299 = .00669 = 0.669%.  Meaning if someone gets a negative result, there is only a 0.669% that they are actually pregnant.

In conditions that are rare, however, something strange ends up happening when looking at probabilities.

Example: The Lupus foundation of America reports that 1.5 million Americans have lupus, a rare auto-immune disorder.  There about 370 million people in the US.  A screening test for lupus reports being 99.5% accurate.  What is the probability a person does NOT have lupus given that they received a positive test result?

To start, let’s set up a contingency table.  Our two sets of conditions are the ACTUAL status and the TEST RESULT status.  We can use the 370 million people in the US as our grand total.  Since 1.5 million people have lupus, that leaves 370-1.5 = 368.5 million Americans do not have lupus.  We can use this information to fill in the bottom row of the table.

 LUPUS – YES LUPUS – NO TOTAL TEST IS POSITIVE TEST IS NEGATIVE TOTAL 1,500,000 368,500,000 370,000,000

We are told that the screening test reports being 99.5% accurate.  Meaning,

99.5% of people who have lupus get a positive result.

99.5% of people who do not have lupus get a negative result.

 LUPUS – YES LUPUS – NO TOTAL TEST IS POSITIVE 0.995 * 1,500,000 = 1,492,500 TEST IS NEGATIVE 0.995 * 368,500,000 = 366,657,500 TOTAL 1,500,000 368,500,000 370,000,000

So, of those with lupus, 1,500,000 – 1,492,500 = 7500 people with the disease were incorrectly diagnosed as negative.

And of those without lupus, 368,500,000 – 366,657,500 = 1,842,500 without the disease were incorrectly diagnosed as positive.

Let’s use this to finish filling in the contingency table — including finding the totals by adding across each row.

 LUPUS – YES LUPUS – NO TOTAL TEST IS POSITIVE 1,492,500 1,842,500 3,335,000 TEST IS NEGATIVE 7500 366,657,500 366,665,000 TOTAL 1,500,000 368,500,000 370,000,000

Now we can use the table to calculate the requested probability: What is the probability a person does NOT have lupus given that they received a positive test result?

Notice the “given that” in the question.  This is another key word indicator that it’s a conditional probability problem. This means we are going to limit our group to whatever comes next — in this case, we are restricting all our work to the “test is positive” row.  So the bottom number of the probability fraction is the “test is positive” total, 3,335,000.  The top of the probability fraction is how many in that row actually have do not have lupus, 1,842,500.

So our probability is 1,842,500/3,335,000 = .5524 = 55.24%.  Meaning if someone gets a positive test result, there is a 55.24% that they actually do not even have this rare disease!

Because tests are not perfect, care needs to be issued in how information is shared with patients and/or other interested parties.  A positive test for a whole lot of people may be incorrect (despite having what is reported as a “very accurate” test).  This is why many times a medical professional will ask you to come in for additional tests before issuing a diagnosis.  Otherwise, there can be a whole lot of unnecessary mental distress (and potentially unnecessary treatments applied) for a patient.

Reflection Activity: Workplace Drug Testing

The “Effectiveness” of Drug Tests

People in many occupations, such as police officers and air traffic controllers, are subject to random drug testing. Such testing is done on the grounds that the employee’s work affects the safety of the general public.  Many jobs also require a potential employee to have a negative drug test result as part of the hiring process.

Drug testing is controversial, however. One objection concerns the potential unfair consequences stemming from the fact that the tests are imperfect. for instance, if a test incorrectly shows someone to be a drug user (a “false positive”), that person could lose his or her job.

PROBLEM: Let’s suppose that only 5 percent of people on the job engage in drug use.  Let’s assume a certain drug test is 98 percent accurate. This means 98 percent of people who used the given drug will test positive and 98 percent of the people who did not use the drug will test negative. If a person tests positive, how likely is it that they actually used drugs?

1. Construct a contingency table. Rather than dealing with percentages, it is easier to consider a large population — let’s say 100,000 total.
2. Find the following probability. If a person tests positive, how likely is it that they actually used drugs?
3. In a paragraph — What are possible repercussions for an employee that tests positive? Based on your probability calculations — how effective do you think drug testing in the workplace is?

RESEARCH: There are many ethical considerations associated with employer drug tests.  Read two of the three articles below.  Based on your reading:

1. Write a paragraph summarizing the benefits and dangers of employer drug tests.
2. In a paragraph, specifically consider your thoughts the following questions. Are any groups unfairly targeted by employer drug tests?  What groups and why?