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.

First, make sure you have your contingency table handy.

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

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

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:
   [latex]P\lgroup neutral\: OR\: automatic\rgroup = \frac{1+136+61}{200}=\frac{198}{200}=0.99=99\%[/latex].

   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.

   [latex]P\lgroup neutral\: OR\: automatic\rgroup = \frac{137+197-136}{200}=\frac{198}{200}=0.99=99\%[/latex].  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

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

   If using the totals method, we find that [latex]P\lgroup manual\: OR\: automatic\rgroup = \frac{3+197-0}{200}=\frac{200}{200}=1=100\%[/latex].  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.

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.

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.

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!