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Marginal Probabilities

The concepts of conditional probability, joint probability and marginal probability are tightly and intricately interwoven in the Theory of Probability. Marginal probabilities are the probabilities on the margins of what is called a "contingency" table, which lists the probabilities of two or more events, their complements in the sample space and their joint probabilities. We will start with the simplest case of two events, A and B, and their complements, AC and BC.

Suppose we consider some statistical data about drunk driving in a mid-sized city in Western Canada. Of the drivers suspected of having consumed too much alcohol, 75% took a breathalyzer test, 30% took a blood test and 22% took both.

Now consider the following contingency table which will, when filled, give a one-shot look at the statistics of these two events in the sample space of drivers that were stopped for drunk driving in this particular city. Fill into the table the total that took the breathalyzer test, the total that took the blood test, and the total that took both.

 
 
Breathalyzer
 
 
   Yes No Total
Blood
 Yes 22%   30%
Test
 No      
 
 Total 75%    

We are now able to fill in the rest of the table by common sense.
Of the 75% of suspected drivers who took a breathalyzer test, 22% also took a blood test and the remainder 75 - 22 = 53% did not.
Of the 30% of suspected drivers who took a blood test, 22% also took a breathalyzer test and the remainder 30 - 22 = 8% did not.

Moreover, the total of all those suspected drivers who did or did not take either test must be the entire sample space. Therefore, the total of the totals must be 100%.   Now the table looks like the following.

 
 
Breathalyzer
 
 
   Yes No Total
Blood
 Yes 22% 8% 30%
Test
 No 53%    
 
 Total 75%   100%

You may be beginning to suspect that the entries in each column must add up to the total in that column (as in the first column, for instance)and the entries in each row must add up to the total in that row (as in the first row).
Consider the "Total" column on the right. Of the 100% of the drivers, if 30% took the blood test, 70% did not.
Consider the bottom row. Of the 100% of the drivers, if 75% took a breathalyzer test, 25% did not.
Look at the table now.

 
 
Breathalyzer
 
 
   Yes No Total
Blood
 Yes 22% 8% 30%
Test
 No 53%   70%
 
 Total 75% 25% 100%

For the remaining blank entry into the table, you can either subtract 53% from 70% (using row 3) or subtract 8% from 25% (using column 3) to get that the percentage of drivers that took neither the breathalyzer nor the blood test was 17%.
The table is now complete.

 
 
Breathalyzer
 
 
   Yes No Total
Blood
 Yes 22% 8% 30%
Test
 No 53% 17% 70%
 
 Total 75% 25% 100%

This table has been filled in using a bit of deductive reasoning from only 3 given quantities. This is possible if the events are "Yes-No" type of events in real life, "exclusive OR" events in logic, or "mutually exclusive" events in statistical probability.

Now let's get back to probability again by filling in this table with probabilities of events instead of simply percentages.
Let S be the sample space of all suspected drivers.
Let A be the event that the driver took a breathalyzer test.
Let B be the event that the driver took a blood test.

Given:   P(A) = 0.75,    P(B) = 0.30,    P(A and B) = 0.22
Deduced:   P(AC) = 1 – 0.75 = 0.25,   P(BC) = 1 – 0.30 = 0.70

By converting all percentages to decimals (divide all entries by 100%), the contingency table becomes filled in with probabilities, and we obtain the following.

 
 
Breathalyzer
 
 
   A AC Total
Blood
 B P(A and B) = .22 P(AC and B) = .08 P(B) = .30
Test
 BC P(A and BC) = .53 P(AC and BC) = .17 P(BC) = .70
 
 Total P(A) = .75 P(AC) = .25 P(S)= 1.00

From the table, we note the following:

  1. The marginal probabilities are those on the margins of the table (those in green). They are the "pure" probabilities of the events A and B and their complements.
  2. The probabilities internal to the table (those in black) are the joint probabilities of the various events.
  3. Once we know the so-called "pure" probabilities and the joint probability, the conditional probabilities may be derived from the Multiplication Rule.

P(A and B) = P(A) · P(B | A) = P(B) · P(A | B)  or 

P(A | B) = P(A and B) / P(B)  and  P(B | A) = P(A and B) / P(A) .

Now we are able to answer the following questions (and others like them).

  1. What is the probability that a driver that has been given a breathalyzer test (event A) will also be given blood test (event B)?
Answer:
P(B | A) = P(A and B) / P(A) = 0.22 / 0.75 = 0.293 .

  1. What is the probability that a driver that has not been given a blood test (event BC) will be given breathalyzer test (event A)?
Answer:
P(A | BC) = P(A and BC) / P(BC) = 0.53 / 0.70 = 0.757 .

  1. Does getting a blood test (event B) change the probability of getting a breathalyzer test (event A)?
Answer:

The question is simply asking whether the two events are independent or not. From the calculation below, we deduce that the events are not independent, and, therefore, getting a blood test does change the probability of getting a breathalyzer test.

P(A | B) = P(A and B) / P(B) = 0.22 / 0.30 = 0.733  is not equal to  0.75 = P(A) .

 

Last Updated by SAS October 07, 2026, 16:00:54