Complementary phenomenon

A complementary phenomenon can sometimes be used when calculating the probability of a phenomenon. A complementary phenomenon to the phenomenon A is the phenomenon A', which contains all possible outcomes that can occur but are not in the phenomenon A.

Definition of

The definitions of random experiment and phenomenon are given in the probability chapter. To briefly recap: a random experiment is, for example, a roll of the dice, the set of all possible outcomes contains those outcomes that can occur, i.e. in the case of a roll of the dice, it is the set $\omega = {1, 2, 3, 4, 5, 6}$, the phenomenon is a subset of $\omega$ and can be "an odd number rolled on the dice", i.e. A = {1, 3, 5}.

If we have the phenomenon A, staying with "an odd number rolled on the dice", i.e. A = {1, 3, 5}, the complementary phenomenon is $A' = \omega \setminus A$. So are all other phenomena; those that are in $\omega$, but not in A. In our case, where we have the phenomenon "an odd number rolled on the dice", the complementary phenomenon would be "an even number rolled on the dice". The complementary phenomenon to the phenomenon "a number 2 or 5 is rolled" is the phenomenon "a number 1, 3, 4 or 6 is rolled".

To stick with set terminology, if $\omega = {1, 2, 3, 4, 5, 6}$ and we have the phenomenon A = {1, 2} ("a number less than three falls"), then $A' = \omega \setminus A = {1, 2, 3, 4, 5, 6} \setminus {1, 2} = {3, 4, 5, 6}$. Verbally, we could describe it as "a number greater than two falls".

For the probability of a complementary event, a simple relation holds: if the probability of A is 25%, what is the probability of A'? In essence, we are asking - what is the probability that the A phenomenon does not occur ? The phenomenon A occurs with probability 25% and so the phenomenon A does not occur with probability 100 − 25 = 75 %. This is always true, so we can say that the complementary phenomenon A' occurs with probability 1 − P(A):

$$ P(A') = 1 - P(A) $$

Solved examples

1. What is the probability that two sixes are not rolled when two dice are rolled? In the chapter on probability, we calculated that there are a total of 36 different possibilities that can occur when two dice are rolled. Now we should calculate how many of these possibilities do not contain two sixes. We can go the other way: we count how many pairs contain two sixes, calculate the probability of this occurring, and then subtract that from one.

   There is only one pair that contains two sixes, [6, 6], so the probability of this phenomenon A is:

   $$ P(A) = \frac{1}{36} $$

   The probability of the complementary phenomenon A', i.e. the phenomenon "two sixes do not fall" is then:

   $$ P(A') = 1 - P(A) = 1 - \frac{1}{36} = \frac{35}{36} = 97{,}2222… \% $$

2. What is the probability of rolling three dice and getting at least one six? Again, we can solve this problem directly and list all the variations in which we roll one six, two sixes and three sixes. However, we can go from the woods and calculate the complementary phenomenon, i.e. the "not even one six fell" phenomenon. First, let's calculate how many total possibilities there are that we can drop. We have three dice, on each of which the numbers 1 to 6 can fall, giving a total of 6³ = 216 possibilities.

   Now we need to count the possibilities where there is not even one six. Each die must have a number from 1 to 5, so there are a total of 5³ = 125 possibilities that do not contain the number 6. Now we just plug that into the formula:

   $$ P(A) = \frac{125}{216}=57{,}87 \% $$

   This was the A phenomenon - "not a single six". We want to know the complementary phenomenon: "at least one six fell". We'll plug that into the formula:

   $$ P(A') = 1 - P(A) = 1 - \frac{125}{216} = \frac{91}{216} = 42{,}1296 \% $$

3. A student draws 3 questions out of 30 questions offered on a math exam. There are 10 questions from algebra, 15 from analysis, and 5 from geometry. What is the probability that he will pull at least two questions from the same subject? Again, it will be easier to solve this problem using the additional phenomenon: what is the probability that the student will pull just one question from each subject?

   First, we calculate how many different triples a student can pull. For this we will need the combinations. We have a total of 30 questions and we pull 3 of them, but it doesn't matter the order. That gives us a combination number

   $$ {30 \choose 3} = 4060 $$

   Next, how many different possibilities are there where a student pulls just one question from each domain? Using the combinatorial product rule, we find that the total is 10 · 15 · 5 = 750. Finally, we plug in the formula to calculate the probability:

   $$ P(A) = \frac{750}{4060}=18{,}47 \% $$

   But we want to know the complementary effect, so we just subtract the result from one:

   $$ P(A') = 1 - \frac{750}{4060} = \frac{3310}{4060} = 81{,}53 \% $$