Complementary event

The complementary event can sometimes be used when calculating the probability of an event. The complement of an event A is the event A^c that contains all the possible outcomes that can occur but are not in the event A.

Definition

The definitions of a probability experiment and an event are given in the chapter on probability. A quick recap: a probability experiment is, for example, rolling a dice, the sample space contains the outcomes that can occur, that is, for a dice it is the set ω = {1, 2, 3, 4, 5, 6}, and an event is a subset of ω, such as “the dice shows an odd number”, that is, A = {1, 3, 5}.

If we have an event A, staying with “the dice shows an odd number”, that is, A = {1, 3, 5}, then by the complementary event we mean the event A^c = ω ∖ A. These are all the other outcomes – the ones that are in ω but not in A. In our case, where we have the event “the dice shows an odd number”, the complementary event would be “the dice shows an even number”. The complement of the event “a 2 or a 5 is rolled” is the event “a 1, 3, 4 or 6 is rolled”.

To stick with the terminology of sets: if ω = {1, 2, 3, 4, 5, 6} and we have the event A = {1, 2} (“a number less than three is rolled”), then A^c = ω ∖ A = {1, 2, 3, 4, 5, 6} ∖ {1, 2} = {3, 4, 5, 6}. In words, we could describe this as “a number greater than two is rolled”.

There is a simple relationship for the probability of the complementary event: if the probability of the event A is 25%, what is the probability of the event A^c? Essentially we are asking – what is the probability that the event A does not happen? The event A happens with probability 25%, so the event A does not happen with probability 100 − 25 = 75%. This is always true, so we can say that the complementary event A^c happens with probability 1 − P(A):

$$P(A^c)=1-P(A)$$

Worked examples

1. What is the probability that, when we roll two dice, we do not get two sixes? In the chapter on probability we worked out that there are 36 different possible outcomes when two dice are rolled. Now we would have to count how many of these outcomes do not contain two sixes. We can go the opposite way instead: we count how many pairs contain two sixes, work out the probability of that event and then subtract it from one.

   There is only one pair that contains two sixes, (6, 6), so the probability of this event A is:

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

   The probability of the complementary event A^c, that is, the event “we do not get two sixes”, is then:

   $$P(A^c)=1-P(A)=1-\frac{1}{36}=\frac{35}{36}=97.2222\dots \mathrm{\%}$$

2. What is the probability that, when we roll three dice, we get at least one six? We could again solve this directly and list all the arrangements in which we got one six, two sixes and three sixes. But we can approach it from the other side and work out the complementary event, that is, the event “no six was rolled”. First we work out how many possibilities there are in total. We have three dice, and each can show the numbers 1 to 6, which gives 6³ = 216 possibilities.

   Now we have to count the possibilities that contain no six at all. Each dice must show a number from 1 to 5, so there are 5³ = 125 possibilities that do not contain the number 6. Now we just substitute into the formula:

   $$P(A)=\frac{125}{216}=57.87\mathrm{\%}$$

   This was the event A – “no six was rolled”. We want the complementary event: “at least one six was rolled”. So we substitute into the formula:

   $$P(A^c)=1-P(A)=1-\frac{125}{216}=\frac{91}{216}=42.1296\mathrm{\%}$$

3. In a maths exam, a student draws 3 questions from 30 available questions. Of these, 10 questions are on algebra, 15 on calculus and 5 on geometry. What is the probability that the student draws at least two questions from the same topic? Again it will be easier to solve this using the complementary event, which asks: what is the probability that the student draws exactly one question from each topic?

   First we work out how many different sets of three the student can draw. For this we will need combinations. We have 30 questions in total and draw 3 of them, and the order does not matter. That gives the binomial coefficient

   $$(\genfrac{}{}{0ex}{}{30}{3})=4060$$

   Next, how many different ways are there for the student to draw exactly one question from each topic? Using the product rule for counting, we find that there are 10 · 15 · 5 = 750. Finally, we substitute into the formula for calculating probability:

   $$P(A)=\frac{750}{4060}=18.47\mathrm{\%}$$

   But we want the complementary event, so we just subtract the result from one:

   $$P(A^c)=1-\frac{750}{4060}=\frac{3310}{4060}=81.53\mathrm{\%}$$