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A final notion that is important to understand is the notion of complement . Just as in geometry when two angles are called 'complementary' if they added upto 90 degrees, (the two angles 'complement' each other to make a right angle), the complement of a set of outcomes $A$ is the set of all outcomes in the sample space but not in $A$ . It is usually denoted ${A}^{\text{'}}$ (or sometimes ${A}^{C}$ ), and called 'the complement of $A$ ' or simply ' $A$ -complement'. Because it refers to the set of everything outside of $A$ , it is also often referred to as 'not- $A$ '. Thus, by definition, if $S$ denotes the entire sample space of possible outcomes, and $A$ is any subset of outcomes that we are interested in (i.e. an event ), then $A\cup {A}^{\text{'}}=S$ is always true, (i.e. ${A}^{\text{'}}$ complements $A$ to form the entire sample space). So in the exercise above, ${P}^{\text{'}}=\{1,4,6,8,9\}$ , while ${E}^{\text{'}}=\{1,3,5,7,9\}$ . So $n\left({P}^{\text{'}}\right)=n\left({E}^{\text{'}}\right)=5$
The probability of a complementary event refers to the probability associated with the complement of an event, i.e. the probability that something other than the event in question will occur. For example, if $P\left(A\right)=0,25$ , then the probability of $A$ not occurring is the probability associated with all other events in $S$ occurring less the probability of $A$ occurring.
In theory, it is very easy to calculate complements, since the number of elementsin the complement of a set is just the total number of outcomes in the sample space minus the outcomes in that set (in the example above, there were 9possible outcomes in the sample space, and 4 possible outcomes in each of the sets we were interested in, thus both complements contained 9-4 = 5 elements).Similarly, it is easy to calculate probabilities of complements of events since they are simply the total probability (e.g. 1 if our total measure is 1) minus the probability of the event in question. So,
Sometimes it is much easier to decide the probability of an event occurring by instead calculating the probability that the complementary event will NOT occur. For example, if the process in question was rolling three dice, and the event we were interested in was that at least one of the faces is a one, it is definitely much easier to figure out the probability that not getting a one will not occur than to try to figure out all the possible combinations of three dice where a one does occur!
If you throw two dice, one red and one blue, what is the probability that at least one of them will be a six?
To solve that kind of question, work out the probability that there will be no six.
The probability that the red dice will not be a six is 5/6, and that the blue one will not be a six is also 5/6.
So the probability that neither will be a six is $5/6\times 5/6=25/36$ .
So the probability that at least one will be a six is $1-25/36=11/36$ .
A bag contains three red balls, five white balls, two green balls and four blue balls:
1. Calculate the probability that a red ball will be drawn from the bag.
2. Calculate the probability that a ball which is not red will be drawn
Let R be the event that a red ball is drawn:
$\therefore $ P(R') = 1 - P(R) = 1 -3/14 = 11/14
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