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There are three important concepts associated with a random experiment: 'outcome', 'sample space', and 'event'. Two examples of experiments will be used to familiarize you with these terms:
An outcome of an experiment is a single result of that experiment.
The sample space of an experiment is the complete set of possible outcomes of the experiment.
An event is any set of outcomes of an experiment. (You can think of it as 'the outcomes we are looking for' or favourable outcomes.)
A Venn diagram can be used to show the relationship between the possible outcomes of a random experiment and the sample space. The Venn diagram in [link] shows the difference between the universal set, a sample space and events and outcomes as subsets of the sample space.
Venn diagrams can also be used to indicate the union and intersection between events in a sample space ( [link] ).
In a box there are pieces of paper with the numbers from 1 to 9 written on them. A piece of paper is drawn from the box and the number on it is noted. Let denote the sample space, let denote the event 'drawing a prime number', and let denote the event 'drawing an even number'. Using appropriate notation, in how many ways is it possible to draw: i) any number? ii) a prime number? iii) an even number? iv) a number that is either prime or even? v) a number that is both prime and even?
The union of and is the set of all elements in or in (or in both). . is also written .
The intersection of and is the set of all elements in both and . . is also written as .
We use to refer to the number of elements in a set , for the number of elements in , etc.
A final notion that is important to understand is the notion of complement . Just as in geometry when two angles were called 'complementary' if they added up to 180 degrees, (the two angles 'complement' each other to make a 'whole' straight line), the complement of a set of outcomes , usually denoted is the set of all outcomes in the sample space but not in (i.e., complements to form the entire sample space). Thus, by definition, is always true. So in the Exercise above, the complement of P (i.e. P^c) = {1,4,6,8,9}, while E^c = {1,3,5,7,9}. So n(P^c) = n(E^c) = 5
In theory, it is very easy to calculate complements, since the number of elements in 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 9 possible 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.
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