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In this article some of the most important theorems and results of the classical interpretation of probability are presented. No attempt is made at mathematical rigour, but instead results are presented through diagrammatic and simple plausibility arguments. |
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Classical Interpretation of Probability
Before we dive into the basic results and theorems of probability theory, it is worth considering what we mean by “classical” probability theory. In essence, the classical interpretation of probability is concerned with undertaking an “experiment” that has a number of possible outcomes, one of which must occur as the outcome of the experiment. For example, when a die is rolled, the outcome of rolling the die must be that either the number 1, 2, 3, 4, 5 or 6 is on the uppermost face of the die. In this case the “experiment” is rolling the die, and the numbers 1 through to 6 form the possible outcomes of the experiment.
At the heart of the classical interpretation of probability is the assumption that each of the possible outcomes of an experiment is equally likely to occur. With this assumption, if there are N possible outcomes of an experiment, then the probability P of any particular outcome O is given by
Thus, the probability of throwing a die and getting a “1” is equal to 1/6, since there are six possible outcomes, each of which is considered to be equally likely to occur.
The assumption of equally likely outcomes lies at the heart of the classical interpretation of probability. However, this is not the only interpretation of probability, and indeed there are difficulties with the classical interpretation. In particular, the classical interpretation can not be used to address questions for which the possible outcomes are not considered equally likely. Examples of this are questions of a subjective nature (e.g. the probability that an acquaintance will buy a new car in the next year), and as such they require a different approach. This is provided by the so-called subjective interpretation of probability. However, this is not considered further in these articles. We will be concerned only with the classical interpretation, under the assumption of equally likely outcomes.
Basic Definitions
In order to proceed, we need to make some basic definitions with respect to an experiment that we undertake, and which leads to a number of possible outcomes.
The Sample Space – symbol S – denotes the set of all possible outcomes of the experiment. For example, when a die is rolled and the uppermost number is noted, the sample space S consists of six elements, written as follows:
An Event – symbol A – represents the subset of outcomes in S that satisfy some criterion for success in the experiment. For example, if a particular event A is that the number obtained by rolling a die is even, then the event A is
Since some events are impossible (e.g. rolling a “7” on a six-sided die), an event may contain no elements at all. A set with no elements is referred to as an Empty Set.
Now consider two events A and B. Two important concepts are the union and intersection of A and B.
The Union of events A and B is the event that contains all the outcomes of A alone, all the outcomes of B alone, and all the outcomes that belong to both A and B. The union of events A and B is written as
and can be displayed diagrammatically as follows:
In this figure, the rectangle represents the total sample space S, and the ellipses represent the subsets of S covered by the events A and B. The event that represents the union of A and B is shaded blue in the above figure.
The Intersection of events A and B is the event that contains all the outcomes that belong to both A and B. The intersection of events A and B is written as
and can be represented diagrammatically as follows:
The event that represents the intersection of A and B is shaded blue in the above figure.
Finally, the Complement of an event A – written AC – is the event that contains all the elements of the sample space S that do not belong to A. The complement of an event A can be represented diagrammatically as follows:
The event that represents the complement of A is shaded blue in the above figure.
As an example, consider the experiment of rolling a die, and let the event A be that the number rolled is even, and let the event B be that the number rolled is three or less. Then:
With these basics, we can begin to think about computing event probabilities.
Probabilities of Events
Consider a sample space that contains N outcomes, such that the probability of any one of the outcomes is 1/N. Now consider an event A that is a subset of S and contains m outcomes. The probability of the event A is given by
(A)
As an example, consider rolling a die and let the event A be that the number rolled is even. Since there are three outcomes in A (2,4 and 6) and six outcomes in S, it follows that m = 3 and N = 6, so that P(A) = 3/6 = 1/2.
Now, two basic axioms of probability theory are as follows:
Although these results are considered to be axioms, they seem plausible on the basis of result (A). The first result follows because m ≥ 0 and the second follows because the sample space S contains a total of N elements, so that m = N in this case. In the first expression above, if A is the empty set then P(A) = 0.
Further important results are as follows:
The last of these results can be visualised by considering the diagram above for the union of two events. The probability is clearly related to the sum of the probabilities of the two events, but the probability of the intersection of the two events needs to be subtracted out, since the intersection is counted twice in the sum P(A) + P(B).
If the events A and B are independent of one another, then
We will return to the notion of one event being conditional on another in a later article on Bayes’ theorem.
Example: Consider the rolling of two dice. Compute the probability that (a) the sum of the rolled numbers is 7 (b) both dice show the same number (c) both dice show an even number.
In this case, the sample space S contains 36 outcomes. This is because the first die can give six values, as can the second die. The sample space is therefore composed of the following 36 pairs of possible outcomes:
To evaluate the required probabilities, we define the appropriate event A and hence work out how many of the 36 outcomes are present in A.
(a) Let the event A be the set of outcomes for which the sum of the numbers is 7. The event A is equal to
and hence contains 6 elements. The required probability is therefore 6/36 = 1/6.
(b) Let the event A be the set of outcomes for which the numbers on the dice are the same. In this case, the event A is equal to
and so the probability of A is once again 6/36 = 1/6.
(c) We could once again define an event A and count all the possible outcomes for which both numbers are even. However, an alternative way to solve the problem is to note that the two throws of the dice are independent. Hence we can define events A and B to be equal to the outcomes that the first die gives an even number, and the second die gives an even number, respectively. Hence P(A) = 1/2 and P(B) = 1/2. Thus the required probability is
In summary, and in preparation for the next article, it can be seen that one of the major techniques to master in probability theory is to count the number of successful outcomes for some given event, and also to count the total number of outcomes in the overall sample space.