Equally Likely Outcomes

Equally Likely Outcomes

The most commonly made assumption for finite sample spaces is that all outcomes are equally likely. This assumption can by no means be taken for granted, however; it must be carefully justified. There are many experiments for which such an assumption is warranted, but there are also many experimental situations in which it would be quite erroneous to make this assumption. For example, it would be quite unrealistic to suppose that it is as likely for no telephone calls to come into a telephone exchange between 1 a.m. and 2 a.m. as between 5 p.m. and 6p.m.
If all the k outcomes are equally likely, it follows that each p; = l/k. For the condition p_i + · · · + p_k=1 becomes kp; = 1 for all i. From this it follows
that for any event A consisting of r outcomes, we have

p(A)=r/k
This method of evaluating P(A) is often stated as follows:

P(A) = (number of ways in which e can occur favorable to A)/(total number of ways in which e can occur)

It is important to realize that the above expression for P(A) is only a consequence
of the assumption that all outcomes are equally likely and is only applicable when
this assumption is fulfilled. It most certainly does not serve as a general definition
of probability.

EXAMPLE 1
A die is tossed and all outcomes are assumed to be equally likely. The event A occurs if and only if a number larger than 4 shows. That is, A = {5, 6}. Hence P(A) = 1/6 +1/6 = 2/6

EXAMPLE 2
 An honest coin is tossed two times. Let A be the event: {one head appears}. In evaluating P(A) one analysis of the problem might be as follows. The sample space is S = {0, 1, 2 }, where each outcome represents the number of heads that occur. Hence P(A) = 1/3! This analysis is obviously incorrect,
since for the sample space considered above, all outcomes are not equally likely. In order to apply the above methods we should consider, instead, the sample space S' = {HH, HT, TH, IT}. In this sample space all outcomes are equally likely, and hence we obtain for the correct solution of our problem, P(A) = 2/4 = 1/2 We could use the sample space S correctly as follows: The outcomes 0 and 2 are equally likely, while the outcome I is twice as likely as either of the others. Hence P(A) = 1/2, which checks with the above answer.
This example illustrates two points. First, we must be quite sure that all outcomes may be assumed to be equally likely before using the above procedure. Second, we can often, by an appropriate choice of the sample space, reduce the problem to one in which all outcomes are equally likely. Whenever possible this should be done, since it usually makes the computation simpler. This point will
be referred to again in subsequent examples.
Quite often the manner in which the experiment is executed determines whether or not the possible outcomes are equally likely. For instance, suppose that we choose one bolt from a box containing three bolts of different sizes. If we choose the bolt by simply reaching into the box and picking the one which we happen to touch first, it is obvious that the largest bolt will have a greater probability of being chosen than the other two. However, by carefully labeling each bolt with a number, writing the number on a tag, and choosing a tag, we can try to ensure that each bolt has in fact the same probability of being chosen. Thus we may have to go to considerable trouble in order to ensure that the mathematical assumption of equally likely outcomes is in fact appropriate. In examples already considered and in many subsequent ones, we are concerned with choosing at random one or more objects from a given collection of objects. Let us define this notion more precisely. Suppose that we have N objects, say a₁, a₂, • • • , a_N.
(a) To choose one object at random from the N objects means that each object has the same probability of being chosen. That is,
                          Prob (choosing a;) = 1/N, i = I, 2, ... , N.
(b) To choose two objects at random from N objects means that each pair of objects (disregarding order) has the same probability of being chosen as any other pair. For example, if we must choose two objects at random from (a₁, a₂, a₃, a₄), then obtaining a₁ and a₂ is just as probable as obtaining a₂ and a₃, etc. This formulation immediately raises the question of how many different pairs there are. For suppose that there are K such pairs. Then the probability of each pair would be l/K. We shall learn shortly how to compute K.
(c) To choose n objects at random (n ≤ N) from the N objects means that each n-tuple, say a_i1, a_i2, • • • , a_in, is as likely to be chosen as any other n-tuple.

Note: We have already suggested above that considerable care must be taken during the experimental procedure to ensure that the mathematical assumption of "choosing at random" is fulfilled.