SECUENCIAS DE IMÁGENES

In this section we define probabilities mathematically as the values of additive set functions. Since the reader is probably most familiar with functions for which the elements of the domain and the range are all numbers, let us first give a very sim-ple examsim-ple where the elements of the domain are sets, while the elements of the range are nonnegative integers, namely, a set function that assigns to each subset A of a finite sample spaceS the number of elements in A, written N(A). Suppose that 500 machine parts are inspected before they are shipped, that I denotes that a ma-chine part is improperly assembled, D denotes that it contains one or more defective

components, and the distribution of the 500 machine parts among the various cate-gories is as shown in the Venn diagram of Figure 3.7.

20 10 5

465 D I

S Figure 3.8

Classification of 500 machine parts

The numbers in Figure 3.8 are N( I∩D ) = 20, N( I ∩D ) = 10, N( I ∩D ) = 5, and N( I∩D ) = 465. Using these values and the fact that the set function is additive (meaning that the number which it assigns to the union of two subsets which have no elements in common is the sum of the numbers assigned to the individual subsets), we can determine the value of N(A) for any other subset A of S. For instance,

N( I )= N( I ∩ D ) + N( I ∩ D ) = 5 + 465 = 470 N( I∪ D) = N( I ∩ D ) + N( I ∩ D ) + N( I ∩ D )

= 20 + 10 + 5 = 35

N( I∪ D) = N( I ∩ D ) + N( I ∩ D ) + N( I ∩ D )

= 10 + 5 + 465 = 480 and

N(D)= N( I ∩ D ) + N( I ∩ D ) = 10 + 5 = 15

Using the concept of an additive set function, let us now explain what we mean by the probability of an event. Given a finite sample spaceS and an event A in S, we define P(A), the probability of A, to be a value of an additive set function that satisfies the following three conditions.

Axiom 1 0≤ P( A ) ≤ 1 for each event A in S.

Axiom 2 P(S ) = 1.

Axiom 3 If A and B are mutually exclusive events inS, then P( A∪ B ) = P( A ) + P( B )

The axioms of probability for a finite sample space

The first axiom states that probabilities are real numbers on the interval from 0 to 1, inclusive. The second axiom states that the sample space as a whole is assigned a probability of 1. SinceS contains all possible outcomes, and one of these must always occur,S is certain to occur. The third axiom states that probability functions must be additive—the probability of the union is the sum of the two probabilities when the two events have no outcomes in common.

Axioms for a mathematical theory require no proof, but if such a theory is to be applied to the physical world, we must show somehow that the axioms are “realistic.”

Thus, let us show that the three postulates are consistent with the classical probability concept and the frequency interpretation.

So far as the first axiom is concerned, fractions of the form s

m, where 0≤ s ≤ m and m is a positive integer, cannot be negative or exceed 1, and the same is true also for the proportion of the time that an event will occur. To show that the second axiom is consistent with the classical probability concept and the frequency interpretation for a long series of repeated experiments, we have only to observe that for the whole sample space

P(S ) = m m = 1

and for the frequency interpretation that some outcome must happen 100% of the time.

So far as the third axiom is concerned, if P( A )= s₁

m, P( B ) = s₂ m

Sec 3.4 The Axioms of Probability 71

and A and B are mutually exclusive, then P( A∪ B ) = s₁+ s2

m = P( A ) + P( B )

Also, if one event occurs in proportion 0.36 or 36% of the time, another event occurs 41% of the time, and the two events are mutually exclusive, then one or the other will occur in proportion 0.36 + 0.41 = 0.77 or 77%.

Before we go any further, it is important to stress the point that the axioms of probability do not tell us how to assign probabilities to the various outcomes of an experiment; they merely restrict the ways in which it can be done. In actual practice, probabilities are assigned on the basis of past experience, on the basis of a careful analysis of conditions underlying the experiment, on the basis of subjective evaluations, or on the basis of assumptions—say, the common assumption that all the outcomes are equiprobable.

EXAMPLE 14 Checking possible assignments of probability

If an experiment has the three possible and mutually exclusive outcomes A, B, and C, check in each case whether the assignment of probabilities is permissible:

(a) P(A)= 1

3, P(B) = 1

3, and P(C) =1 3

(b) P(A)= 0.64, P(B) = 0.38, and P(C) = −0.02 (c) P(A)= 0.35, P(B) = 0.52, and P(C) = 0.26 (d) P(A)= 0.57, P(B) = 0.24, and P(C) = 0.19

Solution (a) The assignment of probabilities is permissible because the values are all on the interval from 0 to 1, and their sum is 1

3+ 1 3+ 1

3 = 1.

(b) The assignment is not permissible because P(C) is negative.

(c) The assignment is not permissible because 0.35 + 0.52 + 0.26 = 1.13, which exceeds 1.

(d) The assignment is permissible because the values are all on the interval from 0

to 1 and their sum is 0.57 + 0.24 + 0.19 = 1. ^j

The approach in the last example extends to any experiment where the sample spaceS is discrete so the outcomes can be arranged in a sequence. An amount of probability p_iis assigned to the ith outcome, where

0≤ pi and 

all outcomes inS p_i= 1

and then the probability of any event A is defined as

P(A)= 

all outcomes in A p_i

When probability is assigned in this manner, the axioms of probability are always satisfied.

Intuitively, we can think of the scientist as starting with a unit amount of clay (probability) and placing a proportion p₁ on the first outcome, p₂ on the second outcome, and so on. Some outcomes can be assigned a large amount and others lesser amounts. The total unit amount of clay (probability) is assigned to the outcomes in the sample space. Then, an event A is assigned the total of all the clay (probability) assigned to each outcome in A.