El proceso de cambio para la mejora organizacional

4.6.1 The addition rule

From the classical definition of probability we deduced the addition theorem, which serves as the addition axiom for the axiomatic definition of probability. Using this axiom we get the following rule:

Addition rule

If the event A1, A2, ..., An are pair wise mutually exclusive events then P(A1+ A2+ ...+An) = P(A1)+P(A2)+ ...+P(An)

In the case of two non-mutually exclusive events A and B we have the formula P(A+B) = P(A) + P(B) – P(AB).

Example 4.9 In a box there are 10 red balls, 20 blue balls, 10 yellow balls and 10 white balls.

At random draw one ball from the box. Find the probability that this ball is color.

Solution Call the event that the ball drawn is red to be R, is blue B, is yellow Y, is white W and is color C. Then P(R) = 10/(10+20+10+10) = 10/50 = 1/5, P(B) = 20/50 = 2/5, P(Y) = 10/50 = 1/5.

Since C = R+B+Y and the events R, B and Y are mutually exclusive , we have P(C) = P(R+B+Y)

= 1/5+2/5+1/5 = 4/5.

In the preceding section we also got the multiplicative theorem. Below for the purpose of computing probability we recall it.

Multiplicative rule

For any two events A and B from the same field of events there holds the formula P(AB) = P(A) P(B|A) = P(B) P(A|B).

If these events are independent then

P(AB) = P(A) P(B).

Now suppose that the event B may occur together with one and only one of n mutually exclusive events A1, A2, ..., An, that is

B = A1B + A2B + ...+AnB. By Addition rule we have

P(B)= P(A1B)+P(A2B)+ ...+P(AnB).

Further, by Multiplicative rule we get a formula, called the formula of total probability.

Formula of total probability

If the event B may occur together with one and only one of n mutually exclusive events A1, A2, ..., An then

P(B)= P(A1)P(B|A1)+P(A2)P(B|A2)+ ...+P(An)P(B|An).

Example 4.10 There are 5 boxes of lamps:

3 boxes with the content A1: 9 good lamps and 1 defective lamp, 2 boxes with the content A2: 4 good lamps and 2 defective lamp.

At random select one box and from this box draw one lamp. Find the probability that the drawn lamp is defective.

Solution Denote by B the event that the drawn lamp is defective and by the same A1, A2 the events that the box with content A1, A2, respectively, is selected. Since the defective lamp may be drawn from a box of either content A1 or content A2 we have B = A1B + A2B. By the formula of total probability P(B) = P(A1)P(B|A1)+P(A2)P(B|A2).

Since P(A1) = 3/5, P(A2) = 2/5, P(B|A1) = 1/10, P(B|A2) = 2/6 = 1/3 we have

P(B) = 3/5 * 1/10 + 2/5 *1/3 = 29/150 = 0.19.

Thus, the probability that the drawn lamp is defective is 0.19.

Now, under the same assumptions and notations as in the formula of total probability, find the probability of the event Ak, given that the event B has occurred.

According to the Multiplicative rule,

P(AkB) = P(B)P(Ak|B) = P(Ak) P(B|Ak)

using the formula of total probability, we then find the following

Bayes’s Formula

The formula of Bayes is sometimes called the formula for probabilities of hypotheses.

Example 4.11 As in Example 4.10, there are 5 boxes of lamps:

3 boxes with the content A1: 9 good lamps and 1 defective lamp, 2 boxes with the content A2: 4 good lamps and 2 defective lamp.

From one of the boxes, chosen at random, a lamp is withdrawn. It turns out to be a defective (event B). What is the probability, after the experiment has been performed (the aposteriori probability), that the lamp was taken from an box of content A1?

Solution We have calculated P(A1) = 3/5, P(A2) = 2/5, P(B|A1) = 1/10, P(B|A2) = 2/6 = 1/3, P(B)

Thus, the probability that the lamp was taken from an box of content A1, given the experiment has been performed, is equal 0.31.

4.7 Summary

In this chapter we introduced the notion of experiment whose outcomes called the events could not be predicted with certainty in advance. The uncertainty associated with these events was measured by their probabilities. But what is the probability? For answer to this question we briefly discussed approaches to probability and gave the classical and statistical definitions of probability. The classical definition of probability reduces the concept of probability to the concept of equiprobability of simple events. According to the classical definition, the probability of an event A is equal to the number of possible simple events favorable to A divided by the total number of possible events of the experiment. In the time, by the statistical definition the probability of an event is approximated by the proportion of times that A occurs when the experiment is repeated very large number of times.

4.8 Exercises

A, B, C are random events.

1) Explain the meaning of the relations:

a) ABC = A; b) A + B + C = A. 2) Simplify the expressions

a) (A+B)(B+C);

b) (A+B)(A+B);

c) (A+B)(A+B)(A+B).

3) A four-volume work is placed on a shelf in random order. What is the probability that the books are in proper order from right to left or left to right?

4) In a lot consisting of N items, M are defective , n items are selected at random from the lot (n<N). Find the probability that m (m ≤N) of them will be prove to be defective.

5) A quality control inspector examines the articles in a lot consisting of m items of first grade and n items of second grade. A check of the first b articles chosen at random from the lot has shown that all of them are of second grade (b<m). Find the probability that of the next two items selected at random from those remaining at least one proves to be second grade.

6) From a box containing m white balls and n black balls (m>n), one ball after another is drawn at random. What is the probability that at some point the number of white balls and black balls drawn will be the same?

7) Two newly designed data base management systems (DBMS), A and B, are being considered for marketing by a large computer software vendor. To determine whether DBMS users have a preference for one of the two systems, four of the vendor’s customers are randomly selected and given the opportunity to evaluate the performances of each of the two systems. After sufficient testing, each user is asked to state which DBMS gave the better performance (measured in terms of CPU utilization, execution time, and disk access).

a) Count the possible outcomes for this marketing experiment.

b) If DBMS users actually have no preference for one system over the other (i.e., performances of the two systems are identical), what is the probability that all four sampled users prefer system A?

c) If all four customers express their preference for system A, can the software vendor infer that DBMS users in general have a preference for one of the two systems?