Overinvolvement Ratios
3.5 B AYES ’ T HEOREM
In this section we introduce an important result that has many applications to manage-ment decision making. Bayes’ theorem provides a way of revising conditional probabil-ities by using available information. It also provides a procedure for determining how probability statements should be adjusted, given additional information.
Reverend Thomas Bayes (1702–1761) developed Bayes’ theorem, originally published in 1763 after his death and again in 1958 (Bayes 1958). Because games of chance—and, hence, probability—were considered to be works of the devil, the results were not widely publicized. Since World War II a major area of statistics and a major area of management decision theory have developed based on the original works of Thomas Bayes. We begin our development with an example problem, followed by a more formal development.
Example 3.23 Drug Screening (Bayes’ Theorem)
A number of amateur and professional sports organizations use routine screening tests to determine if athletes are using performance-enhancing drugs. Jennifer Smith, presi-dent of an amateur athletic union, has asked you to determine the feasibility of screen-ing athletes to determine if they are usscreen-ing performance-enhancscreen-ing drugs. Amateur athletes are increasingly denied participation or deprived of victories if they are found to be users.
As part of the study, you propose the following scenario for analysis. Suppose that 10% of the athletes seeking participation in the athletic union have used performance-enhancing drugs. In addition, suppose that a test is available that correctly identifies an athlete’s drug usage 90% of the time. If an athlete is a drug user, the probability is 0.90 that the athlete is correctly identified by the test as a drug user. Similarly, if the athlete
3.5 Bayes’ Theorem 133 is not a drug user, the probability is 0.90 that the athlete is correctly identified as not using performance-enhancing drugs.
We should note that there are potential ethical and possible legal questions con-cerning the use of these tests. Here, we are concerned about the feasibility of using such a test if one has decided that such a test is proper, given the legal and value systems.
Solution The first step in the analysis is to identify the events in the sample space:
D1: The athlete is a user of performance-enhancing drugs.
D2: The athlete is not a user of performance-enhancing drugs.
The proposed test indicates positive or negative results:
T1: Test says that the athlete is a user of performance-enhancing drugs.
T2: Test says that the athlete is not a user of performance-enhancing drugs.
From the information provided, the following probabilities can be defined:
P1D12 = 0.10 P1D22 = 0.90 P1T1u D12 = 0.90 P1T2u D12 = 0.10 P1T1u D22 = 0.10 P1T2u D22 = 0.90
Using these probabilities, a two-way table, Table 3.11, containing the joint probabilities can be constructed:
P1D1> T12 = P1T1u D12P1D12 = 0.90 * 0.10 = 0.09 P1D1> T22 = P1T2u D12P1D12 = 0.10 * 0.10 = 0.01 P1D2> T12 = P1T1u D22P1D22 = 0.10 * 0.90 = 0.09 P1D2> T22 = P1T2u D22P1D22 = 0.90 * 0.90 = 0.81
Table 3.11 Drug Test Subgroups
T1 (TEST SAYS DRUG USER) T2 (TEST SAYS NOTA DRUG USER) TOTAL
D1 (Drug User) 0.09 0.01 0.10
D2 (Not a Drug User) 0.09 0.81 0.90
Total 0.18 0.82 1.0
From Table 3.11 we can easily determine the conditional probability of a drug user, given that the test says drug user, by dividing the joint probability of D1 and T1 (0.09) by the marginal probability of T1 (0.18):
P1D1u T12 = P1D1> T12
P1T12 = 0.09
0.18 = 0.50
Similarly, the probability of not a drug user, given that the test says not a drug user, can be obtained from the second column:
P1D2u T22 = P1D2> T22
P1T22 = 0.81
0.82 = 0.988
From these results we see that, if the test says an athlete is not a drug user, the probability is very high that the test result is correct. However, if the test says that the athlete is a drug user, the probability is only 0.50 that the athlete is a drug user. This is a large increase over a probability of 0.10 for a randomly selected athlete. However, it is clear that the athletic association would not want to reject athletes merely on the results of this screening test. The potential for unethical actions and serious legal challenge
134 Chapter 3 Elements of Chance: Probability Methods
Given this background, we now provide a more formal development of Bayes’ theo-rem. To begin, we first review the multiplication rule, Equation 3.10:
P1A1> B12 = P1A1u B12P1B12 = P1B1u A12P1A12 Bayes’ theorem follows from this rule.
would be too great. The best strategy would be to use a second independent test to further screen the athlete identified as a drug user by the first test. We stress again that there may be serious ethical and medical concerns if athletes are rejected on the basis of only the first test!
Bayes’ Theorem
Let A1 and B1 be two events. Then Bayes’ theorem states that P1B1u A12 = P1A1u B12P1B12
P1A12 (3.14)
and
P1A1u B12 = P1B1u A12P1A12 P1B12
Solution Steps for Bayes’ Theorem 1.Define the subset events from the problem.
2.Define the probabilities and conditional probabilities for the events defined in Step 1.
3.Compute the complements of the probabilities.
4.Formally state and apply Bayes’ theorem to compute the solution probability.
Here, we apply these solution steps to a problem that requires careful analysis. Con-sider Example 3.23 again. The first task is to identify the events in the sample space. The sample space in Example 3.23 consists of athletes separated into D1, users of performance-enhancing drugs, and D2, nonusers of the drugs. This required an independent diagnosis to determine which athletes were actually drug users and which were not. These events cover the sample space. Athletes were also identified by their test classification, T1, the test indicates drug user, and T2, the test indicates not a drug user. These events also cover the sample space. Note that a test result T1, which indicates drug user, does not guarantee that the person is a drug user.
After the events have been defined, we need to determine the capability of the pro-cedure to predict, using the data. Thus, in Example 3.23 the test was given to a group of known users of performance-enhancing drugs and to a group of known non–drug users.
These test results provided the conditional probabilities of the test results, given either drug user or not. The data were converted to information concerning the quality of the screening test predictions by using Bayes’ theorem. The final task is to express one or more questions in the form of Bayes’ theorem. In Example 3.23 we were interested in the probability that an athlete was a drug user, given that the athlete obtained a positive re-sult on the test. We also realized that it was important to know the probability that an athlete was not a drug user, given a positive test result.
Bayes’ theorem is often expressed in a different, but equivalent, form that uses more detailed information. Let E1, E2, . . . , EK be K mutually exclusive and collectively exhaustive
3.5 Bayes’ Theorem 135 events, and let A1 be some other event. We can find the probability of Ei, given A1, by using Bayes’ theorem:
P1Eiu A12 = P1A1u Ei2P1Ei2 P1A12
The denominator can be expressed in terms of the probabilities of A1, given the various Eis, by using the intersections and the multiplication rule:
P1A12 = P1A1> E12 + P1A1> E22 + g + P1A1> EK2
= P1A1u E12P1E12 + P1A1u E22P1E22 + g + P1A1u EK2P1EK2 These results can be combined to provide a second form of Bayes’ theorem.
Bayes’ Theorem (Alternative Statement)
Let E1, E2, . . . , EK be K mutually exclusive and collectively exhaustive events, and let A be some other event. The conditional probability of Ei, given A, can be expressed as Bayes’ theorem:
P1Eiu A12 = P1A1u Ei2P1Ei2 P1A12
P1Eiu A12 = P1A1u Ei2P1Ei2
P1A1u E12P1E12 + P1A1u E22P1E22 + g + P1A1u EK2P1EK2 (3.15) where
P1A12 = P1A1> E12 + P1A1> E22 + g + P1A1> EK2
= P1A1u E12P1E12 + P1A1u E22P1E22 + g + P1A1u EK2P1EK2
The advantage of this restatement of the theorem lies in the fact that the probabilities it involves are often precisely those that are directly available.
This process for solving conditional probability and>or Bayes’ problems is summa-rized in Example 3.24.
Example 3.24 Automobile Sales Incentive (Bayes’ Theorem)
A car dealership knows from past experience that 10% of the people who come into the showroom and talk to a salesperson will eventually purchase a car. To increase the chances of success, you propose to offer a free dinner with a salesperson for all people who agree to listen to a complete sales presentation. You know that some people will do anything for a free dinner, even if they do not intend to purchase a car. However, some people would rather not spend a dinner with a car salesperson. Thus, you wish to test the effectiveness of this sales promotion incentive. The project is conducted for 6 months, and 40% of the people who purchased cars had a free dinner. In addition, 10%
of the people who did not purchase cars had a free dinner.
The specific questions to be answered are the following:
a. Do people who accept the dinner have a higher probability of purchasing a new car?
b. What is the probability that a person who does not accept a free dinner will pur-chase a car?
Solution