Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc. 6.1 Chapter Six Probability

Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc. 6.2 There is a 99% chance we’ll do… …in today’s class !

Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc. 6.3 Random Experiment… …a random experiment is an action or process that leads to one of several possible outcomes. For example: ExperimentOutcomes Flip a coinHeads, Tails Exam MarksNumbers: 0, 1, 2,..., 100 Assembly Timet > 0 seconds Course GradesF, D, C, B, A, A+

Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc. 6.4 Requirements of Probabilities… Given a sample space S = {O 1, O 2, …, O k }, the probabilities assigned to the outcome must satisfy these requirements: (1)The probability of any outcome is between 0 and 1 i.e. 0 ≤ P(O i ) ≤ 1 for each i, and (2)The sum of the probabilities of all the outcomes equals 1 i.e. P(O 1 ) + P(O 2 ) + … + P(O k ) = 1 P(O i ) represents the probability of outcome i

Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc. 6.5 Approaches to Assigning Probabilities… There are three ways to assign a probability, P(O i ), to an outcome, O i, namely: Classical approach: make certain assumptions (such as equally likely, independence) about situation. Relative frequency: assigning probabilities based on experimentation or historical data. Subjective approach: Assigning probabilities based on the assignor’s judgment.

Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc. 6.6 Events & Probabilities… An individual outcome of a sample space is called a simple event, while An event is a collection or set of one or more simple events in a sample space. Roll of a die: S = {1, 2, 3, 4, 5, 6} Simple event: the number “3” will be rolled Event: an even number (one of 2, 4, or 6) will be rolled

Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc. 6.7 Joint, Marginal, Conditional Probability… We study methods to determine probabilities of events that result from combining other events in various ways. There are several types of combinations and relationships between events: Complement of an event Intersection of two events Union of two events

Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc. 6.8 Example 6.1… Why are some mutual fund managers more successful than others? One possible factor is where the manager earned his or her MBA. The following table compares mutual fund performance against the ranking of the school where the fund manager earned their MBA: Venn Diagrams Mutual fund outperforms the market B1 Mutual fund doesn’t outperform the market B2 A1 - Top 20 MBA program A2 - Not top 20 MBA program E.g. This is the probability that a mutual fund outperforms AND the manager was in a top- 20 MBA program; it’s a joint probability.

Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc. 6.9 Conditional Probability… Conditional probability is used to determine how two events are related; that is, we can determine the probability of one event given the occurrence of another related event. Conditional probabilities are written as P(A | B) and read as “the probability of A given B” and is calculated as:

Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc Conditional Probability… We want to calculate P(B 1 | A 1 ) Thus, there is a 27.5% chance that that a fund will outperform the market given that the manager graduated from a top-20 MBA program. B1B1 B2B2 P(A i ) A1A A2A P(B j )

Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc Problems: Contingency Table Probabilities The costs of medical care is increasing very fast. The following table list the joint probabilities associated with drinking beer and liver disease among year old men. If one year old male is randomly selected from the population, calculate the following probabilities. HOW DO YOU GET THIS DATA? * Probability he is a drinker * Probability he does not have liver disease * Probability he is a drinker and has liver disease * Probability he has liver disease given he is a drinker

Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc Problems: Contingency Tables The following table lists the joint probabilities associated with smoking and lung disease among 60 – 65 year old men. HOW DO YOU GET THIS DATA? One 60 – 65 year old man is selected at random. What is the probability of the following events? *He is a Smoker *He has lung disease *He has lung disease given he is a smoker *He is a Smoker “and” has lung disease *He is a Smoker “or” has lung disease

Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc Independence… One of the objectives of calculating conditional probability is to determine whether two events are related. In particular, we would like to know whether they are independent, that is, if the probability of one event is not affected by the occurrence of the other event. Two events A and B are said to be independent if P(A|B) = P(A) or P(B|A) = P(B)

Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc Probability Rules and Trees… We introduce three rules that enable us to calculate the probability of more complex events from the probability of simpler events… The Complement Rule The Multiplication Rule The Addition Rule

Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc Complement Rule… The complement of an event A is the event that occurs when A does not occur. The complement rule gives us the probability of an event NOT occurring. That is: P(A C ) = 1 – P(A) For example, in the simple roll of a die, the probability of the number “1” being rolled is 1/6. The probability that some number other than “1” will be rolled is 1 – 1/6 = 5/6.

Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc Multiplication Rule… The multiplication rule is used to calculate the probability of two events both occurring: P(A and B) P(A and B) = P(A | B)P(B) Likewise, P(A and B) = P(B | A) P(A) If A and B are independent events, then P(A and B) = P(A)P(B)

Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc Addition Rule… Addition rule provides a way to compute the probability of event A or B or both A and B occurring; i.e. the union of A and B. P(A or B) = P(A) + P(B) – P(A and B) Why do we subtract the joint probability P(A and B) from the sum of the probabilities of A and B?

Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc Addition Rule… P(A 1 ) = =.40 P(B 1 ) = =.17 By adding P(A) plus P(B) we add P(A and B) twice. To correct we subtract P(A and B) from P(A) + P(B) B1B1 B2B2 P(A i ) A1A A2A P(B j ) P(A 1 or B 1 ) = P(A) + P(B) –P(A and B) =.46 B1B1 A1A1

Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc Addition Rule for Mutually Excusive Events If and A and B are mutually exclusive the occurrence of one event makes the other one impossible. This means that P(A and B) = 0 The addition rule for mutually exclusive events is P(A or B) = P(A) + P(B)

Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc Problelms: Probabilities Suppose there are 20 students in this class, of whom 8 are operations management majors. Two students are drawn at random [without replacement –that means you do not put the first student selected back in population for the second draw]. Calculate the following probabilities. *Probability both are operations management majors *Probability only one is an operations management major.

Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc Bayes’ Law… Bayes’ Law is named for Thomas Bayes, an eighteenth century mathematician or had your instructor been born before the eighteenth century it would have been called Bakers’ Law In its most basic form, if we know P(B | A), we can apply Bayes’ Law to determine P(A | B) P(B|A) P(A|B) for example …

Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc Breaking News: New test for early detection of cancer has been developed. Let C = event that patient has cancer C c = event that patient does not have cancer + = event that the test indicates a patient has cancer - = event that the test indicates that patient does not have cancer Clinical trials indicate that the test is accurate 95% of the time in detecting cancer for those patients who actually have cancer: P(+/C) =.95 but unfortunately will give a “+” 8% of the time for those patients who are known not to have cancer: P(+/ C c ) =.08 It has also been estimated that approximately 10% of the population have cancer and don’t know it yet: P(C) =.10 You take the test and receive a “+” test results. Should you be worried? P(C/+) = ?????

Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc P(+/C) =.95 P(+/ C c ) =.08 P(C) =.10

Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc Bayesian Terminology… The probabilities P(C) and P(C C ) are called prior probabilities because they are determined prior to the test being conducted. The conditional probability P(C | +) is called a posterior probability (or revised probability), because the prior probability is revised after the results of the cancer test.

Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc Problems: Bayes’ Law The Rapid Test is used to determine whether someone has HIV. The false positive and false negative rates are 0.05 and 0.09 respectively. The doctor just received a positive test results on one of their patients [assumed to be in a low risk group for HIV]. The low risk group is known to have a 6% probability of having HIV. What is the probability that this patient actually has HIV [after they tested positive]. Feel free to use a table to work this problem [there are also other ways to work the problem]

Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc Problem: Bayes’ Law Transplant operations for hearts have the risk that the body may reject the organ. A new test has been developed to detect early warning signs that the body may be rejecting the heart. However, the test is not perfect. When the test is conducted on someone whose heart will be rejected, approximately two out of ten tests will be negative (the test is wrong). When the test is conducted on a person whose heart will not be rejected, 10% will show a positive test result (another incorrect result). Doctors know that in about 50% of heart transplants the body tries to reject the organ. *Suppose the test was performed on my mother and the test is positive (indicating early warning signs of rejection). What is the probability that the body is attempting to reject the heart? *Suppose the test was performed on my mother and the test is negative (indicating no signs of rejection). What is the probability that the body is attempting to reject the heart?