Slide 1 Math 1520, Lecture 21

Transcription

1 Slide 1 Math 1520, Lecture 21 This lecture is concerned with a posteriori probability, which is the probability that a previous event had occurred given the outcome of a later event.

2 Slide 2 Conditional Probability Definition: Suppose A and B are events. The probability that event B happens (or happened) given that the event A occurrs, is called the conditional probability of B given A and is written using the notation: P (B A) There are two scenarios in which the conditional probability P (B A) may need to be calculated 1. In a priori probability, the event A has occurred before event B 2. In a posteriori probability, the event A has occurred after event B For examples of each type, consider drawing two cards (without replacement) from a well-shuffled deck. 1. (a priori) What is the probability that the second card drawn is a King given that the first card was a King? 2. (a posteriori) What is the probability that the first card drawn was a King given that the second card was a King?

3 Slide 3 Example Suppose a manufacturer makes widgets in three plants, A,B,C. The production schedule has 30% of the widgets made in plant A, 60% in plant B and 10% in plant C. Testing has shown that 2% of the widgets from plant A are defective, and 4% of the widgets from plant B are defective and 1% of the widgets from plant C are defective. All the widgets are combined for sale. 1. Define events, A, B, C, D and their complements. 2. Draw a tree diagram to represent this situation. 3. For each of the following questions: Write notation for the probability that is desired. Determine the probability. 1. (a priori) If a widget from plant A is drawn at random. What is the probability that it is defective? 2. A widget is drawn at random from all 3 lots. What is the probability that it is defective and from plant A? 3. A widget is drawn at random from all 3 lots. What is the probability that it is defective? 4. (a posteriori) A widget is drawn at random from all 3 lots and found to be defective. What is the probability that it is from plant A?

4 Slide 4 iclicker A disease occurs in 1% of the population. A blood test can be given to test for the disease. If a person has the disease, the test indicates the presence of the disease in 95% of the cases. If a person does not have the disease, the test indicates the presence of the disease in 5% of the cases (5% false positives). Question Suppose a person has the disease. What is the probability that the blood test will detect it? A. 95 B..95 C D

5 Answer to Question Suppose a person has the disease. What is the probability that the blood test will detect it? A. 95 B..95 is the correct answer. C D

6 Slide 5 iclicker A disease occurs in 1% of the population. A blood test can be given to test for the disease. If a person has the disease, the test indicates the presence of the disease in 95% of the cases. If a person does not have the disease, the test indicates the presence of the disease in 5% of the cases (5% false positives). Question A person takes the blood test. What is the probability that the person has the disease and the blood test detects it? A. 95 B..95 C D

7 Answer to Question A person takes the blood test. What is the probability that the person has the disease and the blood test detects it? A. 95 B..95 C is the correct answer. D The answer simplifies to This is just slightly less the probability of having the disease.

8 Slide 6 iclicker A disease occurs in 1% of the population. A blood test can be given to test for the disease. If a person has the disease, the test indicates the presence of the disease in 95% of the cases. If a person does not have the disease, the test indicates the presence of the disease in 5% of the cases (5% false positives). Question A person takes the blood test. What is the probability that the test is positive, indicating a disease? A. B C D

9 Answer to Question A person takes the blood test. What is the probability that the test is positive, indicating a disease? A. is the correct answer. B C D The answer simplifies to.0590, so this happens a little over 5% of the time.

10 Slide 7 iclicker A disease occurs in 1% of the population. A blood test can be given to test for the disease. If a person has the disease, the test indicates the presence of the disease in 95% of the cases. If a person does not have the disease, the test indicates the presence of the disease in 5% of the cases (5% false positives). Question A person takes the blood test and the test is positive indicating the presence of the disease. What is the probability that the person actually has the disease? A B C D.

11 Answer to Question A person takes the blood test and the test is positive indicating the presence of the disease. What is the probability that the person actually has the disease? A B C D. is the correct answer. The answer simplifies to approximately.161, which might be lower than one original expected for the test. It is so low because the number of false positives is so large relative to the number of true positives. Since there are so few people with the disease, even a false positive rate of 5% leads to this situation.

12 Slide 8 Bayes Theorem The technique we have been using to calculate a-posteriori probability is encapsulated in a theorem called Bayes Theorem. Here is one case of Bayes Theorem. Theorem Let A 1, A 2, A 3 be a partition of a sample space S and let E be an event such that P (E) 0 and P (A 1 ) 0, P (A 2 ) 0, P (A 3 ) 0. The the a-posteriori probability P (A 1 E) is given by P (A 1 E) = P (A 1 ) P (E A 1 ) P (A 1 ) P (E A 1 ) + P (A 2 ) P (E A 2 ) + P (A 3 ) P (E A 3 ) 1. By a partition of S we mean that A 1, A 2, A 3 are mutually exclusive and A 1 A 2 A 3 = S. 2. Illustrate Bayes Theorem with a tree diagram. 3. The same Theorem holds true given any number of A s.

13 Slide 9 iclicker Question Let A 1, A 2 be a partition of a sample space S and let E be an event such that P (E) 0 and P (A 1 ) 0, P (A 2 ) 0. The the a-posteriori probability P (A 2 E) is given by : A. P (A 2 E) = B. P (A 2 E) = C. P (A 2 E) = D. P (A 2 E) = P (A 1 ) P (E A 1 ) P (A 1 ) P (E A 1 ) + P (A 2 ) P (E A 2 ) + P (A 3 ) P (E A 3 ) P (A 2 ) P (E A 2 ) P (A 1 ) P (E A 1 ) + P (A 2 ) P (E A 2 ) + P (A 3 ) P (E A 3 ) P (A 1 ) P (E A 1 ) P (A 1 ) P (E A 1 ) + P (A 2 ) P (E A 2 ) P (A 2 ) P (E A 2 ) P (A 1 ) P (E A 1 ) + P (A 2 ) P (E A 2 )

14 Answer to Question Let A 1, A 2 be a partition of a sample space S and let E be an event such that P (E) 0 and P (A 1 ) 0, P (A 2 ) 0. The the a-posteriori probability P (A 2 E) is given by : A. P (A 2 E) = B. P (A 2 E) = C. P (A 2 E) = D. P (A 2 E) = answer. P (A 1 ) P (E A 1 ) P (A 1 ) P (E A 1 ) + P (A 2 ) P (E A 2 ) + P (A 3 ) P (E A 3 ) P (A 2 ) P (E A 2 ) P (A 1 ) P (E A 1 ) + P (A 2 ) P (E A 2 ) + P (A 3 ) P (E A 3 ) P (A 1 ) P (E A 1 ) P (A 1 ) P (E A 1 ) + P (A 2 ) P (E A 2 ) P (A 2 ) P (E A 2 ) P (A 1 ) P (E A 1 ) + P (A 2 ) P (E A 2 ) is the correct