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Bayes’ Theorem -- Partitions Given two events, R and S, if P(R  S) =1 P(R  S) =0 then we say that R and S partition the sample space. More than 2 events.

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Presentation on theme: "Bayes’ Theorem -- Partitions Given two events, R and S, if P(R  S) =1 P(R  S) =0 then we say that R and S partition the sample space. More than 2 events."— Presentation transcript:

1 Bayes’ Theorem -- Partitions Given two events, R and S, if P(R  S) =1 P(R  S) =0 then we say that R and S partition the sample space. More than 2 events can partition the sample space

2 Bayes’ Theorem Bayes’ theorem: Suppose that B 1, B 2, B 3,..., B n partition the outcomes of an experiment (the sample space) and that A is another event. For any number, k, with 1≤k ≤n, we have the formula:

3 Example A Jar I contains 4 red balls and 2 blue balls and Jar II contains 3 red balls and 2 blue. The experiment is to chose a jar at random and from this jar select a ball and note the color of the ball. What is the probability that a blue ball is drawn from Jar II?

4 Example 2 All tractors made by a company are produced on one of three assembly lines, named Red, White, and Blue. The chances that a tractor will not start when it rolls off of a line are 6%, 11%, and 8% for lines Red, White, and Blue, respectively. 48% of the company’s tractors are made on the Red line and 31% are made on the Blue line. What fraction of the company’s tractors that do not start come from assembly line White?

5 Example 3 Thirty percent of the population have a certain disease. Of those that have the disease, 89% will test positive for the disease. Of those that do not have the disease, 5% will test positive. What is the probability that a person has the disease, given that they test positive for the disease?

6 Example 4 A test attempts to recognize the presence of a certain disease. Records show that 10% of adults have a strong form of the disease, 18% have a mild form of the disease, and the rest have no form of the disease. A person with a strong form of the disease has a 15% chance of testing negative. A person with a mild form has a 10% chance of testing negative. A person who does not have the disease has a 13% chance of testing positive, thereby falsely indicating that the person has the disease.

7 Example 4 con’t What is the probability that a person who has the disease will test positive? What is the probability that a person who is disease free will test negative? What is the probability that a person tests negative given that he does not the disease?

8 Focus on the Project We would like to find P(S | Y  T  C) and P(F | Y  T  C) but couldn’t find them directly because of the records We can, however, find P(Y  T  C | S) and P(Y  T  C |F) Now that we know Bayes’ Theorem we can find it “indirectly”

9 Focus on the Project How can we use the two conditional probabilites that we can find to help us get what we want? Well, using Bayes’ Theorem we have the following:

10 Focus on the Project We have a similar formula for the probability of failure given the 3 conditions What is our last step to finding the probability that we want?

11 Focus on the Project Now that we have our probabilites, we can cacluate the expected value and make a determination of foreclosure or workout We are not done yet! Do the same exact calculations but look at a range of years instead of one specific year


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