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Conditional Probability and Bayes’ Theorem

EMIS 7370 STAT 5340. Department of Engineering Management, Information and Systems. Probability and Statistics for Scientists and Engineers. Conditional Probability and Bayes’ Theorem. Dr. Jerrell T. Stracener, SAE Fellow. Leadership in Engineering. Conditional Probability. Basic Concept

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Conditional Probability and Bayes’ Theorem

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  1. EMIS 7370 STAT 5340 Department of Engineering Management, Information and Systems Probability and Statistics for Scientists and Engineers Conditional Probability and Bayes’ Theorem Dr. Jerrell T. Stracener, SAE Fellow Leadership in Engineering

  2. Conditional Probability • Basic Concept • Definition • Reduced Sample Space • Rules • Bayes’ Rule

  3. Conditional Probability If A and B are any events in S and P(B)  0, the conditional probability of A, given that B has occurred is denoted by P(A | B), and Note: The given event is called the reduced sample space.

  4. Conditional Probability: Rules • Rule • If A and B are any events in S, then • P(A  B) = P(A)P(B|A) if P(A)  0 • = P(B)P(A|B) if P(B)  0 • Rule • Two events A and B are independent if • P(A|B) = P(A), • and are dependent otherwise.

  5. Product Rule continued • Rule • If A, B and C are events in S for which P(A) > 0, P(B) > 0, • and P(C) > 0 , then • P(A  B  C) = P(A)P(B|A)P(C|A  B) • Rule • For events A1, A2, ... An in S, can occur, then • P(A1  A2 ...  An) = P(A1)P(A2|A1)P(A3|A1 A2)… • P(An|A1... An-1)

  6. Let {B1, B2, ..., Bn} be a set of events forming a partition of the sample space S, where P(Bi)  0, for i = 1, 2, ... , n. Let A be any event of S such that P(A)  0. Then, for k = 1, 2, ..., n, Bayes Theorem

  7. In a sense, Bayes’ Rule is updating or revising the prior probability P(B) by incorporating the observed information contained within event A into the model.

  8. Example A chain of video stores sells three different brands of VCR’s. Of its VCR sales, 50% are Brand 1 (the least expensive), 30% are Brand 2, and 20% are Brand 3. Each manufacturer offers a 1-year warranty on parts and labor. It is known that 25% of Brand 1’s VCR’s require warranty repair work, whereas the corresponding percentages for Brands 2 and 3 are 20% and 10% respectively. 1. What is the probability that a randomly selected purchaser has bought a Brand 1 VCR that will need repair while under warranty? 2. What is the probability that a randomly selected purchaser has a VCR that will need repair while under warranty? 3. If a customer returns to the store with a VCR that needs warranty repair work, what is the probability that it is a Brand 1 VCR? A Brand 2 VCR? A Brand 3 VCR?

  9. Example: solution The probability that a VCR sold will be Brand 1 is P(B1) = 0.50, the probability it will be Brand 2 is P(B2) = 0.30, and the probability that it will be Brand 3 P(B3) = 0.20. Once a Brand of VCR has been selected R represents that the VCR needs repair R’ represents that the VCR does not need repair The probability that a Brand 1 VCR needs repair, P(R|B1) = 0.25 The probability that a Brand 2 VCR needs repair, P(R|B2) = 0.20 The probability that a Brand 3 VCR needs repair, P(R|B3) = 0.10

  10. Example: solution Outcome Probability B1R 0.125 B1R' 0.375 B2R 0.060 B2R' 0.240 B3R 0.020 B3R' 0.180 1.000 0.25 repair Brand 1 0.75 no repair 0.50 repair 0.20 Brand 2 0.30 0.80 no repair 0.20 repair Brand 3 0.10 no repair 0.90

  11. Example: solution 1. P(B1 and R) = P(B1 R) = P(B1)P(R|B1) = (0.50)(0.25) = 0.125 or, by inspection from the tree diagram P(B1 and R) = 0.125 2. Since R = (B1 R)  (B2 R)  (B3 R), P(R) = P(B1 R) + P(B2 R) + P(B3 R) = 0.125 + 0.060 + 0.020 = 0.205

  12. Example: solution 3. P(B1|R) = P(B1 R)/P(R) = 0.125/0.205 = 0.61 P(B2|R) = P(B2 R)/P(R) = 0.060/0.205 = 0.29 P(B3|R) = P(B2 R)/P(R) = 0.020/0.205 = 0.1 Note: P(B3|R) = 1 - P(B1|R) - P(B2|R) = 0. 10

  13. Example An electrical system consists of four components whose reliability configuration is C A B D The system works if components A and B work and either of the components C or D work. The reliability (probability of working) of each component is 0.9.

  14. Find the probability that: • The entire system works • The component C does not work, given that the entire system works. • Assume that four components work independently.

  15. Solution • In this configuration of the system, A, B, and the subsystem C and D constitute a serial circuit system, whereas the subsystem C and D itself is a parallel circuit system. • Clearly the probability that the entire system works can be calculated as the following: The equalities above hold because of the independence among the four components

  16. = P(the system works) (b) To calculate the conditional probability in this case, notice that P= P(the system works but C does not work ) P(the system works) (0.9)(0.9)(1-0.9)(0.9) = =0.090909 0.8019

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