1. Chapter 4, part E Download this file. Download this file.

2. VI. Bayes’ Theorem The Reverend Thomas Bayes (1702-1761) is credited with the original work. The theorem is used as a means for updating probability calculations when new information becomes available.Thomas Bayes Why must I stand for this?

3. The idea. The idea is that you have prior probabilities for events, you observe something which gives you more information than you had before, you apply Bayes’ theorem and compute modified (posterior) probabilities. Did you say “posterior”? Oooh, behave!

4. An example. As a manufacturer, you receive parts from 2 suppliers (A1 and A2). 65% of your parts come from supplier #1. P(A1)=.65 35% of your parts come from supplier #2. P(A2)=.35

5. Historically, 2% of the parts from A1 are defective, and 5% of the parts from A2 are defective. Events: Good part (G), defective part (B) Thus: P(G|A1) =.98P(B|A1) =.02 P(G|A2) =.95P(B|A2) =.05

6. A probability tree A1 (.65) A2 (.35) G (.98) B (.02) B (.05) G (.95) Experimental Outcomes (A1,G) (A1,B) (A2,G) (A2,B) Ouch!

7. Probability of each outcome You just multiply your way down the probability tree! Do you recognize the multiplication law in here? P(A1,G)=P(A1  G)=P(A1)*P(G  A1)=.65*.98=.6370 Likewise you can calculate: P(A1,B)=.65*.02=.0130 P(A2,G)=.35*.95=.3325 P(A2,B)=.35*.05=.0175 These are all prior probabilities.

8. Posterior probabilities Now suppose your production manager tells you that they’ve found a defective part (B). What’s the probability that it came from supplier A1? P(A1|B)=P(A1)*P(B|A1)/P(B) The numerator is specific enough, but the denominator can happen two different ways: (A1,B) and (A2,B) Not my problem

9. Bayes’ Theorem In our example we have two events that satisfy the condition in the denominator, so we substitute for P(B) and get Bayes’ Theorem for a 2-event case. P(A1|B)= P(A1)*P(B|A1)/{P(A1)*P(B|A1)+P(A2)* P(B|A2) Bring specific questions with you to class and we’ll look at more practice with Bayes’ Theorem.