1 / 11

6.3 Bayes Theorem

6.3 Bayes Theorem. We can use Bayes Theorem…. …when we know some conditional probabilities, but wish to know others. For example: We know P(test positive|have disease), and we wish to know P(have disease|test positive ). Ex. 1 (book ex 2- p. 419).

jerome
Download Presentation

6.3 Bayes Theorem

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. 6.3 Bayes Theorem

  2. We can use Bayes Theorem… • …when we know some conditional probabilities, but wish to know others. • For example: We know P(test positive|have disease), and we wish to know P(have disease|test positive)

  3. Ex. 1 (book ex 2- p. 419) • Suppose that one person in 100,000 has a particular rare disease for which there is a fairly accurate diagnostic test. This test is correct 99% of the times for someone who has the disease and 99.5%% of the time for someone who does not.

  4. Define E, F, E’, F’ Let F=event one has the disease E=event one tests positive We know that P(F) = 1/100,000 = .00001 P(E|F)= P(positive|disease) = .99 and P(E’ |F’ ) = P(negative| don’t have disease) = .995 Determine P(F|E) = P(has disease|test positive) = ___ and P ( F’ |E’ )= P(does not have disease |test negative)= ___

  5. Draw tree diagram starting with F, F’

  6. Find P(F|E) and P(F’| E’) • P(F|E) = = 0.002 • P(F’ | E’)= = 0.9999999

  7. Ex. 2: F=studied for final, E=passed class Assume: P(F) = P(studied)=.8 P(E|F)= P(passed|studied)=.9 and P(E|F ’ ) = P(passed|didn’t study)=.2 Find P(F|E) = P(studied|passed)= ___ P (F’ | E’ )= P(didn’t study | failed) = ___

  8. Tree diagram, starting with F, F’

  9. Spam filters Ex. 3: Spam filters Idea: spam has words like “offer”, “special”, “opportunity”, “Rolex”, … Non-spam has words like “mom”, “lunch”,… False negatives: when we fail to detect spam False positives: when non-spam is seen as spam Let S=spam E=has a certain word Assume P(S)=0.5

  10. Tree diagram starting with S, S’

  11. Given a message says “Rolex”, find probability it is spam Consider that “Rolex” occurs in 250/2000= .125 spam messages and in 5/1000=.005 non-spam messages. Assume P(S)=0.5 Ex: P(S|uses word “Rolex”) = = .962

More Related