1 / 16

Section 4.1 (cont.) Probability Trees

Section 4.1 (cont.) Probability Trees. A Graphical Method for Complicated Probability Problems. Example: Southwest Energy. A Southwest Energy Company pipeline has 3 safety shutoff valves in case the line starts to leak. The valves are designed to operate independently of one another:

ocorbin
Télécharger la présentation

Section 4.1 (cont.) Probability Trees

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. Section 4.1 (cont.) Probability Trees A Graphical Method for Complicated Probability Problems

  2. Example: Southwest Energy • A Southwest Energy Company pipeline has 3 safety shutoff valves in case the line starts to leak. • The valves are designed to operate independently of one another: • 7% chance that valve 1 will fail • 10% chance that valve 2 will fail • 5% chance that valve 3 will fail • If there is a leak in the line, find the following probabilities: • That all three valves operate correctly • That all three valves fail • That only one valve operates correctly • That at least one valve operates correctly

  3. A: P(all three valves operate correctly) P(all three valves work) = .93*.90*.95 = .79515

  4. B: P(all three valves fail) P(all three valves fail) = .07*.10*.05 = .00035

  5. C: P(only one valve operates correctly) P(only one valve operates correctly = P(only V1 works) +P(only V2 works) +P(only V3 works) = .93*.10*.05 +.07*.90*.05 +.07*.10*.95 = .01445

  6. D: P(at least one valve operates correctly) 7 paths P(at least one valve operates correctly = 1 – P(no valves operate correctly) = 1 - .00035 = .99965 1 path

  7. Example: AIDS Testing • V={person has HIV}; CDC: Pr(V)=.006 • P : test outcome is positive (test indicates HIV present) • N : test outcome is negative • clinical reliabilities for a new HIV test: • If a person has the virus, the test result will be positive with probability .999 • If a person does not have the virus, the test result will be negative with probability .990

  8. Question 1 • What is the probability that a randomly selected person will test positive?

  9. Probability Tree Approach • A probability tree is a useful way to visualize this problem and to find the desired probability.

  10. Probability Tree Multiply branch probs clinical reliability clinical reliability

  11. Question 1 Answer • What is the probability that a randomly selected person will test positive? • Pr(P )= .00599 + .00994 = .01593

  12. Question 2 • If your test comes back positive, what is the probability that you have HIV? (Remember: we know that if a person has the virus, the test result will be positive with probability .999; if a person does not have the virus, the test result will be negative with probability .990). • Looks very reliable

  13. Question 2 Answer Answer two sequences of branches lead to positive test; only 1 sequence represented people who have HIV. Pr(person has HIV given that test is positive) =.00599/(.00599+.00994) = .376

  14. Summary • Question 1: • Pr(P ) = .00599 + .00994 = .01593 • Question 2: two sequences of branches lead to positive test; only 1 sequence represented people who have HIV. Pr(person has HIV given that test is positive) =.00599/(.00599+.00994) = .376

  15. Recap • We have a test with very high clinical reliabilities: • If a person has the virus, the test result will be positive with probability .999 • If a person does not have the virus, the test result will be negative with probability .990 • But we have extremely poor performance when the test is positive: Pr(person has HIV given that test is positive) =.376 • In other words, 62.4% of the positives are false positives! Why? • When the characteristic the test is looking for is rare, most positives will be false.

  16. examples 1. P(A)=.3, P(B)=.4; if A and B are mutually exclusive events, then P(AB)=? A B = , P(A B) = 0 2. 15 entries in pie baking contest at state fair. Judge must determine 1st, 2nd, 3rd place winners. How many ways can judge make the awards? 15P3 = 2730

More Related