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: a.That all three valves operate correctly b.That all three valves fail c.That only one valve operates correctly d.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) P(at least one valve operates correctly = 1 – P(no valves operate correctly) = 1 -.00035 =.99965 7 paths 1 path

7. Example: AIDS Testing zV={person has HIV}; CDC: Pr(V)=.006 zP : test outcome is positive (test indicates HIV present) zN : test outcome is negative zclinical reliabilities for a new HIV test: 1.If a person has the virus, the test result will be positive with probability.999 2.If a person does not have the virus, the test result will be negative with probability.990

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

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

10. Probability Tree Multiply branch probs clinical reliability

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

12. Question 2 zIf 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). zLooks 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 zQuestion 1: zPr(P ) =.00599 +.00994 =.01593 zQuestion 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 zWe have a test with very high clinical reliabilities: 1.If a person has the virus, the test result will be positive with probability.999 2.If a person does not have the virus, the test result will be negative with probability.990 zBut we have extremely poor performance when the test is positive: Pr(person has HIV given that test is positive) =.376 zIn other words, 62.4% of the positives are false positives! Why? zWhen 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 1 st, 2 nd, 3 rd place winners. How many ways can judge make the awards? 15 P 3 = 2730