A Graphical Method for Complicated Probability Problems 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: 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

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

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

C: 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

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

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

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

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

Probability Tree Multiply clinical reliability branch probs

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

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

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

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

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.