1. A Graphical Method for Complicated Probability Problems
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 =

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

5. 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 =

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 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 clinical reliability branch probs

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

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/( ) = .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/( ) = .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.