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:• 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 correctlyb. That all three valves failc. That only one valve operates correctlyd. 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)

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

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

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

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:

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

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 TreeMultiply

branch probsclinical reliability

clinical reliability

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

Answertwo 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 = .01593Question 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:1. If a person has the virus, the test result will be

positive with probability .9992. 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.

examples1. P(A)=.3, P(B)=.4; if A and B are

mutually exclusive events, then P(AB)=?

A B = , P(A B) = 02. 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