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7/28/2019 Lecture 05b. Conditional Probability
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Statistics
ST 361: Introduction to Statistics
Introduction to Probability:
Conditional Probability
Kimberly [email protected]
5260 SAS Hall
7/28/2019 Lecture 05b. Conditional Probability
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Statistics
Recall Example: Southwest Energy
A Southwest Energy Company pipeline has 3 safety shutoffvalves 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
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Statistics
C: P(only one valve operates correctly)
P(only one valve operates correctly)= P[(only V1 works) OR (only V2 works) OR (only V3 works)]
= P(V1 works & V2 fails & V3 fails) + P(V1 fails & V2 works & V3
fails) + P(V1 fails & V2 fails & V3 works)
=P(V1 works * V2 fails * V3 fails) + P(V1 fails * V2 works * V3
fails) + P(V1 fails * V2 fails * V3 works)
Find corresponding paths and multiply along branches
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Statistics
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
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Statistics
C: P(only one valve operates correctly)
P(only one valve operates correctly)= P[(only V1 works) OR (only V2 works) OR (only V3 works)]
= P(V1 works & V2 fails & V3 fails) + P(V1 fails & V2 works & V3
fails) + P(V1 fails & V2 fails & V3 works)
=P(V1 works * V2 fails * V3 fails) + P(V1 fails * V2 works * V3
fails) + P(V1 fails * V2 fails * V3 works)
Find corresponding paths and multiply along branches
= (.93*.10*.05) + (.07*.90*.05) + (.07*.10*.95)
= .01445
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Statistics
Conditional Probability
Suppose the probability of A depends onwhether B has occurred
Use Conditional Probability
If A and B are any two events with P(B) >0,the probability of A given B is denoted by
P(A|B) = P(A and B)/P(B)
7/28/2019 Lecture 05b. Conditional Probability
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Recall Example: AIDS Testing
V={person has HIV}; CDC: P(V)=.006
+: test outcome is positive (test indicates
HIV present)
-: 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
7/28/2019 Lecture 05b. Conditional Probability
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Statistics
Recall Example: AIDS Testing
Clinical reliabilities for a new HIV test (Now letswrite as conditional probability):
1. If a person has the virus, the test result will be
positive with probability .999,
i.e. P(+|V) = .999
2. If a person does not have the virus, the test result
will be negative with probability .990,
i.e. P(-|V) = .990
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Statistics
Recall AIDS Testing Question 2
If your test comes back positive, what is theprobability that you have HIV?
P(V|+) = P(V and +)/P(+)
Note by rearrangement of cond prob formula
P(V and +) = P(V)*P(+|V)
We can see this in a probability tree.
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7/28/2019 Lecture 05b. Conditional Probability
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Statistics
Recall AIDS Testing Question 2
If your test comes back positive, what is theprobability that you have HIV?
P(V|+) = P(V and +)/P(+)
Use Law of Total Probability to find P(+)
P(+) = P(+|V)P(V) + P(+|V)P(V)
=(.999)(.006) + (.01)(.994) = .015934
Note: P(+|V) = 1P(-|V) = 1 - .990 = .01
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Statistics
Recall AIDS Testing Question 2
If your test comes back positive, what is theprobability that you have HIV?
P(V|+) = P(V and +)/P(+)
So P(V|+) = .00599/.015934 = .376
Compare with previous result.
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Statistics
Independent Events
Two events are said to be independent if theoccurrence of either event has no effect
whatsoever on the likelihood of occurrence of
the other.
P(A and B) =P(A)P(B)
P(A|B) = P(A)
Can extend to multiple events
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Statistics
Application: Reliability of Systems
If A and B comprise a 2-component series system,then the system works only if each A and B work.
If A and B comprise a 2-component parallel system,
then the system works if either A or B works.
Common assumption: A and B are independent
Idea extends to multiple components.