Lecture 05b. Conditional Probability

7/28/2019 Lecture 05b. Conditional Probability

1/14

Statistics

ST 361: Introduction to Statistics

Introduction to Probability:

Conditional Probability

Kimberly [email protected]

5260 SAS Hall


2/14

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


3/14

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


4/14

Statistics


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


5/14

Statistics


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


6/14

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)


7/14Statistics

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


8/14

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


9/14

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.


10/14


12/14

Statistics



P(V|+) = P(V and +)/P(+)

So P(V|+) = .00599/.015934 = .376

Compare with previous result.


13/14

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


14/14

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.

Lecture 05b. Conditional Probability

Documents

Transcript of Lecture 05b. Conditional Probability