Transcript For Friday, Oct 4 - Arizona State University

Bayes’ Theorem -- Partitions
Given two events, R and S, if
P(R  S) =1
P(R  S) =0
then we say that R and S partition the
sample space.
More than 2 events can partition the
sample space
Bayes’ Theorem
Bayes’ theorem:
Suppose that B1, B2, B3,. . . , Bn partition the
outcomes of an experiment (the sample
space) and that A is another event. For any
number, k, with 1≤k ≤n, we have the
formula:
P( Bk A) 
P( A Bk )  P( Bk )
n
 P( A B )  P( B )
i 1
i
i
Example
A Jar I contains 4 red balls and 2 blue
balls and Jar II contains 3 red balls and
2 blue. The experiment is to chose a
jar at random and from this jar select a
ball and note the color of the ball.
What is the probability that a blue ball
is drawn from Jar II?
Example 2
All tractors made by a company are produced on one of
three assembly lines, named Red, White, and Blue.
The chances that a tractor will not start when it rolls
off of a line are 6%, 11%, and 8% for lines Red,
White, and Blue, respectively. 48% of the company’s
tractors are made on the Red line and 31% are made
on the Blue line.
• What fraction of the company’s tractors that do not
start come from assembly line White?
Example 3
Thirty percent of the population have a
certain disease. Of those that have the
disease, 89% will test positive for the
disease. Of those that do not have the
disease, 5% will test positive.
What is the probability that a person
has the disease, given that they test
positive for the disease?
Example 4
A test attempts to recognize the presence of a
certain disease. Records show that 10% of adults
have a strong form of the disease, 18% have a mild
form of the disease, and the rest have no form of the
disease. A person with a strong form of the disease
has a 15% chance of testing negative. A person with
a mild form has a 10% chance of testing negative. A
person who does not have the disease has a 13%
chance of testing positive, thereby falsely indicating
that the person has the disease.
Example 4 con’t
What is the probability that a person
who has the disease will test positive?
What is the probability that a person
who is disease free will test negative?
What is the probability that a person
tests negative given that he does not
the disease?
Focus on the Project
We would like to find P(S | Y  T  C)
and P(F | Y  T  C) but couldn’t find
them directly because of the records
We can, however, find P(Y  T  C | S)
and P(Y  T  C |F)
Now that we know Bayes’ Theorem we
can find it “indirectly”
Focus on the Project
How can we use the two conditional
probabilites that we can find to help us
get what we want?
Well, using Bayes’ Theorem we have
the following:
P(Y  T  C | S ) P( S )
P( S | Y  T  C ) 
P(Y  T  C | S ) P( S )  P(Y  T  C | F ) P( F )
Focus on the Project
We have a similar formula for the
probability of failure given the 3
conditions
What is our last step to finding the
probability that we want?
Focus on the Project
Now that we have our probabilites, we
can cacluate the expected value and
make a determination of foreclosure or
workout
We are not done yet!
Do the same exact calculations but look
at a range of years instead of one
specific year