CS4600  Introduction to Intelligent Systems
Fall 2000
Homework 9  Sample Solution
Problem 1
Of the entire population, 2% has a certain disease X.
A test Y, which indicates whether or not a
person has the disease, is not 100% accurate. If a person has the disease, there is a 6% chance that
it will go undetected by the test.
However, there is also a 9% chance of "false alarm" (meaning
that the person does not have the disease but the test indicates otherwise). A person Z takes a test
which later comes out positive (meaning that the test says he has the disease).
What is the proba
bility of this person having the disease in reality?
Let D be"having the disease"
+ be "test positive"
We are given the following information:
P(D) = 0.02
which implies P(not D) = 0.98
P(not +  D) = 0.06
which implies P(+  D) = 0.94
P(+  not D) = 0.09
First, we compute P(+)
= P(+ AND D) + P(+ AND (not D))
= P(+  D) P(D) + P(+  not D) P(not D)
= 0.94 x 0.02 + 0.09 x 0.98
= 0.107
We would like to know P(D  +)
= P(+  D) x P(D) / P(+)
= 0.94 x 0.02 / 0.107
~= 0.1757
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View Full DocumentProblem 2
Consider the following Bayesian network:
a) Are D and E necessarily independent given evidence about both A and B?
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 Fall '08
 SvenKoenig
 Following, 6%, 2%, Bayesian probability, Bayesian network, independent given evidence

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