Advanced Math/Conditional probability
Expert: Paul Klarreich - 5/16/2008
QuestionA test correctly identifies a disease in 95% of people who have it. It correctly identifies no disease in 94% of people who do not have it. In the population, 3% of the people have the disease. What is the probability that you have the disease if you tested positive?
Answer
Questioner: Wendy
Category: Advanced Math
Private: No
Subject: Probability and Statistics
Question: A test correctly identifies a disease in 95% of people who have it. It correctly identifies no disease in 94% of people who do not have it. In the population, 3% of the people have the disease. What is the probability that you have the disease if you tested positive?
=================================================
Hi, Wendy,
Let D and T represent these sets:
D = A person has the disease.
D' = A person is well.
T = the test is positive for the disease.
T'= the test is negative for the disease.
You are told:
p(D) = 0.03
p(T, if D) = 0.95
p(T', if D') = 0.94
There are four sections: [See attached two-ring diagram.]
p(D and T) : p(#3)
p(D and T') : p(#2)
p(D' and T) : p(#4)
p(D' and T') : p(#1)
which add up to 1.000
...........................................
A test correctly identifies a disease in 95% of people who have it.
That means #3 is 95% of #2+#3
But #2 + #3 = 0.03 = % of people with disease.
And #1 + #4 = 0.97 = % of people without disease.
So #3 = 0.03 * 0.95 = 0.0285 = % of people with disease and positive test.
And #2 = 0.0015 = % of people with disease and NEGAtive test.
..................................
It correctly identifies no disease in 94% of people who do not have it.
That means #1 is 94% of #1 + #4
And #1 + #4 = 0.97 = % of people without disease.
So #1 = 0.97 * 0.94 = 0.9118 = % of people withOUT disease and negative test.
And #4 = 0.97 * 0.06 = 0.0582 = % of people with disease and positive test.
Summary:
#1 = 0.9118
#2 = 0.0015
#3 = 0.0285
#4 = 0.0582
-------------
1.0000
Now we are ready for your question:
What is the probability that you have the disease if you tested positive?
That means: what is p(D and T, given T)
That is p(D and T)/p(T) = p(#3)/[p(#3) + p(#4)]
= 0.0285/[0.0285 + 0.0582]
= 0.0285/[0.0867] = 0.3287