Question

Use Bayesian analysis to determine the following. A disorder has a test that gives no false negatives, but will produce a false positive 1 in 5,000 tests. This disorder only affects 1/200,000 people. What is the probability that a person has the rare disease if they receive 1 positive test, 2 positive tests in a row, and 3 positive tests in a row? Please show all work.

Answer 1

We are given here that:

P( + | not disorder ) = 1/5000 = 0.0002
P( - | not disorder ) = 0.9998

P( - | disorder ) = 0
P( + | disorder) = 1

Also, we are given that: P( disorder ) = 1/200000 = 0.000005,
Therefore P(no disorder) = 0.999995

P(+) = P( + | not disorder )P(no disorder) + P( + | disorder)P(disorder)
P(+) = 0.0002*0.999995 + 1*0.000005 = 0.000204999

a) Probability of having disorder given there is a + test:
This is computed using bayes theorem as:

P( disorder | +) = P(+ | disorder)P(disorder) / P(+) = 1*0.000005 / 0.000204999 = 0.024390
Therefore 0.024390 is the required probability here.

b) P( ++ | disorder ) = 1
P( ++ | no disorder ) = 0.0002*0.0002 = 0.00000004

P(++) = P( ++ | not disorder )P(no disorder) + P( ++ | disorder)P(disorder)
P(++) = 0.00000004*0.999995 + 1*0.000005 = 0.0000050399998

Therefore, P(disorder | ++) = P( ++ | disorder)P(disorder) / P(++)
P(disorder | ++) = 1*0.000005/0.0000050399998 = 0.992064

Therefore 0.992064 is the required probability here.

c) P( +++ | disorder ) = 1
P( +++ | no disorder ) = 0.0002*0.0002*0.0002 = 0.000000000008

P(+++) = P( +++ | not disorder )P(no disorder) + P( +++ | disorder)P(disorder)
P(+++) = 0.000000000008*0.999995 + 1*0.000005 = 0.000005000008

Therefore, P(disorder | +++) = P( +++ | disorder)P(disorder) / P(+++)
P(disorder | +++) = 1*0.000005/0.000005000008 = 0.999998

Therefore 0.999998 is the required probability here.