In: Statistics and Probability
PROBABILITY QUESTION
Assume that, as a clinic worker, you are asked to conduct lab
tests for diagnosis of a disease. From ex-
periments, it is known that any person in the population is either
has the disease (positive), or has not
(negative), i.e. there is no carrier. Over the entire population of
people only 0.005 have this disease and the
lab test returns a correct positive result in only 97% of the cases
in which the disease is actually present
and a correct negative result in only 98% of the cases in which the
disease is not present. The state of the
patient is presented with random variable S and the test results
are presented with random variable T.
Question 2.1 [5 pts] Obtain the probabilities P(S = disease),
P(S = healthy), P(T = positive j S =
disease), P(T = negative j S = disease), P(T = positive j S =
healthy) and P(T = negative j S = healthy).
Question 2.2 [5 pts] Suppose a new patient comes to the clinic
and the test returns a positive result.
Show whether the patient should be diagnosed as having the disease
or not.
1) P (Disease) = 0.005 P (Healthy) = 1- 0.005 = 0.995
P (Positive / Disease) = 0.97 (given) which means P (Negative / Disease) = 0.03
P (Negative / Healthy) = 0.98 which means P (positive / healthy) = 0.02
P (Test is positive j Disease) = P( Test is positive / Disease) * p (Disease) = 0.97* 0.005 = 0.00485 (by definition of conditional probability)
P (Test is negative j Disease) = P(Test is negative / Disease) * p (Disease) = 0.03*0.005= 0.00015
P ( Test is positive j Healthy) = P (Test is positive / Healthy) * p (healthy) = .02*0.995 = 0.0199
P (test is negative j healthy) = P (test is negative / healthy) * p(healthy) = 0.98*0.995 = 0.9751
2) We will apply Bayes Formula here. It is required to find :
P (Disease is there / Test is positive)
By Bayes Theorom , P (B/A) = P (BA)/ P(A)
Other words
P (Disease is there / Test is positive) = P (Disease j positive) / P (positive)
P (positive) = P ( Positive and Disease) + P (Positive and healthy)
Hence Required probability = 0.00485 / (0.00485 + 0.0199) = 0.00485 /0.02475 = 0.1959
As the false positives are very high (close to 2% of the population),this test is not a very reliable measure of the disease.