FINITE MATH HELP

Question

A test for a genetic abnormality was developed recently. When the test is administered to a person who actually possesses the abnormality, the test will correctly identify the abnormality with probability .9 . When the test is administered to a person who does not possess the abnormality, the test will correctly indicate the absence of the abnormality with probability .4 . If it is known that 1 out of 10 people in a given population do in fact possess this genetic abnormality, then what is the probability that a person selected at random from this population will actually possesses the genetic abnormality given that the test indicates the genetic abnormality?

Answer 1

This is a pretty straightforward application of Bayes' rule.

We want P(A|T_+) = P(T_+|A) * P(A) / P(T_+)

Where, P(A) represents the probability that someone has the abnormality

P(T_+) represents the probability that a test will return a positive result.

For P(T_+), we can calculate it as follows:

P(T_+) = P(T_+ and A) + P(T_+ and !A)

= P(T_+|A)P(A) + P(T_+|!A)P(!A)

= .9*.1 + .6(probability of T_+ for someone who doesn't have the anomaly)*.9 ~ .63

So, P(A|T_+) = P(T_+|A)P(A) / P(T_+)

= .9*.1/.63 ~ .143 or about 14%

In other words, the test isn't good enough to significantly add more information than the base rate of the population is about 10%

This formulation is assuming the test is binary (positive or negative, no unclear results, so that P(T_+) = 1-P(T_-)).