Question

A diagnostic test for a disease is such that it (correctly) detects the disease in 90% of the individuals who actually have the disease. Also, if a person does not have the disease, the test will report that he or she does not have it with probability 0.9. Only 2% of the population has the disease in question. If a person is chosen at random from the population and the diagnostic test indicates that she has the disease, what is the conditional probability that she does, in fact, have the disease? (Round your answer to four decimal places.)

Answer 1

Answer:

0.1552

Step-by-step explanation:

Hi!

Give a random person, lets call:

D = the person has the disease

¬D = the person doesnt have the disease

T = the test on this person is positive (detects the disease)

¬T = the test on this person is negative

We are give the conditional probabilities:

P(T | D) = 0.9

P(¬T | ¬D) = 0.9

P(D) = 0.02

From those we know that:

P(T | ¬D) = 0.1

P(¬T | D) = 0.1

We must find P(D | T). For that we use Bayes theorem:

$P(D|T) = \frac{ P(T|D)P(D)}{P(T|D)P(D) +P(T| \neg D)P(\neg D) }$

$P(D | T) = \frac{0.9*0.02}{0.9*0.02 + 0.1*0.98} = 0.1552$