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 3% 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

Answer 1

Answer:

0.2177 = 21.77% conditional probability that she does, in fact, have the disease

Step-by-step explanation:

Conditional Probability

We use the conditional probability formula to solve this question. It is

$P(B|A) = \frac{P(A \cap B)}{P(A)}$

In which

P(B|A) is the probability of event B happening, given that A happened.

$P(A \cap B)$ is the probability of both A and B happening.

P(A) is the probability of A happening.

In this question:

Event A: Test positive

Event B: Has the disease

Probability of a positive test:

90% of 3%(has the disease).

1 - 0.9 = 0.1 = 10% of 97%(does not have the disease). So

$P(A) = 0.90*0.03 + 0.1*0.97 = 0.124$

Intersection of A and B:

Positive test and has the disease, so 90% of 3%

$P(A \cap B) = 0.9*0.03 = 0.027$

What is the conditional probability that she does, in fact, have the disease

$P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.027}{0.124} = 0.2177$

0.2177 = 21.77% conditional probability that she does, in fact, have the disease