Question

a test for diabetes results in a positive test in 95% of the cases where the disease is present and a negative test in 07% of the cases where the disease is absent. if 10% of the population has diabetes, what is the probability that a randomly selected person has diabetes, given that his test is positive

Answer 1

Answer:

0.9378 = 93.78% probability that a randomly selected person has diabetes, given that his test is positive.

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: Positive test

Event B: Person has diabetes.

Probability of a positive test:

0.95 out of 0.1(person has diabetes).

0.007 out of 1 - 0.1 = 0.9(person does not has diabetes). So

$P(A) = 0.95*0.1 + 0.007*0.9 = 0.1013$

Probability of a positive test and having diabetes:

0.95 out of 0.1. So

$P(A \cap B) = 0.95*0.1 = 0.095$

What is the probability that a randomly selected person has diabetes, given that his test is positive?

$P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.095}{0.1013} = 0.9378$

0.9378 = 93.78% probability that a randomly selected person has diabetes, given that his test is positive.