Question

A new medical test has been designed to detect the presence of the mysterious Brainlesserian disease. Among those who have the disease, the probability that the disease will be detected by the new test is 0.9. However, the probability that the test will erroneously indicate the presence of the disease in those who do not actually have it is 0.06. It is estimated that 16 % of the population who take this test have the disease. If the test administered to an individual is positive, what is the probability that the person actually has the disease?

Answer 1

To find the probability that a person actually has the disease given a positive test result, we can use Bayes' theorem. Given the probabilities of a positive test result given the person has the disease and does not have the disease, as well as the probability of having the disease, we can calculate the conditional probability.

Step-by-step explanation:

To find the probability that a person actually has the disease given a positive test result, we can use Bayes' theorem. Let's define the events: A = person has the disease, B = person tests positive. We are given that P(B|A) = 0.9 (probability of a positive test given the person has the disease), P(B|A') = 0.06 (probability of a positive test given the person does not have the disease), and P(A) = 0.16 (probability that a person has the disease).

Bayes' theorem states: P(A|B) = (P(B|A) * P(A)) / P(B).

Substituting the given values, we have: P(A|B) = (0.9 * 0.16) / P(B).

We don't have the value of P(B), but we can calculate it as follows: P(B) = (P(B|A) * P(A)) + (P(B|A') * P(A')). Plugging in the values, we have: P(B) = (0.9 * 0.16) + (0.06 * 0.84).

Now we can substitute the value of P(B) into the formula for P(A|B) to calculate the probability.

Therefore, the probability that the person actually has the disease given a positive test result is approximately 0.231.

Answer 2

Let D denote the event that a particular person has the disease. P (D) = 0.16. Let N indicate the event that the person doesn't have the disease.
Let M denote the event that the medical test turns positive. We need to compute P (D l M) .

We know that P (D) = 0.16, P (N) = 0.84. And, P (D l M) = 0.9, P ( D l N) = 0.06.

When we plug this information into the Bayes' Theorem equation, we get the following:


`P ((D∩M))/(P(M))`


=
`(0.9x0.16)/(0.9x0.16+0.06x0.84)`


= 0.74

This means that the probability the person actually has the disease is 0.74, or 74%.