Stats bases on Bayes theorem

Question

People with the disease D have a 90% probability of testing positive on the D-test. If they do not have disease D, they have a 99% probability that they will test negative. We know that 4% of all people test positive.

1. What is the probability of having disease D?
2. A person just tested positive on the D-test. What is the probability that they are actually healthy?

Answer 1

Let's define the events:

1. A: The person has disease D.
2. A^c: The person does not have disease D.
3. B: The person tests positive for the disease.

We are given:

1. P(B∣A) = 0.90 (Probability of testing positive given having the disease).
2. P(B∣A^c) = 1 − 0.99 = 0.01 (Probability of testing positive given not having the disease).
3. P(B) = 0.04 (Overall probability of testing positive).

P(B) = P(B∣A)P(A) + P(B∣A^c)P(A^c) using the law of total probability

P(A) = 1 - P(A^c)

0.04 = 0.90P(A) + 0.01(1 − P(A))

P(A) = 0.0337

P(A^c∣B) =P(B∣A^c)P(A^c) / P(B)​ using Bayes' Theorem

We can find P(B) now that we know what P(A) and P(A^c) are. After plugging in the values, you should get P(A^c∣B) = 0.2416

Hope this was helpful.