NOTE: This is a multi-part question. Once an answer is submitted, you will be unable to return to this part. Suppose that one person in 10,000 people has a rare genetic disease. There is an excellent test for the disease; 99.6% of people with the disease test positive and only 0.05% who do not have the disease test positive. What is the probability that someone who tests positive has the genetic disease

The probability that someone who tests positive has the genetic disease is approximately 0.166 or 16.6% who tests positive actually has the disease

Step-by-step explanation:

The number of persons in 10,000 people that have the disease = One

The percentage of people with the disease that test positive = 99.6%

The percentage of people who do not have the disease test positive = 0.05%

The probability that someone who tests positive has the genetic disease

P(disease) = 0.01% = 0.0001

P(positive disease)  = 99.6% = 0.996

P(nodisease) = (10,000,000 - 100)/10,000,000 = 99.99% = 0.9999

P(positivenodisease) = 0.05% = 0.0005

Whereby 10,000,000 people are tested, 1000 out of the 1,000,000 will have the disease, 1,000 - 0.996 × 1000 = 4 people out of the 1,000 will test negative while 996 will test positive. From the 9,999,000 people who do not have the disease, 9,999,000 × 0.0005 = 4999.5 will give positive test results.

Therefore, the total number of people that tests positive = 4,999.5 + 996 = 5,995.5

Therefore, out of the 5,995.5 that test positive for the disease, 996 will test positive

The probability that someone who tests positive has the genetic disease, P(disease positive) = 996/5,995.5 ≈ 0.166 or 16.6%

Therefore, approximately 16.6% of the people that test positive for the disease actually has the disease

We have;