Question

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

Answer 1

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

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 | positive) = (P(disease) \cdot P(positive | disease))/(P(disease) \cdot P(positive | disease) + P(nodisease) \cdot P(positive | nodisease))`

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;

`P(disease | positive) = (0.0001 * 0.996)/(0.0001 * 0.996 + 0.9999 * 0.0005) = 0.16612459344`