Question

suppose that one person in 10,000 people has a rare genetic disease. there is an excellent test for the disease; 99.7% of people with the disease test positive and only 0.04% who do not have the disease test positive. what is the probability that someone who tests negative does not have the disease? (enter the value of the probability in decimal format and round the final answer to three decimal places.)

Answer 1

The probability that someone who tests negative does not have the disease is approximately 0.999 (upto three decimal places).

Step-by-step explanation:

To find the probability that someone who tests negative does not have the disease, we can use Bayes' theorem.

Let's define the events: A = person has the disease, B = person tests positive.

From the given information, we can calculate the probability of testing positive given that a person has the disease, which is 99.7% or 0.997.

The probability of testing positive given that a person does not have the disease is 0.04% or 0.0004. The probability of having the disease is 1 in 10,000 or 0.0001.

Applying Bayes' theorem, the probability that someone who tests negative does not have the disease is:

P(A' | B') = P(A') * P(B' | A') / (P(A') * P(B' | A') + P(A) * P(B' | A))

= (1 - P(A)) * (1 - P(B | A)) / ((1 - P(A)) * (1 - P(B | A)) + P(A) * (1 - P(B | A')))

= 0.9999 * 0.9996 / (0.9999 * 0.9996 + 0.0001 * 0.9996)

≈ 0.9999

Therefore, the probability that someone who tests negative does not have the disease is approximately 0.999 (upto three decimal places).