After your yearly checkup, the doctor has bad news and good news. The bad news is that you tested positive for a serious disease, and that the test is 99% accurate (i. e., the probability of testing positive given that you have the disease is 0.99, as is the probability of testing negative given that you dont have the disease). The good news is that this is a rare disease, striking only one in 10,000 people. What are the chances that you actually have the disease? (Show your calculations as well as giving the final result.)

The probability that you have the disease, given that your test is positive is ≈ 0.0098

Explanation:

This is a conditional probability problem.

Let P (A|B) denote the conditional probability of A given B and it satisfies the equation

(1) P (A|B) = P (A) * P (B|A) / P (B)

We have the the probabilities:

P (Testing Positive | Having Disease) = 0.99 P (Testing Negative | Not Having Disease) = 0.99 P (Testing Positive | Not Having Disease) = 1-0.99=0.01 P (Having Disease) = 0.0001 (striking only one in 10,000 people) P (Not Having Disease) = 1 - 0.0001 = 0.9999

We can calculate:

P (Testing Positive) =

P (Having Disease) * P (Testing Positive | Having Disease) + P (Not Having Disease) * P (Testing positive | Not Having Disease) = 0.0001*0.99 + 0.9999*0.01 = 0.010098

from (1) we have the equation:

P (Having Disease|Testing Positive) = P (Having Disease) * P (Testing Positive | Having Disease) / P (Testing Positive) = 0.0001*0.99/0.010098≈0.0098

Thus, the probability that you have the disease, given that your test is positive is ≈ 0.0098