Question: One in two hundred people in a population have a particular disease. A diagnosis test gives a false positive $3$% of the time, and a false negative $2$% of the time. Ross takes the test and the report comes positive. Find the probability that Ross has the disease.
What I solved: Assume the probability that a person has disease is $\frac{1}{200}$. Then, we have the following probabilities:
- person has disease and test is negative: $\frac{2}{100}$
- person does not have disease and test is positive: $\frac{3}{100}$
- person has disease and test is positive: $\frac{98}{100}$
- test is positive: $\frac{98}{100}+\frac{3}{100}=\frac{101}{100}$
I am not sure about the above line the probability is $\frac{101}{100}$ I think this is not correct can any one tell me where I went wrong.
The probability that the person has the disease and the test is negative is not $2/100$, but $1/200\cdot 2/100 = 1/10000$.
The likelihood that the person has the disease and the test is positive is $2/100\cdot 98/100 = 196/10000 = 0.0196$, not $98/100$.
The likelihood that the person does not have the disease and test is positive is $199/200\cdot 3/100 = 597/20000 = 0.02985$.
So the likelihood of having a positive test is then $196/10000 + 597/20000 = 9989/20000 = 0.04945$.
I guess Bayes' theorem comes in when computing the likelihood that a person with a positive test actually has the disease, although that was not stated.