# Probability Bayes' theorem

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.

• It is hard to follow your work, you string the sentences together so we can't see where one equation ends and the next begins. I could try to edit it for you, but I fear that I'd introduce errors. Can you clarify? Insert Line breaks in between each equation.
– lulu
Aug 8 '17 at 18:39
• Probability never gets over 1, for anything - ever. Aug 8 '17 at 18:48
• You must have missed the class where Baye's Theorem was discussed, and not done any reading. You can find how to approach problems like this by looking at the "Related" Q & A. Aug 8 '17 at 19:00
• Related. perhaps a duplicate: math.stackexchange.com/questions/2279851/… Aug 8 '17 at 19:13

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 $196/10000 + 597/20000 = 9989/20000 = 0.04945$.