In a community 1/2 % of all people has a certain disease. If a person has the di
ID: 3297965 • Letter: I
Question
In a community 1/2 % of all people has a certain disease. If a person has the disease then it is 99 % probabilty that the person is diagnosed in a test (that is the test is positive). If a person does not have the diesease, there is still 1 % probabilty that the test will give a positve sign. It is possible to take the test ofter than twice and we will assume that the tests are undependent. Let Jmark the event that the result that the test is positive and let D mark the event that the person has a disease
a) If a person gets two times a positive result from the test, what is the probability that the person has the disease ?
b) What is the probabilty that two tests will give a positve result if the person has not the disease ?
Explanation / Answer
a) Given, J mark the event that the result that the test is positive and let D mark the event that the person has a disease.
Let ~J mark the event that the result that the test is not positive and let ~D mark the event that the person does not has a disease.
Given, P(D) = 0.5 % = 0.005
P(J | D) = 0.99
P(J | ~D) = 0.01
We need to calculate P(D | J), the conditional probability that the person has a disease given the test is positive.
P(~D) = 1 - P(D) = 1 - 0.005 = 0.995
By law of total probability,
P(J) = P(D) P(J | D) + P(~D) P(J | ~D)
= 0.005 * 0.99 + 0.995 * 0.01 = 0.0149
By Bayes theorem,
P(D | J) = P(J | D) P(D) / P(J)
= 0.99 * 0.005 / 0.0149 = 0.3322
So, the probability that the person has a disease given the test is positive is 0.3322
Given all tests are independent, then if a person gets two times a positive result from the test, the probability that the person has the disease
= 0.3322 * 0.3322 = 0.1104
b)
The conditional probability that the test is positive given that the person does not has a disease is P(J | ~D)
Given, P(J | ~D) = 0.01
The probabilty that two tests will give a positve result if the person has not the disease = 0.01 * 0.01 = 0.0001