# Probability of choosing 4 sick people from 100 where there is a probability of 8.6% for each of them to have a disease.

#### boburvik

##### New Member
I have the following task: There is 4% of sick people in a population. There is 5% chance for a positive test for a healthy person. There is 2% chance for a negative test for a a sick person. We tested 100 people and 4 had a positive test. What is the probability of such an outcome?

So far I did this:
+ means positive test, D means diseased
D = 0.04, P(+|D) = 0.98, P(+|!D) = 0.02
P(+|D) = P(+∩D) / P(D) => P(+∩D) = P(D) * P(+|D) = 0.0392
P(+|!D) = P(+∩!D) / P(!D) => P(+∩!D) = P(!D) * P(+|!D) = 0.048
P(+) = P(+∩D) + P(+∩!D) = ‭0,0872‬

So I found that the probability for a test to be positive is ‭0,0872‬, but I am not sure what to do next.

Last edited:

#### fed2

##### Active Member
assuming your calc of P(+) is right, apply binomial distribution the get prob 4 + in 100 trials. I tried to follow your calcs but my brain stopped. I recommend just draw the 2 x 2 table, makes it easier to follow.