Question

the data represent the results for a test for a certain disease. assume one individual from the group is randomly selected. find the probability of getting someone who tested positive, given that he or she did not have the disease. the individual actually had the disease
yes no
positive 120 5
negative 33 142
the probability is approximately (round to three decimal places as needed.)

Explanation:

Step1: Identify relevant values

We want the probability of positive - test given no disease. The number of people who tested positive and did not have the disease is 5, and the total number of people who did not have the disease is \(5 + 142=147\).

Step2: Calculate the conditional probability

The formula for conditional probability \(P(A|B)=\frac{P(A\cap B)}{P(B)}\). In the case of frequency data, if \(A\) is the event of testing positive and \(B\) is the event of not having the disease, the probability \(P=\frac{\text{Number of positive - test and no - disease}}{\text{Total number of no - disease}}\). So \(P=\frac{5}{147}\).

Answer:

\(0.034\)