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
negative	19	121

the probability is approximately (round to three decimal places as needed.)

Explanation:

Step1: Identify relevant values

We want $P(\text{Positive}|\text{No disease})$. The number of positive - test results among those without the disease is 25, and the total number of people without the disease is $25 + 121$.

Step2: Apply conditional - probability formula

The formula for conditional probability $P(A|B)=\frac{P(A\cap B)}{P(B)}$. In terms of counts, $P(\text{Positive}|\text{No disease})=\frac{\text{Number of positive and no disease}}{\text{Number of no disease}}=\frac{25}{25 + 121}$.

Step3: Calculate the probability

$\frac{25}{25+121}=\frac{25}{146}\approx0.171$

Answer:

$0.171$