Question

The test to detect the presence of a liver disorder is 98% accurate for a person who has the disease and 97% accurate for a person who does not have the disease. If 3.5% of the people in a given population actually have the disorder, what is the probability that a randomly chosen person tests positive?

Answer 1

We are given the following statements
P(test positive | diseased) = .98
P(test negative | not diseased= .97
P(diseased)= 0.35
Hope this helps

Answer 2

Answer:

The probability that a randomly chosen person tests positive is 0.06325

Step-by-step explanation:

if we chose a person randomly and the test is positive, there are two possibilities:

1. The person actually have a disease: the probability of this case is given by the multiplication of the probability of having the disease and the probability that the test detect the disorder. That can be written as:

0.035*0.98=0.0343

where 0.98 is the probability that the test detect the disease given that a person have a disease.

2. The person doesn't have a disease: the probability of this case is given by the multiplication of the probability of doesn't have the disease and the probability that the test said that the person have the disorder. That can be written as:

0.965*(1-0.97)=0.02895

Where (1-0.97) is the probability that the test detect the disease given that a person don't have a disease.

Finally the probability that a randomly chosen person tests positive is the sum of both cases and it calculated as:

P=0.0343+0.02895=0.06325