A medical researcher selects a random sample 1,000 adults and finds that 2% have this type of cancer. Each if the 1000 adults is given the test and it is found that the test indicates cancer 99% of those who do not. Based on the results, what is the probability of a ramdomly chosen person having cancer given that the test indicates cancer? Of a person having cancer given that the test does not indicate cancer?
Use conditional probability: [tex]P(A|B) = \frac{P(AB)}{P(B)}[/tex]
For part 1) A = Has cancer B = Test indicates cancer We know that P(A) = 0.02 and the test has 0.99 success rate. The test will be positive for 99% of those with cancer and 1% of those without. P(B) = (.02)(.99) + (.98)(.01) P(AB) is only for those who both have cancer and test positive, (.02)(.99)
Part 2 is similar except B is now Test is negative. The is true for 1% for those with cancer, 99% for those without. P(B) = (.02)(.01)+(.98)(.99) P(AB) is if you both have cancer and test negative, (.02)(.01)
