Question

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?

Answer 1

Use conditional probability:
$P(A|B) = \frac{P(AB)}{P(B)}$

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)

$P(A|B) = \frac{.02*.99}{.02*.99 + .98*.01} = 0.6689$

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)

$P(A|B) = \frac{.02*.01}{.02*.01+.98*.99} = 0.0002$