Anew test has been developed to detect a particular type of cancer. the test must be evaluated before it is put into use. a medical researcher selects a random sample of 1,000 adults and finds (by other means) that 4% have this type of cancer. each of the 1,000 adults is given the new test, and it is found that the test indicates cancer in 99% of those who have it and in 1% of those who do not.
a) based on these results, what is the probability of a randomly chosen person having cancer given that the test indicates cancer?
b) what is the probability of a person having cancer given that the test does not indicate cancer?

Part A)
Probability you both have cancer and test indicates you have cancer is:
.04*.99
Probability test indicates you have cancer. You have to look at 2 cases:
The one above and the one where you do not have cancer but test says you do. ---> .01*.96
The total probability then is the sum of the 2 cases:
.04*.99 + .01*.96

Finally, the conditional probability is first case divided by total probability:
(.04*.99)/(.04*.99 + .01*.96)

Part B)
Same as part a, except we are looking at when test says you do Not have cancer. If you have cancer, this occurs 1% of the time. If you dont, then 99%.
The probabilities are flipped.

The conditional probability is:
(.04*.01)/(.04*.01 + .99*.96)

The answer is mathg.a.y