A new, simple 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 2 comma 000 adults and finds (by other means) that 2% have this type of cancer. Each of the 2 comma 000 adults is given the 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. Based on these results, what is the probability of a randomly chosen person having cancer given that the test indicates cancer? Of a person having cancer given that the test does not indicate cancer?
Solution :
Given that,
P(no cancer positive) = 0.99 * 0.01 = 0.0099
P( cancer positive) = 0.02 * 0.99 = 0.0198
The probability that the positive test = 0.0099 + 0.0198 = 0.0297
(a)
the probability of a randomly chosen person having cancer given that the test indicates cancer is
= 0.0099 / 0.0297
= 0.33
Probability = 0.33
(b)
The probability of a person without cancer having a negative test is ,
= 0.99 * 0.99 = 0.9801
The probability of a person with cancer having a negative test is,
= 0.02 * 0.02 = 0.0004
A person having cancer given that the test does not indicate cancer is,
= 0.004 / 0.0004 + 0.9801
= 0.0004 / 0.9805
= 0.0004
Probability = 0.0004
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