Question

In a specific population of​ people, cancer is found in 0.8​% of the people. If a person does have the​ disease, a diagnostic procedure will identify the presence of cancer in 93​% of such people. If a person does not have​ cancer, this diagnostic procedure will give a​ false-positive result 0.3% of time. What is the probability that a person with a positive test result has cancer.

Answer 1

Solution:

Given:

In a specific population of​ people, cancer is found in 0.8​% of the people.

Let C = person has Cancer , NC = Person do not have cancer

thus

P(C) = 0.8% = 0.008

thus

P(NC) = 1 - P(C) = 1 - 0.008 = 0.992

If a person does have the​ disease, a diagnostic procedure will identify the presence of cancer in 93​% of such people.

Let TP = Test Positive and TN = Test Negative

Thus we have:

P(TP | C) =93% = 0.93

If a person does not have​ cancer, this diagnostic procedure will give a​ false-positive result 0.3% of time.

False-Positive means showing Test positive result in fact person does not have cancer.

that is we are given:

P( TP | NC) = 0.3% = 0.003.

We have to find: the probability that a person with a positive test result has cancer.

that is we have to find:

P(C | TP) = ..............?

Using Bayes rule of probability, we get:

Thus the probability that a person with a positive test result has cancer is 0.7143