Question

In a clinical study of a test devised to detect colorectal cancer it was found that 13% of people without cancer received a positive result (false positive) and 8% of people with cancer received a negative result (false negative). According to the American Cancer Society, the lifetime risk of developing colorectal cancer is about 1 in 22 (4:49%) for men and 1 in 24 (4:15%) for women. According to the World Bank the population of the U. S. is 50:52% female. If we simplify the model assuming that there are only two genders, male and female, (a) what is the probability that a person with undisclosed gender will develop colorectal cancer in their lifetime? (b) what is the probability that a person with undisclosed gender has colorectal cancer, given that they took the test twice and the results were positive the first time and negative the second time?

Answer 1

a)

P(MC) = probability of men developing colorectal cancer = 0.0449

P(FC) = probability of women developing colorectal cancer = 0.0415

P(F) = proportion of a person being female = 50.52% = 0.5052

P(M) = proportion of a person being female = 49.48% = 0.4942

Overall probability of developing cancer be P(C)

Then P(C) = P(M)*P(MC) + P(F)*P(FC)

=0.4942*0.0449 +  0.5052*0.0415

= 0.04316

The probability that a person with undisclosed gender will develop colorectal cancer in their lifetime is 0.04316 or 4.316%.

b)

Let P(+ve) represent probability of positive result

and P(-ve) represent probability of negative result

Also, Let P(C) =overall probability of any person having cancer

and P(NC) =overall probability of any person having cancer

From given information, we have

P(-ve / C) = 8% = 0.08

P(+ve / C) = 1 - 0.08 = 0.92

P(+ve / NC) = 13% = 0.13

P(+ve / C) = 1 - 0.13 = 0.87

Let E =event that results were positive the first time and negative the second time for any person.

Then,

We need to find the probability that the person has cancer given that event E has occured.

From, Bayes' theorem

P (C | E) = P(E | C) *P(C) /P(E)

P(C) = 0.04316

P(NC) = 1-0.04316 = 0.9568

P(E given that person has cancer) = P(E | C)

= P(+ve / C)*P(-ve / C)

=0.92*0.08

=0.0736

P(E given that person does not have cancer) = P(E | NC)

= P(+ve / NC)*P(-ve / NC)

=0.13*0.87

=0.1131

Therefore,

P (E) = P(E | C) * P(C)+ P(E | NC)*P(C)

=0.0736* 0.04316 + 0.1131* 0.9568

=0.1114

P (C | E) = P(E | C) *P(C) /P(E)

=0.0736* 0.04316 / 0.1114

=0.0285

Therefore, the probability that a person with undisclosed gender has colorectal cancer, given that they took the test twice and the results were positive the first time and negative the second time is 0.0285.