Question

Lung cancer is the leading cause of cancer-related deaths in the United States. Researchers examined the idea of testing all Medicare-enrolled heavy smokers for lung cancer with a computed tomography (CT) scan every year. In this population, the lifetime chance of developing lung cancer is high. In any given year, approximately 3% of heavy smokers develop lung cancer. The CT scan positively identifies lung cancer 89% of the time, and it gives a negative result for 93% of individuals who do not have lung cancer.
a. Use probability notation to express the three values cited.
b. Make a probability tree representing the outcomes in the sample space of Medicare-enrolled heavy smokers. Be sure to place probability values on all the branches of your tree.
c. What percent of CT scans in this target population would be positive?
d. What is the probability that a Medicare-enrolled heavy smoker who gets a positive CT scan actually has lung cancer, (cancer ? positive)?

Answer 1

Answer:

a. P(C) = 0.03

   P(test+ | C) = 0.89

   P(test- | no C) = 0.93

b. Tree diagram is attached.

c. P(test+) = 0.0946

d. P(C | test+) = 0.2822  

Step-by-step explanation:

a. Let C denote Lung Cancer The probability that a heavy smoker will develop lung cancer is 3% i.e. 0.03.

If lung cancer is present, the CT scan result comes out to be positive 89% of the times while if lung cancer is not present, the results are negative 93% of the time. Let test+ denote that the test result is positive and test- denote that the test result is negative.

P(C) = 0.03

P(test+ | C) = 0.89

P(test- | no C) = 0.93

b. The first pair of branches represent whether the individual has cancer or not.

P(C) = 0.03

P(no C) = 1 - 0.03 = 0.97

The second pair of branches will represent if the test result came out positive or negative.

For patients with cancer, P(test+) = 0.89, P(test-) = 1 - 0.89 = 0.11

For patients with no cancer, P(test -) = 0.93, P(test+)= 1 - 0.93 = 0.07.

The tree diagram is attached as an image file.

c. P(test+) = P(C)*P(test+| C) + P(No C)*P(test+| No C)

                                = (0.03)(0.89) + (0.97)(0.07)

                                = 0.0267 + 0.0679

    P(test+) = 0.0946

d. P(C | test+) = P(C ∩ test+) / P(test+)

                       = P(C)*P(test+ | C) / P(test+)

                       = (0.03)*(0.89) / 0.0946

    P(C | test+) = 0.2822