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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
