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Here's the question

To diagnose colorectal cancer, the hemoccult test --- among others --- is conducted to detect occult blood in the stool. This test is used from a particular age on, but also in routine screening for early detection of colorectal cancer. Imagine you conduct a screening using a hemoccult test in a certain region. For symptom-free people over 50 years old who participate in screening using the hemoccult test, the following information is available for this region.

The probability that one of these people has colorectal cancer is 0.3 percent. If a person has colorectal cancer, the probability is 50 percent that this person will have a positive hemoccult test. If a person does not have colorectal cancer, the probability is 3 percent that this person will still have a positive hemoccult test. Imagine a person (over age 50, no symptoms) who has a positive hemoccult test in your screening. What is the probability that this person actually has colorectal cancer.

Here's my solution

Tree

(0.7 * 0.03) / ((0.7 * 0.03) + (0.3 * 0.5)) = 0.021 / (0.15 + 0.021) = 0.1228

Am I correct? If not, could you point out where I'm off at?

EDIT 1:

I wasn't correct, see answer below. Correct answer is Okay so 0.003 * 0.5 / (0.003 * 0.5 + 0.997 * 0.03) = 0.0477554919.

Here's an updated graphic

tree2

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The probability a person has the cancer is $0.3$ percent. This is $0.003$, not $0.3$. The probability of not having it is $0.997$, not $0.7$.

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  • $\begingroup$ Okay so 0.003 * 0.5 / (0.003 * 0.5 + 0.997 * 0.03) = 0.0477554919? $\endgroup$ – CamHart Nov 5 '14 at 5:22
  • $\begingroup$ The stuff up to the $=$ sign is right. I have not checked the calculator work. $\endgroup$ – André Nicolas Nov 5 '14 at 5:26

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