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# I HAVE NO IDEA WHAT THESE PROBABILITY QUESTIONS MEAN?

1) Suppose we have a diagnostic test for lung cancer that is 99% accurate both on those who have cancer and those who don't. .2 percent of the population have lung cancer, and we want to figure out the probability that a tested person has lung cancer given that the test result indicates that he or she has lung cancer.

This is like the drug case in the text. We will use the long form of Bayes' Theorem.

First, let L = has lung cancer, + = has tested positive for lung cancer. Notice that Prob(L) = .002.

Now what is Prob(+ | not L)?

Your answer should be a percentage (no decimals), e.g. 23% not 23.1%

2) Suppose we have a diagnostic test for lung cancer that is 99% accurate both on those who have cancer and those who don't. .2 percent of the population have lung cancer, and we want to figure out the probability that a tested person has lung cancer given that the test result indicates that he or she has lung cancer.

This is like the drug case in the text. We will use the long form of Bayes' Theorem.

First, let L = has lung cancer, + = has tested positive for lung cancer. Notice that Prob(L) = .002.

Now what is Prob(not L)?

Your answer should be a percentage (with one decimal point), e.g. 23.1%

3) Suppose we have a diagnostic test for lung cancer that is 99% accurate both on those who have cancer and those who don't. If .2 percent of the population have lung cancer, what is the probability that a tested person has lung cancer given that the test result indicates that he or she has lung cancer?

This is like the drug case in the text. Use the long form of Bayes' Theorem. Your answer should be a percentage (no decimals), e.g. 23% not 23.1%

### 1 Answer

- PhiloLv 71 decade agoFavorite Answer
1. prob( + | not L) is the fraction of the population that does not have cancer but tests positive because of the imperfection of the test. These are called false positives. 1% of the test results are wrong, 99.8% of the pop does not have cancer, so 1% of 99.8% = 0.998% ≈ 1%

2. prob( not L) we did in 1, 99.8%. by the way, you lose a lot of accuracy if you round off your decimals early.

3. no question is asked here after the 3rd copy of the same background info.

you answer the question this way. 99% of the 0.2% of the pop with cancer will test positive, 0.198%. and 1% of the 99.8% who don't have cancer will test positive, 0.998%. total is 1.196%. so of those, the fraction that really do have cancer, p( L | +), is 0.198% / 1.196% = 16.6%

this is easier with a tree diagram and almost impossible with just a formula.

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