I have a question. Exactly 1/5th of the people in a town have Beaver Fever . There are two tests for Beaver Fever, TEST1 and TEST2. When a person goes to a doctor to test for Beaver Fever, with probability 2/3 the doctor conducts TEST1 on him and with probability 1/3 the doctor conducts TEST2 on him. When TEST1 is done on a person, the outcome is as follows: If the person has the disease, the result is positive with probability 3/4. If the person does not have the disease, the result is positive with probability 1/4. When TEST2 is done on a person, the outcome is as follows: If the person has the disease, the result is positive with probability 1. If the person does not have the disease, the result is positive with probability 1/2. A person is picked uniformly at random from the town and is sent to a doctor to test for Beaver Fever. The result comes out positive. What is the probability that the person has the disease?

I think this is an excellent question, so I am passing it along to the readers of this blog. One suggestion: you might want to use my world famous Bayesian update worksheet.

Hint: This question is similar to one I wrote about last year. In that article, I started with a problem that was underspecified; it took a while for me to realize that there were several ways to formulate the problem, with different answers.

Fortunately, the problem posed by Riya is completely specified; it is an example of what I called Scenario A, where there are two tests with different properties, and we don't know which test was used.

There are several ways to proceed, but I recommend writing four hypotheses that specify the test and the status of the patient:

TEST1 and sick

TEST1 and not sick

TEST2 and sick

TEST2 and not sick

For each of these hypotheses, it is straightforward to compute the prior probability and the likelihood of a positive test. From there, it's just arithmetic.

Here's what it looks like using my world famous Bayesian update worksheet:

After the update, the total probability that the patient is sick is 10/26 or about 38%. That's up from the prior, which was 1/5 or 20%. So the positive test is evidence that the patient is sick, but it is not very strong evidence.

Interestingly, the total posterior probability of TEST2 is 12/26 or about 46%. That's up from the prior, which was 33%. So the positive test provides some evidence that TEST2 was used.

Hey thanks Mr. Allen for taking up the question.I would just like to mention that I came across this question from the MIT OCW final exam for computer science.

ReplyDeleteI have thought of two approaches and of course they're giving different answers, although I'm a little biased towards one of the approaches. I would be really glad if I could finally understand what's happening in this question.

Thanks for letting me know about the source of the problem. If you have a link, I'll include it in the article.

DeleteI enjoyed the problem, but I got a different answer. I thought the prior for TEST1 was 2/3.

ReplyDeleteI have corrected that error. Thanks!

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ReplyDeleteI have a question though. I mentioned that this question was taken from the MIT 2006 OCW final exam for computer science. The solution was also available on the internet. Even though they have mentioned the answer to be 5/13 I.e. 10/26 but there's some calculation error in their method. The correct answer according to their method is 25/63. But I really don't understand why they did the question that way. I'm adding the links to both the question and the answer here. It's the fourth question.

ReplyDeleteanswer

question

Hi Riya,

DeleteNot sure if you are still checking this blog, but the computational error in the answer of the MIT course is just confined to the labels of two branches (test1, disease absent). They switched the labels with the probs of test1 detecting or not detecting the disease. However, the final probs at the leafs of their table are correct (they assume proper labels in their computations). And once those are correct, then the answer is in fact identical to 5/13 (as they report).

Long story short, the results of the MIT course and the results presented here are identical, they are both correct, and the only mistake is the wrong label on two branches.

And finally, I punched these numbers in the Netica program (not that it's really needed beyond the worksheet), and also get the correct 5/13 answer. Unfortunately I don't know how to share Netica worksheets in blog post, otherwise I would post it...