This is a problem offered to our students this week in the monthly Florida Math league contest. Lots of schools across the state participate. Not many of our kids got it, but surely you are smarter that a high school student, right??

The Surfboard Store sells Special Surfboards that

are so carefully made that only 1 in 1 thousand

is bad. The store tests all Special Surfboards

using a test that is 99% accurate. If you buy

a Special Surfboard that tests bad, what is the

probability that it really is bad?

Think about it a while and put your answer in the comments here or at the forum where this is posted. (or you can message me if you're shy) I will try to explain the answer tomorrow.

I think that about 9% of the boards that test bad are actually bad.

ReplyDeleteP(board tests bad and is bad)/P(board tests bad)

P(board tests bad) = P(good board and tests bad)+P(bad board and tests bad)

P(board tests bad) = (999/1000)(.01) + (1/1000)(.99) = 0.01098

The .01 in the first part is the probability that a test which is 99% accurate incorrectly says a good board is bad.

The .99 in the second part is the probability that a bad board is correctly identified.

P(board tests bad and is bad) = [(1/1000)(.99)]/0.01098 = 0.09016 or 9.016%

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DeleteThanks so much Neil and Kyle for your thoughts. I wonder if part of the problem with questions like these is decifering the language. Of course that is the difficulty in a lot of problems isn't it?? You can check out my thoughts and solution here. http://debcostello.blogspot.com/2013/10/surfs-up-solution.html

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DeleteNeil,

DeleteWhen you are looking at conditional probabilities like this you have to take into account all of the options. It helps if you draw a tree diagram so that you can see all of the possibilities. There are four possible outcomes in this scenario.

Board Good and Tests good

Board Good and Tests bad

Board Bad and Tests Bad

Board Bad and Tests Good

In order to find the probabilities of each of these four options, you have to multiply the probabilities of each of the components together (i.e. (probability of getting a good board (.999) times the probability of getting a positive test (.99) and so on)

When you have more than one thing happening, it is often helpful to draw a diagram. In this case, my students would probably opt for a tree diagram so that they could identify which boards tested bad (which from above would include two separate groups of boards).

I know that this seems counter intuitive, but that is the reason that most students will miss this problem. In reality, this has applications to medical tests for rare diseases. No test is 100% accurate (at least not any that I am aware of) and so there are always some false positives. If the disease is very rare, than it should make sense that it is more likely that the test was wrong than someone actually have the rare disease.

I hope this helps. Just as a bit of background, after looking at Deb's explanation, my answer has a lot more math involved because I would have had my students use my explanation on a Free Response problem on the AP Statistics test. I like Deb's answer better in terms of explaining the reasoning behind why my answer was correct to my students. Hopefully they will be able to understand the concept and then apply their understanding in a way which shows their understanding of probabilities. We'll see.

Thanks for your insights, Kyle. I do not teach AP Stats, but I am teaching a regular version of the class and have found it very challenging to try to help students understand how to think about the probability of conditional events. This question just appeared and seemed a really good example of the ideas we struggle with. I actually got it wrong at first too. I have been thinking a lot lately about the role of statistics and probability in our popular culture and the ongoing misunderstanding that is clear in most areas. Even math teachers, long dedicated to the pursuit of calculus, are loathe to address these questions in their courses. It's a worthy conversation and will continue to be so.

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