Part I You are not feeling well, so you go to the doctor and you test positive for a disease effecting .1% of the population. The test you took correctly identifies 99% of the people that have the disease and is only wrong 1% of the time What is the probability that you actually have the disease Part II Now assume you go to another doctor and take the same test and test positive again what is the probability that you have the disease now???
Spoiler After the first test it is not very high. After the second test it is high. For more information see this Youtube
99% 99-100 99.99% 99.99 - 100 ETA: The above was written before watching the video. I watched 1/2 of the video, and I'll stick with my answer. It doesn't matter what per cent of the population may or may not ever get the disease. The question states that the test is 99% accurate. 1,000,000 people take the test Statistically only .1% have the disease. If the test was 100% accurate and 1,000,000 get tested, the results would show 1000 people have the disease (1,000,000 * .001) However, since the test only correctly identifies 99% of the people that have the disease, the results would show 990 people have the disease.
If the test is 99% accurate, there's a 1% chance for a false positive, which is why it's important to know the distribution of the disease in the population. So, in 1 000 000 people, with 99% accuracy, 990 people would accurately be proven to have the disease, but 10 000 people would inaccurately be proven to have the disease too.
Probabilities always range from 0 to 1. In percentages, this is expressed from 0% to 100%. "90,834% probability" shows you do not remotely understand probability analysis. P1 = .99 P2 = .9999 Incidentally, I recently invented a new knot. I call it the Probably Knot. (Is it strong?) Probably not.
There is a distinction between a test that fails to correctly identify the disease 1% of the time and a test that will give a false positive in healthy patients 1% of the time. The natural reading of your statement implies the former (intentionally I’d suggest, given it’s part of the “trick” to the question).
Edit: I still haven't watched the video, but may later. This is the sort of statistics problem commonly found on USMLE exams for medical students. It sounds like we're assuming sensitivity is 99%, and specificity is 99%. And we're asking for positive predictive value. I always construct a 3x3 table with totals on the bottom and sides. Populate the table using known population prevalence multiplied by sensitivity and specificity. The smallest total you can use with whole numbers is 100,000 for this example. Start with the bottom total row to fill in the rest of the table: _______Reality+_______Reality-________Totals Test+______99__________999_________1098 Test- _______1________98901________98902 Totals_____100________99900_______100000 Part I: 0.09 Part II: 0.17 (just probabilities multiplied, assuming error is random) As one would expect with a low population prevalence, the positive predictive value is still bad even when sensitivity and specificity are 99%. The negative predictive value, in contrast, would be very good. However, presumably the doctor is ordering the test because he has a heightened pre-test probability compared to the wider population (consider relative risk)... so the number we would want isn't 0.1% of the population (which would be what you would use for totally random screening), but rather the proportion of patients who present with similar symptoms who have the disease. This percentage is usually going to be on the order of 10%-90% even for very rare diseases, but will vary widely depending upon how common the symptoms/signs are. It's the rarity of the symptoms that makes the difference. Commons things like just a fever may put it below 10%, while an elaborate story with a constellation of symptoms that together are very specific to the disease may put it above 90%. That is why one of the first things they tell you in medical school is that taking a good history is the most important thing.
I got these numbers by using the Bayes' theorem. As one should do. Then looked at the video, and we pretty much got the same answers. With some difference after the commas, which could very well do with how I rounded up and down the numbers while doing the calculation. If your answer to 1 and 2 is .99, and .9999, I really hope you're not a chemical engineer. You don't even need to watch the video, just know Bayes' theorem, and you'll have everything you need to solve this: P(A|B)=(P(B|A)*P(A))/P(B) In this case, A=sick, and B=positive test result. So, the probability of sickness, given a positive test result, equals the ((probability of a positive test result given sickness) multiplied with the (probability of sickness)) all divided on the probability of a positive test result.
Yes it does. Just add all those false positives, and true positives together. You end up with a population of 10 990. Of those, 990 have the disease. 990/10990= 0,09. And 10 000 do not. 10 000/10990 = 0,91.
Not sure I understand what you mean here. The probability of an event occurring can't be over 100%. This probability piece would be easier for most people to grasp if the accuracy of the test had different value for its sensitivity and specificity. I just used the Bayes theorem, with 99% sensitive, and 99% specific. Your first assumption of 99% chance of having the disease would be correct if the test had 99% sensitivity, and 100% specificity. That is, absolutely no chance of false positives. But since the OP states it's 99% accurate, this also means it's 99% accurate on non-diseased people, leaving 1% to get a false positive.
I think I know the reason for the confusion. Otern means part 2 = 90.834% which is under 100%. He uses a comma instead of a period as that is the convention in his part of the world. Edit. More on this issue here https://en.wikipedia.org/wiki/Decimal_mark
Yes. But I don't think most of this is confusion, but more of a lack of will to understand. Of course I don't mean there's a 90 834% probability of an event happening. I probably should've rounded the numbers to 91% though. That is 0,91. Or 0.91. Or .91.
I knew that some countries use spaces (1 000 000). I never knew there were as many different conventions as shown in your link. Thanks.
Not a lack of will to understand. As shown by Robot's link there are at least a dozen different conventions. We may be behind in the use of the Metric System, but I think we have commas and periods right. 1,234,567.89 is a lot easier to comprehend than 1 234 567.89 or 1 234 567,89
I don't know which is the best way, but none of these methods looks confusing to me. Especially not in the context of probability, where all answers will be something between 0 and 1.
If you two would rather score pointless points off each other than increase your knowledge then you just make an ass of each other. Either one of you could have made the post I made and increased your own knowledge. But you did not question your own assumptions. Please remember members can come from anywhere in the world. Their way of doing things may not be the same as yours. You could try to learn what they are.
Part 1 0.1% because the probability of actually having the disease is not changed by the test. If you are asking what is the probability that the test is correct then that is 99%. The probability that the test is correct is not contingent on the probability of having the disease. Part 2 The same as before. Repeating the test does not change the probability. As I see it, this is the problem with using data derived from populations for single subject in isolation.
When you multiply, your answer should have the same amount of significant figures as your least precise measurement It looks like the least precise measurement has two significant figures???
Just thinking about this. Maybe the answer is not so simple. If you have some symptoms that are typical of the disease then you are more likely than the general population to have that disease. The more unusual and specific the symptoms then the greater the chance you have the disease. It is only if you have NO symptoms and they test you then the calculations are correct. For example you may have an accident involving your arm. You hear a bang and your arm then huts. You suspect you have broken it. Only 0.1% of the population has a broken arm and x-rays are only 99% accurate. You have one and it says the arm is broken. The probability of you have a broken arm is not 0.1% * .99%. Before the test you may be 95% sure you have a broken arm so the probability is .95% * .99%. I have made up the numbers, just for illustration purposes.