<p>A certain disease occurs in 1 out of 101 people. A test for the disease is 100% accurate for patients with the disease and 99% accurate for patients with out it. That is it gives a “false positive” 1% of the time even if the person tested does not have the disease. If you take this test and it returns a positive result, what is the probability that you have this disease?</p>
<p>Consider this population of 101 people. Theoretically, one of them should be sick, and 100 of them should be healthy. The sick person will have a positive test. In addition, one of the healthy people will have a positive test, too.</p>
<p>That’s two positives. Out of the two positives, how many of them actually have the disease? One.</p>
<p>So if you have a positive result, the probability that you’re sick is 1/2.</p>
<p>But you know what makes this such a hard SAT question? The fact that it wouldn’t be on the SAT!</p>
<p>The thing about this question is, it’s not really that hard if you know some probability, but it’ll probably never appear on the SAT.</p>
<p>Unfortunately many professionals in business, in medicine and in law, while often encountering real-world problems such as this, do not know how to address them. As a result they draw wrong conclusions and potentially they make bad decisions.</p>
<p>The general statement of this problem is “Given that the probability of a false positive in the diagnosis of a rare disease is low – what is the probability that you have the disease when the diagnosis is positive?” Intuitively if the probability of the false positive is high compared to the rareness of the disease then you probably don’t have the disease!</p>
<p>This is a conditional probability problem. When the test result is positive either “you do not have the disease AND the diagnosis is a false positive” or “you have the disease.” </p>
<p>Consider N total trials where N is a large number. Of these N trials, only (100/101)xNx(1/100) + (1/101)xN result in a positive – So of the N trials (100/101)x(1/100)xN are false positives and (1/101)xN are true positive, a total of (1/101 + 1/101)xN trials. We need consider just these. Of these (1/101)xN are true positives – or 1/2 of the total. So the probability that you have the disease is 1/2.</p>
<p>Another Example:
Suppose that in the general population 1/5000 people have a mutation M. A test to determine whether or not the mutation is present has a 1/500 false error rate. Suppose you are diagnosed with the mutation. What is the probability that this is a valid diagnosis?</p>
<p>I am sorry,but i don’t get ur explanations to the questions… :/</p>
<p>and for the other example: is it 499/500??</p>
<p>oh wait… after i reread sikorsky’s explanation, I think it might be 1/2 as well… but i m not sure at all…</p>
<p>The knowledge of the test result (i.e. positive) limits the relevant sample space to that of positive test results. Consider only these. You do not consider negative test results because you know that these are excluded – the test result is positive! Within this limited sample space compute the relative number of true positives to ALL of the positives (true or false). Then you have the probability of a true positive.</p>
<p>In the exercise the computation is:</p>
<p>(1/5000)/[(1/5000) + (1/500)(4999/5000)]</p>
<p>Note that the denominator is the measure of the space of “positive test results – either positive or negative”. The numerator is the measure of the true positives among these. The answer is about 1/11. Not surprising – not too likely that you have the mutation since the false positive rate is pretty high relative to the rareness of the mutation.</p>
<p>What if the mutation occurs less frequently – e.g. one in 5,000,000?</p>
<p>Once you get the gist of the analysis you’ve would have an intuitive understanding of conditional probability – an important and very common application of probability to real life situations.</p>