1
$\begingroup$

Suppose that you are diagnosed with microscopic hematuria (blood in the urine that is only visible under a microscope). This symptom occurs in 10 percent of all people and 100 percent of people with kidney cancer. You would like to know the probability that you have kidney cancer, which occurs in 0.0002 percent of all people.

Now, when I solve it with Bayes' Theorem, the answer is as follows;

[1.0][0.000002]/[0.1] = .00002

But the thing is, what if the chance of the symptom occuring increased to 20%? Then the answer would be .00001, but shouldn't the answer also increase? What am I missing?

$\endgroup$
  • 1
    $\begingroup$ I haven't checked your work, but just on the face of it, if there are more "red herrings" in the general populace, I would expect there to be more of a chance that it is a false positive and less of a chance that it is a true positive, wouldn't you? $\endgroup$ – MPW Mar 23 '14 at 14:38
  • $\begingroup$ To expand on the previous correct comment, consider what would happen if the symptom occured in 100% of people. Then the answer would be .0002 percent, i.e. the same as the chance of having cancer, i.e. the test would tell you nothing, which is what you would expect if everyone tested positive. $\endgroup$ – Tom Collinge Mar 23 '14 at 15:18
  • $\begingroup$ Oh, thanks guys. I don't know why made me think like that. Shame I can't upvote your comments due to my being newbie. $\endgroup$ – user137387 Mar 23 '14 at 15:54
  • $\begingroup$ Good luck from here on. $\endgroup$ – Tom Collinge Mar 23 '14 at 16:30
1
$\begingroup$

Going through your calculations, we have

Probability of hematuria $= P(H) = 0.10$,

Probability of kidney cancer $= P(C)=0.000002$,

Probability of hematuria given kidney cancer = $P(H\mid C) = 1$,

So, you working (which appears to be correct) shows that the probability of kidney cancer given hematuria is

$$ P(C\mid H)=\frac{P(H\mid C)p(C)}{p(H)}=\frac{1\times 0.000002}{0.1}=0.00002$$

If, however, $P(H)=0.20$, then $P(C\mid H)$ is halved to $0.00001$ which is expected.

This is because the probability of kidney cancer is the same, but you have twice as many people exhibiting the symptom of hematuria. In other words you will have more people diagnosed with hematuria but who do not have kidney cancer.

So the probability of someone having kidney cancer given that they exhibit hematuria will reduce - it will be halved, as you have twice the number diagnosed with hematuria.

$\endgroup$
  • $\begingroup$ @Michael Hardy - many thanks for improving the formatting. $\endgroup$ – Alijah Ahmed Mar 23 '14 at 15:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy