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Suppose after a set of tests you are 75% sure you have the antigen in your body. Then you run one more test. It returns positive. Now you are 95% sure. The false positive rate is 10%. What can you infer about the false negative rate?

Is there enough information to find the false negative rate? Im a bit lost on how to apply Bayes theorem to this question.

$P(Negative | has A) = \frac{P(has A | Negative)P(has A)}{P(Negative)}$

Is the 75% supposed to be treated as $P(has Antigen)?$ And how do i figure out $P(Negative)?$

CHTM
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  • Possibly helpful: https://math.stackexchange.com/questions/2279851/applied-probability-bayes-theorem/2279888#2279888 – Ethan Bolker Oct 22 '22 at 20:22
  • You don't mention any information about how rare is this disease which is necessary for this kind of problems. – Constantinos Pisimisis Oct 22 '22 at 20:30
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    There are two ways to test positive: either you have the antigen and get a true positive or you don't and you get a false positive. The probability of getting a false negative is linked to the probability of getting a true positive. – lulu Oct 22 '22 at 20:31
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    Should say: these problems are always a bit frustrating because the assumptions are unclear. Presumably we are to imagine that the tests are "independent" in the sense that errors (false positives and false negatives) occur or don't regardless of past trials. This isn't terribly realistic, however. Perhaps the false positives are produced by an antigen similar too but distinct from the one you want to test for...in which case you might expect a string of false positives. And so on. – lulu Oct 22 '22 at 20:34
  • @lulu So From what you said $P(Positive) = P(Positive|has Antigen)P(has Antigen) + P(Positive | not Antigen)P(not Antigen)$. But i dont know what $P(Positive | has Antigen)$ is? – CHTM Oct 22 '22 at 20:50
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    Not following. We are told the a priori probability that you have the antigen, that's $.75$ Now, the probability that you have the antigen and test positive is $.75\times P(\text {True Positive})$. That's your numerator. The denominator is that plus the probability that you don't have it but test positive anyway, which is $.25\times .1$ – lulu Oct 22 '22 at 20:55
  • @lulu See my answer. – user2661923 Oct 22 '22 at 20:59
  • user2661923's answer below is correct; consider Accepting(green-ticking) it? This diagram and glossary may be helpful. – ryang Feb 03 '23 at 17:34

1 Answers1

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While I agree with the real life concerns expressed in lulu's comment, I (also) think that the intent of the problem composer is clear.

Let $F$ denote the probability of a false negative, which is the variable that you are trying to solve for.

You are given the following premises:

  • After the first test, you are required to assume that, given no other information, the probability of having the antigen is $~\dfrac{3}{4}.$

  • The 2nd test showed positive.

  • As a result of the 2nd test showing positive, the new Mathematically computed probability that you have the Antigen is $~\dfrac{19}{20}.$

  • The probability of a false positive on the 2nd test is $~\dfrac{1}{10}.$

The intent of the problem composer seems clear to me. You are supposed to algebraically derive $F$, based on the above premises.

There are two distinct ways that the second test could show positive, which it did.

  • Possibility-1
    You actually have the Antigen, and the test was accurate.
    Under the assumption that you have the Antigen, you are given that the probability of a false negative is $(F)$. This implies that when you have the Antigen, the probability that the Antigen will be correctly diagnosed is $(1 - F).$

    Therefore, the probability of Possibility-1 occurring is
    $~\displaystyle \frac{3}{4} \times (1 - F).$

  • Possibility-2
    You do not have the Antigen and the test incorrectly reported that you do have the Antigen.

    The probability of this occuring is
    $\displaystyle \frac{1}{4} \times \frac{1}{10} = \frac{1}{40}.$

To simplify the Math, make the change of variable that

$$T = \frac{3}{4}\left[1 - F\right].$$

Then, using Bayes Theorem, the derived probability, after the 2nd test, of having the Antigen, may be expressed as

$$\frac{19}{20} = \frac{T}{T + \frac{1}{40}} \implies $$

$$\frac{3}{4} \left[1 - F\right] = T = \frac{19}{40} \implies $$

$$1 - F = \frac{76}{120} \implies F = \frac{44}{120} = \frac{11}{30}.$$

user2661923
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