Chance of having covid given negative test

Question

I was reading an article (https://www.healthline.com/health/how-accurate-are-rapid-covid-tests#at-home-tests) on the accuracy of Rapid Lateral flow tests and was interested in finding out what my chance of having covid are given that I have tested negative.

The article stated that people without symptons .....

• If they have covid they correctly test positive 58.1% So from this can we assume that ..
• If they have covid they incorrectly test negative 41.9%?

So if I want to work out p(c|neg) then this is worked out by formula p(c,neg) / (p(c,neg)+p(notC,neg)

The article didn't say what the real % of people in population who actually have covid is. I guess because this is something that we don't really know for certain. But I read one estimate being 39.42 out of 10000 people, so 0.003942

I read in another article that the specitivity of Rapid flow tests ranged from 0.924 to 1 so if I take the middle of the range as being 0.962

So the probabilities are .. p(c) = 0.003942 p(nc) = 0.9961 p(neg|c) = 0.419 p(neg|nc) = 0.962

So does this mean that p(c|neg) = (0.003942x0.419)/(0.003942x0.419)+(0.9961x0.962)) = 0.0017 or 0.17%. I feel like I must have made a mistake in my calculation somewhere.

Also, if I take 5 tests all on the same night and they all return negative results does this mean that the chance that I have covid is 0.0017^5 or 1.419857e-14

I wanted to work out what my chance of having covid anytime over the last year would be given I have had say 8 negative test results, but this all depends on when the tests were taken i.e. if I take one test and get a negative result and then take another test and get a negative result the chance of the second test being accurate is higher than if it was taken 3 weeks later. And it would all depend on the rate of covid within population when the tests were taken?

Answer 1

$S:\quad$ symptomatic
$C:\quad$ COVID
$+:\quad$ positive test result

What you are interested in called the false omission rate $P(C|-),$ which I've written about it in some detail here and here. The following answer will be more satisfying after having returned from those two.

The article stated that people without symptoms.....

• If they have covid they correctly test positive 58.1% So from this can we assume that ..
• If they have covid they incorrectly test negative 41.9%?

The original passage

The researchers found that people without COVID-19 symptoms correctly tested positive in 58.1 percent of rapid tests.

is ambiguous, as the $58.1\%$ could refer to $$P(+|S^c\cap C)$$ or $$P(C\cap+|S^c)$$ or $$P(C|+\cap S^c).$$

I'm guessing the intended meaning is the first option, which agrees with your suggestion about the $41.9\%.$

So the probabilities are ..
p(c) = 0.003942
p(nc) = 0.9961
p(neg|c) = 0.419
p(neg|nc) = 0.962

So does this mean that p(c|neg) = (0.003942x0.419)/(0.003942x0.419)+(0.9961x0.962)) = 0.0017 or 0.17%. I feel like I must have made a mistake in my calculation somewhere.

Your derivation is correct, but according to our interpretation above, it is $P(-|S^c\cap C)$—not $P(-|C)$— that equals $41.9\%.$ Thus, the required value isn't $0.17\%.$

Also, if I take 5 tests all on the same night and they all return negative results does this mean that the chance that I have covid is 0.0017^5 or 1.419857e-14

No, you want $P(C|-\cap-\cap-\cap-\cap-)$ instead of $\big[P(C|-)\big]^5.$ Refer to the bottom of the first linked answer above for an explanation.

And it would all depend on the rate of covid within population when the tests were taken?

Yes. This is also addressed in the linked answers.

Answer 2

This is a very classical exercise on Bayes theorem. And the particular instance you are studying is known as "False Positive Paradox" or "Base Rate Fallacy." Check this interactive calculator by no other than the British Medical Journal (i.e. perhaps the most prestigious medical journal out there): https://www.bmj.com/content/373/bmj.n1411/rr

Let $+$ and $-$ denote a positive and negative test, respectively, and denote by $\mathrm{C}$ the event of having covid; we use $\mathrm{Not\ C}$ for the complement of this event. By definition, the sensitivity is $P(+ \mid \mathrm{C})$ and the specificity is $P(- \mid \mathrm{Not\ C}).$ (In other words, the sensitivity is the true positive rate and the specificity is the true negative rate. Intuitively, you want both the specificity and sensitivity to be close to 1.) Law of total probability (misnomered "Bayes Theorem", which by the way, Bayes never formulated) states that $$ P(\mathrm{C} \mid +) = \dfrac{P(\mathrm{C}, +)}{P(+)} = \dfrac{P(+ \mid \mathrm{C}) P(\mathrm{C})}{P(+ \mid \mathrm{C}) P(\mathrm{C}) + P(+ \mid \mathrm{Not\ C}) P(\mathrm{Not\ C})}. $$ We can further use that $P(\mathrm{Not\ C}) = 1 - P(\mathrm{C})$ so that if you know the sensitivity an specificity, to calculate $P(\mathrm{C} \mid +)$ you only require the number $P(\mathrm{C}).$ This number is typically approximated by the fraction of people who have the disease (and are infectious). The "fallacy" or "paradox" I was talking about comes from the case of rare diseases; if $P(\mathrm{C})$ is very small and both the sensitivity and specificity are high, at around 0.9 say, then $$ P(\mathrm{C} \mid +) = \dfrac{(0.9)10^{-k}}{(0.9)10^{-k} + (0.1)(1-10^{-k})} \approx \dfrac{10^{-k}}{10^{-k} + 10^{-1}} \approx 10^{-k+1} $$ and this approximation will be much less than the sensitivity if $k$ is large. In other words, if a disease is rare, even a 90% accurate test will yield a true positive about 10 times more frequently than the proportion of infectios people.

Some important considerations:

• If you tested everyone (as they do in China when a 5million city has 50 cases of covid, for instance), then a massive number of positive test will be false positives. Massive testing is a control measure and not a health measure for most people who test positive will not be positive.

• Reexposure to COVID is different than this as you are in contact time and again with the same and new people. So, this method does not solve how likely are you to have covid. In fact, your chances of contracting covid are quite difficult to pin-point as it seriously depends on the infectiousness of the virus itself, who has it, how much exposure you where subjected to and how vulnerable is your body to the disease. For instance, with Omicron, the chances of catching it seems as of yet to be close to 1 (as long as you are in contact with other people).

• To be clear, what this shows is that if you tested everyone, then you'd expect a much smaller proportion of people to have covid than those who tested positive.

• This method also applied for someone selected at random (if you have symptoms, this is already not at random and you'd need to redefine your sample space appropriately; in other words, you'd want the probability of actually having covid given that you tested positive and have symptoms, the probability found this way is likely to not be "paradoxical"). For example, you one day, "for peace of mind", decided to get tested. It would be more likely to reach quite the contrary (i.e. some anxiety instead) as if you tested positive, you may actually be negative.