Bayes' Theorem in Clincal Settings: Testing For Cancer a Second Time

In a previous problem on a patient who receives a positive result in a cancer screen , we determined what the odds were that this patient has cancer.

But what are the odds that the patient has cancer if they are tested a second time and receive a positive result for the second test as well?

From before, you should assume there is a 1% prevalence rate (that 1% of the population has breast cancer), a hit rate (sensitivity) of 80%, and a false positive rate of 9.6% on the initial test.

Please provide this percentage answer as an integer with no remainder, for instance, 11 for 11.2137%, as opposed to entering 0.11.

Hint: You will need the answer to the previous problem, linked to again here.


The answer is 41.

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1 solution

In the problem from before the probability that a patient who tested positive once on this cancer test actually having cancer is 8 / 103 8 /103 or 7.8 7.8 %.

Bayes theorem is P ( A B ) = P ( B A ) × P ( A ) P ( B ) = P ( B A ) × P ( A ) P ( B A ) × P ( A ) + P ( B A ) × P ( A ) P(A\mid B) = \frac{P(B \mid A) \times P(A)} {P(B) } = \frac{ P(B\mid A) \times P(A)} {P(B\mid A) \times P(A) + P(B\mid -A) \times P(-A)}

Let's use similar notation to before P ( c a p o s ) = P ( p o s c a ) × P ( c a ) P ( p o s c a ) × P ( c a ) + P ( p o s c a ) × P ( c a ) P(ca\mid pos) = \frac{ P(pos \mid ca) \times P(ca)} {P(pos\mid ca) \times P(ca) + P(pos\mid -ca) \times P(-ca)}
Where: P ( c a p o s ) P(ca\mid pos) is the probability that the patient has cancer given a second positive result on this second test.
P ( p o s c a ) P(pos \mid ca) is the probability that if a positive result appears on the test, then patient has cancer.
P ( c a ) P(ca) is the probability that any patient has cancer given that the patient already tested positive once. This is where the previous problem comes in. It is no longer a 1% change of having cancer, but a 7.8 % 7.8\% chance.
P ( p o s c a ) P(pos\mid -ca) is the probability that if a a positive result appears on the test, then patient does not have cancer.
P ( c a ) P(-ca) is the probability that any patient does not have cancer given that they tested positive one the first test. Which is now no longer 99 99 % but 100 100 % 7.8 - 7.8 % = 92.2 = 92.2 %


Now we can plug in the given variables:
P ( c a p o s ) = P ( p o s c a ) × P ( c a ) P ( p o s c a ) × P ( c a ) + P ( p o s c a ) × P ( c a ) = ( 0.80 ) × ( 0.078 ) ( 0.80 × 0.078 ) + ( 0.096 × 0.922 ) P(ca\mid pos) = \frac{ P(pos \mid ca) \times P(ca)} {P(pos\mid ca) \times P(ca) + P(pos\mid -ca) \times P(-ca)} = \frac{(0.80) \times (0.078)} {(0.80 \times 0.078) + (0.096 \times 0.922)} = .41... = 41%

Which means, that after two positive results, the odds of the patient having cancer increase from 7.8% to 41%.

I know that this requires the two sequential tests to be independent....if the tests are correlated then the post-test odds following the second positive test will be <41%.

I suspect that the latter can be shown mathematically but I don't know how.

Any ideas?

Cello Sabbatical - 4 years, 5 months ago

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Actually these two tests are dependent. The first test is "What are the odds a patient has cancer if they test positive?" This one is "What are the odds that a patient has cancer if they get two positive tests in a row?". This incorporates the additional evidence of the first test being positive into determining if the second test confirms cancer. In fact, in our solution, we changed the odds from 1 % 1\% to 7.8 % 7.8\% because the first test was positive. It's no longer 1 % 1\% of the population that has cancer, it's 7.8 % 7.8\% of the population who gets a positive result on their first test has cancer.

I think what you're asking is about two independent tests. We can imagine this patient running to a new Bayesian doctor and not telling them about the first test. That second doctor unknowingly repeats the first test, gets a positive result, what would the doctor say the patient's odds of actually having cancer are? Well, if he did the Bayesian analysis we did in the first problem he'd give them the same answer: 7.8 % 7.8\% . But we know this would be based in ignorance, ignorance that there was a first test that had also come back positive. The key point with Bayes' theorem is incorporating evidence to update the statistical likelihood of some outcome.

Christopher Williams - 4 years, 4 months ago

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This is Much better of an amazing explanation, thanks!

Hans Henry - 3 years, 9 months ago

I think that what Cello Sabatical was proposing is an scenario where both tests are run in the same frame of time, same machine, same personal... This should imply that your second test is less independent from the first, so its validity should be reduced from that 41% to somewhat lower.

Now that's on the spotlight the COVID-19 we could make an analogy with a "quick test". If you are tested inmediatly with the same batch of tests, same person, etc. one new positive is less reliable than if it were to be done with another batch of test for COVID (even if with another "brand" if both technologies are identical, but only changes the maker).

Félix Pérez Haoñie - 1 year, 1 month ago

Maybe 0.99 insted of 0.922? It is no longer a 1% change of having cancer, but a chance. Why?

Igor Kolupaev - 3 years, 3 months ago

It would be helpful if it were stated that the two tests are independent events. The probability is incomputable without this fact.

Daniel Pendergast - 1 year, 5 months ago

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