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.
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In the problem from before the probability that a patient who tested positive once on this cancer test actually having cancer is 8 / 1 0 3 or 7 . 8 %.
Bayes theorem is P ( A ∣ B ) = P ( B ) P ( B ∣ A ) × P ( A ) = P ( B ∣ A ) × P ( A ) + P ( B ∣ − A ) × P ( − A ) P ( B ∣ A ) × 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 )
Where: P ( c a ∣ p o s ) is the probability that the patient has cancer given a second positive result on this second test.
P ( p o s ∣ c a ) is the probability that if a positive result appears on the test, then patient has cancer.
P ( c a ) 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 % chance.
P ( p o s ∣ − c a ) is the probability that if a a positive result appears on the test, then patient does not have cancer.
P ( − c a ) is the probability that any patient does not have cancer given that they tested positive one the first test. Which is now no longer 9 9 % but 1 0 0 % − 7 . 8 % = 9 2 . 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 . 8 0 × 0 . 0 7 8 ) + ( 0 . 0 9 6 × 0 . 9 2 2 ) ( 0 . 8 0 ) × ( 0 . 0 7 8 ) = .41... = 41%
Which means, that after two positive results, the odds of the patient having cancer increase from 7.8% to 41%.