0

I learned about Bayes' theorem recently and hoped to apply it to the following problem:

1% of a country's population has a rare disease. A resident takes a disease test with a true positive rate of 80% and a true negative rate of 90%. If the test result is positive, what is the chance that the resident actually has the disease?

I only saw problems involving a true positive rate but now struggle to see how to also integrate a true negative rate into the equation. As I didn't find an understandable explanation in similar posts, I'll appreciate any help or explanations.

Alexanderfrs
  • 3
  • 3
  • 2
    Suppose you have $1$ million people in the country. How many do you expect to have the disease? How many of those having the disease do you expect to test positive and how many negative? How many do you expect not to have the disease? How many of those not having the disease do you expect to test positive and how many negative? So how many in total do you expect to test positive? What proportion of those testing positive do you expect to actually have the disease? – Henry Oct 26 '21 at 12:41
  • Please do not re-post [the same question](https://stats.stackexchange.com/questions/549651/bayes-theorem-for-a-rare-disease) after it was put on hold. Instead you can always edit your question adequately to the on-hold message: tell us what kind of help you need, what you don't understand, etc as very similar question was already given a detailed answer on the site. – Tim Oct 28 '21 at 06:55

1 Answers1

1

I will try to get you started with a formula for computation:

Bayes' Theorem for a partition consisting of sets $D$ and $D^c:$

$$P(D|+) = \frac{P(D\cap +)}{P(+)} = \frac{P(D)P(+|D)}{P(D\cap +) + P(D^c\cap +)}\\ =\frac{P(D)P(+|D)}{P(D)P(+|D) + P(D^c)P(+|D^c)}$$

Sometimes the denominator of the final term is said to be due to the 'Law of Total Probability'.

You know $P(D) = 0.01$ so that $P(D^c)=1-P(D) = 0.99.$ Then from the 'true positive' and 'true negative' rates given in the statement of the problem, you have (or can easily find by complementation) numerical values for the conditional probabilities to plug into the numerator and denominator. Finally, compute $P(D|+).$

Notes: Your answer should be consistent with @Henry's helpful Comment. Finally, I hope you can match my statement of Bayes' Theorem with a similar statement in your text or class notes.

Nowadays during the Covid pandemic, such computations are crucial. In case you are interested, this Q&A shows more advanced computations along these lines. Also, the Wikipedia article on Bayes' Theorem may be of interest. (It is a very long article, but you may find selected parts of it interesting.)

BruceET
  • 47,896
  • 2
  • 28
  • 76