I am stuck at this question. I am trying to use Bayes Theorem for it and the try imagining 10,000 technique but still can't seem to solve it.

A test for a disease is 95% accurate if the disease is present, and 99% accurate if it is absent. 5% of the population have the disease.

(a) In a sample of 4 people, what is the probability that one actually has the disease? (b) For the same sample, what is the probability that at least one person actually has the disease?


closed as off-topic by user99914, NCh, max_zorn, Micah, A. Pongrácz Sep 11 '18 at 6:04

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  • 1
    $\begingroup$ I think you are missing something in the wording of your question, probably about testing positive $\endgroup$ – XRBtoTheMOON Nov 29 '17 at 16:45
  • $\begingroup$ That is what I have been provided with. $\endgroup$ – user508281 Nov 29 '17 at 16:49
  • $\begingroup$ But then the information about the test's accuracy is irrelevant. Are you sure this is exactly the problem's wording? $\endgroup$ – Francisco José Letterio Nov 29 '17 at 20:41
  • $\begingroup$ Unfortunately, I feel I had to flag this. This isn’t a site that will solve problems for you, and some effort or examples of your own thoughts must be shown. That way, we can help you—and others—the best we can. For more information, please read how to ask a good question. At any rate, I wish the best of luck to you, and please continue to contribute to our wonderful site! $\endgroup$ – let's have a breakdown Nov 29 '17 at 23:30
  • $\begingroup$ I am new to this site and won't repeat the same mistake again. $\endgroup$ – user508281 Nov 30 '17 at 17:26

well assuming its worded correctly (it's not)

a) With 4 people, the probability that one has the disease, is 3 of them not having it, and 1 having it. Multiplied by the number of ways you could pick one of them to have it

$\binom{4}{1}.95^3.05 = .17$

b) The probability that at least 1 person has the disease, is 1 - (probability no one has disease)

$1 - .95^4 = .8145$

  • $\begingroup$ Many thanks, I think it can be solved via binomial formula. $\endgroup$ – user508281 Nov 30 '17 at 17:27

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