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**tttcomrader** Consider a test for a certain illness. Say that the probability of a false-negative is 0.2 and a false-positive is 0.01. Suppose that a disease carrier and 4 non-carriers get tested, what is the probability that we will get at least 2 positive result out of them? And what is the expected number of positive tests?

My idea:

So to get at least 2 positive result is saying the probability of having either:

i. The disease carrier get positive and at least one of the non-carrier get false positive.

ii. The disease carrier gets false negative and at least two of the non-carrier get false positive.

Now, I believe I can use the binomial distribution for the non carrier part, implies:

P( disease carrier pos and at least one non-carrier get pos) = $\displaystyle (.8) b(X \geq 1;4,0.01) = (.8) [ 1 - \frac {4!}{4!} (0.01)^0(0.04)^4 ]$

P( disease carrier neg and at least two non-carrier get pos) = $\displaystyle (.2) b(X \geq 2;4,0.01) = (.8) [ 1 - b(1;4,0.01) - b(0;4,0.01) ] $

Then do I just add them up?

And how do I find the expected value? I know I need to do $\displaystyle E[X] = np $, but I'm not so sure about the extra stuff that comes with the disease carrier.

Thank you!