# Math Help - Disease carrier problem with binomial dist.

1. ## Disease carrier problem with binomial dist.

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) = $(.8) b(X \geq 1;4,0.01) = (.8) [ 1 - \frac {4!}{4!} (0.01)^0(0.04)^4 ]$

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

Then do I just add them up?

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

Thank you!

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) = $(.8) b(X \geq 1;4,0.01) = (.8) [ 1 - \frac {4!}{4!} (0.01)^0(0.04)^4 ]$

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

Then do I just add them up?

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

Thank you!
It looks to me like there's missing information. eg. Pr(disease carrier) is unknown but required.