A box contains 5 electric devices, 2 of them are damaged. We check them randomly each one at a time, until we find out which ones are damaged. Calculate the mean of the number of checks we need to make.
I thought it's a Negative Binomial distribution, thus, it should be:
E(X)=r/p=2/0.4.
but the correct answer should be 3.5.
I assume you don't replace the ones you inspect, hence it's a hypergeo
If you do replace them, then it's a binomial, not neg binomial
And I'm not sure what your random variable is here.
Is X=number you must observe in order to find the second defective?
Then x=2,3,4,5 only.
P(X=2)=P(defective on first and second)=(2/5)(1/4).
P(X=3)=P((one defective on first or second pick) and (defective on third))....
I think x = 6 will also be needed.
eg. first selected device could be damaged, next four selected could be undamaged. So to observe the second damaged device a sixth selection is needed. So a maximum of six checks will be needed.
Edit: I'm not so sure the hypergeometric distribution can be directly used here - because the order is important, the last selection has to be a damaged device eg. *, *, D, *, D versus *, *, D, D, *. In each case there're been five selections that would get counted by the hypergeometric distribution, but only the first is valid in this question.