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:
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))....
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.