1. ## Negative Binomial ?

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.

2. Originally Posted by WeeG
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.
Let N be the random variable number of devices removed until two defectives are obtained.

Calculate Pr(N = 2), Pr(N = 3), .... Pr(N = 6) and then calculate the expected value of N in the usual way.

3. 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))....

4. Originally Posted by matheagle
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.
[snip]
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.

5. But there are only 5 devices.

6. Originally Posted by matheagle
But there are only 5 devices.
Quite right. My mistake. I'd misread as 5 devices and two defectives (= 7 devices).

But I'm still not convinced the hypergeometric distribution is directly applicable (I await being exposed as an ignoramus for the second time in my life).

7. I don't think it's hyper either, only my misbehaving children are hyper