I have a population of N=100 and I am taking a simple random sample without replacement of n=10. I am trying to figure out the inclusion probability of 1 unit (we can call it unit one) of the 100 possible units for the sample. My idea is that after selecting unit one, the other 9 units are represented by 99 choose 9, meaning there are 99 choose 9 random sample that include unit one. Then you take that and divide by the total number of possible sample, which is represented by 100 choose 10. So the answer is

99 choose 9/ 100 choose 10 = .1

Can someone verify that my answer is correct or incorrect?

  • 1
    $\begingroup$ It's actually easier than that. The chance the first one you take is unit one is 1/100, and it's the same for the other nine (doesn't matter that the sampling is without replacement). Now add up these probabilities to get 0.1. $\endgroup$ – dsaxton Sep 17 '17 at 1:36
  • 1
    $\begingroup$ @dsaxton, Christopher's argument is correct, and yours isn't. You have to enumerate the samples for each event of interest (along with their probabilities if they aren't equal -- does not matter in this particular case though), and take the ratio (in this case, the denominator is all of the probability space). $\endgroup$ – StasK Sep 17 '17 at 14:51
  • 1
    $\begingroup$ @StasK The argument is fine. Write the event where the item is selected as $\cup_{i=1}^{10} A_i$ where $A_i$ is the event that the $i^\text{th}$ element is the one we want. The $A_i$ are disjoint so the probability we're after is $\sum_{i=1}^{10} P(A_i) = 1/10$ since $P(A_i)$ is 1/100 for each $i$. $\endgroup$ – dsaxton Sep 17 '17 at 18:19
  • 1
    $\begingroup$ This argument is even wronger. If in doubt, use the probability of elementary outcomes (each specific sample), and that's what Christopher has demonstrated. You are adding some marginal probabilities, and by some wild luck you may have this right -- but this is not a generalizable argument. $\endgroup$ – StasK Sep 19 '17 at 3:52
  • $\begingroup$ @StasK I think you're both correct in your approaches, but in order to do so I have to imagine you are misinterpreting the argument by dsaxton, which looks convincing. $\endgroup$ – whuber Oct 8 '17 at 21:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.