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I start with 100 eggs, 10 of them being broken. I randomly select eggs without replacement until they are all split into baskets of 10 eggs each.

Here's what I know:

  • Best case scenario all 10 bad eggs go into 1 basket, giving me 9 good baskets out of 10. Worst case is each basket carries 1 bad egg giving me 0 good baskets.
  • Probability of any one basket being good, p, is 0.90^10=0.349.
  • Mean number of good baskets would be 10*0.349=3.49? However this calculation assumes replacements, or at least requires large number of baskets/eggs to be a good approximation. Every good basket created decreases the probability that the next basket(s) will be good.

For large numbers of eggs, say 50,000, is the non-replacement a big issue? Which distribution would allow me to calculate a confidence interval for the mean number of good baskets? Given that there are a large number of eggs (thousands), with known defect rate (around 1%) and fixed number of eggs per basket.

gregsan
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  • For a large numbers of eggs, the form of a reasonable approximation depends on whether you have a large number of baskets or a small number of baskets. – Henry Dec 17 '14 at 07:06
  • I might be wrong but in my intuition asking for the probability that a basket is full of good eggs is like asking for the prob. of hitting 0 winning tickets in a lottery with a total of 100 tickets and 10 winning tickets, i.e. a hypergeometric distribution. – j4GGy Dec 17 '14 at 07:10
  • @GenericNickname: Indeed the probability of any one basket being good is ${90 \choose 10}/{100 \choose 10}\approx 0.33$ so the expected number of good baskets is about $3.3$ – Henry Dec 17 '14 at 07:12
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    @Henry Can one really compute the expected number of good baskets like this? I mean the prob of two baskets being good is not just $0.33 \cdot 0.33$ since the events are not independent. – j4GGy Dec 17 '14 at 07:18
  • @GenericNickname: Yes, but linearity of expectation holds for dependent random variables too. Henry is computing the expectation of the sum of random variables $X_i = {\text{$1$ if basket $i$ is good, $0$ otherwise}}$. –  Dec 17 '14 at 07:25
  • @Rahul Right, I always seem to forget about the easiest solutions :) Thanks for clearing that up! – j4GGy Dec 17 '14 at 07:35
  • {90 \choose 10}/{100 \choose 10}\approx 0.33 makes sense to me. Is it possible to get a confidence interval around this mean? – gregsan Dec 17 '14 at 07:55

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