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I have essentially infinite number of balls, ⅓ each of 3 colors. How many must I blindly draw to be 95% confident I have 1 or each color? I think the answer is 11, but I'd like the equation.

geneticist
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  • Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking. – Community Apr 26 '22 at 15:53
  • Why do you think the answer is 11? Can you please add the work you did to get to 11 in your post? That is what we expect a quality question post to do. And that will help us focus our feedback. – Bram28 Apr 26 '22 at 16:03
  • Suppose that you draw exactly $5$ balls. What is the exact probability that the $5$ balls do not contain at least $1$ of each color? Suppose that the $3$ colors are red, blue, green. What is the probability that (for example) neither of the 1st $2$ balls drawn are red? – user2661923 Apr 26 '22 at 16:50
  • I could be mistaken, but it looks like you are going to have to be concerned about Inclusion-Exclusion. See this article for an introduction to Inclusion-Exclusion. Then, see this answer for an explanation of and justification for the Inclusion-Exclusion formula. – user2661923 Apr 26 '22 at 16:52
  • To answer Bram28: This problem was solved many years ago, I can't recall it correctly. As I recall the left side of the equation began "1-alpha = ", alpha representing the confidence interval or 0.95. So the equation was determining the 0.05 chance that the balls drawn did not contain all three colors. The equation did not involve combinations, or factorials. I recall the equation, once solved, being something like 10.43???, meaning I had to select 11 balls. – geneticist Apr 27 '22 at 19:22
  • Response to user2661923: I recall the answer was a straightforward equation, nothing with all those set functions. It just had (1-n)^⅔ * (1-n)^⅓. or something like that. Something that even I could solve with algebra – geneticist Apr 27 '22 at 19:26
  • @geneticist Consider $$\frac{N}{D} ~: D = 3^n$$ and $N$ is the number of choices possible, (where order of selection is regarded as pertinent), and where, after $n$ selections, there are at least one of each of the $3$ colors. Let $S_1,S_2,S_3$ denote the respective ways of choosing $n$ items, that are missing a Red ball, Green ball, or Blue ball, respectively. Then $N = D - |S_1 \cup S_2 \cup S_3|.$ ...see next comment – user2661923 Apr 28 '22 at 01:25
  • For this particular problem, this approach is extremely painless: see also this answer for an explanation of and justification for the Inclusion-Exclusion formula. – user2661923 Apr 28 '22 at 01:25

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