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There are 10 differente figures (A,B,C,D,E,F,G,H,I,J), all equally probable to get.

We bought 25 pack which each contains only 1 figure

What is the porbability if we bought 25 packs we got all the differente figures, we found all the figures?

I figured that the total numbers of outcomes will be based on repetitive combination so (25-1+10 10) but i don't know how i should calculate the numbers of possible good outcomes

No_Name
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  • What have you tried? – David G. Stork May 12 '20 at 17:46
  • Please show your efforts on the question to avoid it being closed or heavily downvoted. See here to know how to ask a good question on this website. (Please take note that this website does not do your homework for you) – sai-kartik May 12 '20 at 17:52
  • well i figured out, that the sum of the outcomes will be based on repetitive combination so (25-1+10 10), but i can't get around my head that how i should calculate the number of ways this event can happen. – No_Name May 12 '20 at 17:52
  • i don't understand is it given that we found all the figures after opening the packs? is it a conditional probability? what is the second sentence meant to be "we found all the figures?" – CSch of x May 12 '20 at 17:55
  • @OUR i just wanted to emphasis that we find all the 10 differente figures after opening all of the 25 packs, so it doesn't matter if we found 1 figure multiple times as long as we found at least one out of the remaining 9 - this was just an example – No_Name May 12 '20 at 17:58
  • Have you read about the coupon collector's problem? This is not the same, but the logic is useful. – Ross Millikan May 12 '20 at 17:58
  • If ${25-1+10\choose 10}$ corresponds to stars and bars method, you can assume that in good outcome each bar has a star 'glued' to it. Hope that helps. – Alexey Burdin May 12 '20 at 18:02
  • @Alexey Burdin so you are saying that each bar has one star glued to it, which means 25 - 10 = 15, which means we have 15 figures left to place, so then (15−1+10 10) or am i wrong? – No_Name May 12 '20 at 18:25
  • Seems right, but I'm not that sure, just wanted to point the approach if it makes sense. – Alexey Burdin May 12 '20 at 18:28
  • I dont think stars & bars are applicable here. That counts the number of ways to distribute 25 indistinguishable packs into 10 distinguishable figures, but not every such way is equi-probable if the underlying model is that each pack is equi-probable to be any of the 10. – antkam May 12 '20 at 19:54
  • @joriki: only that it wants the probability of all coupons with a given number of draws, not the expected number of draws to get them all. I didn't know about the one you link to. I agree it is a duplicate – Ross Millikan May 12 '20 at 23:51
  • @RossMillikan: OK, thanks, I closed it. I collected the questions about the coupon collector problem that often come up as duplicates in this answer in the List of Generalizations of Common Questions. (Does that make me a coupon collectors' collector? :-) – joriki May 12 '20 at 23:55
  • @joriki Since you closed my question and said it was a duplicate, i looked into the "Probability distribution in the coupon collector's problem" which you gave me, but i don't understand that how it connects to my problem? I see what you dod there but in the m∑j=1m1/j : how should i use this function with my problem? – No_Name May 13 '20 at 12:11
  • @No_Name: That's hard to read; please use MathJax for formatting; here's a tutorial and reference. I'm guessing that you're referring to the expected value at the end of the answer. I specifically linked to that question because it doesn't just give the well-known expected value, but the entire probaility distribution, which is what you'd asked about. I'm not sure why you're focussing on the expected value. – joriki May 13 '20 at 14:05
  • @joriki Well i found a formula online which is called poincare formula and using this formula my final result looks like this : $1-\sum_{i=1}^{10}(-1)^{i-1}\binom{10}{i}(\frac{10 - i}{10})^{25}$ . So is it wrong, or am i using it wrong? – No_Name May 13 '20 at 18:23
  • @No_Name: Did you look at the linked answer? – joriki May 13 '20 at 19:43

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