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klimenkov Group Title

There are \(n\) lottery tickets in sale. \(m\) of them are the winning tickets. You have bought \(r\) tickets. What is the probability for you to win in the lottery. P.S. You win if you have one or more winning tickets.

  • 2 years ago
  • 2 years ago

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  1. wio Group Title
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    The probability that a ticket wins is: \[\frac{m}{n}\] The probability that it loses is thus: \[1 - \frac{m}{n}\] The probability that you have \(r\) loosing tickets is thus: \[\left(1 - \frac{m}{n}\right)^r\] The probability that NOT all tickets loose (at least one tickets wins) is thus: \[1 - \left(1 - \frac{m}{n}\right)^r\]

    • 2 years ago
  2. klimenkov Group Title
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    If I have bought more than \(n-m\) tickets, so \(r>n-m\) I will 100% win. But your probability is not equal to 1 even if \(r=n\). Think again.

    • 2 years ago
  3. wio Group Title
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    Right, so we have to take into account the fact that with each loss, the probability of a ticket failing decreases. So the probability that a ticket loses at the start is \[1-\frac{m}{n}\] and for the second one it is: \[1-\frac{m}{n-1}\] So we get \[1 - \prod_{k=0}^{r-1} \left(1 - \frac{m}{n-k}\right)\] Umm how about that?

    • 2 years ago
  4. klimenkov Group Title
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    Holy ... . Let me think about your answers...

    • 2 years ago
  5. klimenkov Group Title
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    @experimentX you was so close...

    • 2 years ago
  6. wio Group Title
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    Am I still missing something?

    • 2 years ago
  7. klimenkov Group Title
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    @wio I think about your answer. I think it must be right.

    • 2 years ago
  8. klimenkov Group Title
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    Yes. You are right.

    • 2 years ago
  9. experimentX Group Title
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    why was it ?? \[ \huge {\sum\limits_{k=1}^r \binom{m}{k}\binom{n-m}{r-k}\over \binom{n}{r}} \]

    • 2 years ago
  10. klimenkov Group Title
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    My answer was \[1-\frac{C_{n-m}^r}{C_n^r}=1-\frac{A_{n-m}^r}{A_n^r}\]I don't know what letters are used in other countries, but we use this letters. I hope you will get it. \(C_n^k=\left(\begin{matrix}n \\ k\end{matrix}\right)\) and \(A_n^k=k!C_n^k\).

    • 2 years ago
  11. klimenkov Group Title
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    Actually it equals \[\frac1{C_n^r}\cdot\sum_{i=0}^{r-1}C_{n-m}^iC_m^{r-i}\]

    • 2 years ago
  12. experimentX Group Title
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    yeah ... just confused about that index. Damn indexes. http://en.wikipedia.org/wiki/Vandermonde's_identity I would look funny if it were for real.

    • 2 years ago
  13. wio Group Title
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    There is more than one way to do it. You can do it the way where you add up every way you can win (win with 1, win with 2, win with r) since they are mutually exclusive events.

    • 2 years ago
  14. klimenkov Group Title
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    Yes you are both right! You can get the Vandermonde's identity solving this problem! Just put the "=" sign between 2 solutions! \[\frac1{C_n^r}\cdot\sum_{i=0}^{r-1}C_{n-m}^iC_m^{r-i}=1-\frac{C_{n-m}^r}{C_n^r}\]If you do some simplifications you will get what Wiki says about - the Vandermonde's identity!

    • 2 years ago
  15. klimenkov Group Title
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    It is a pity that this article is in English. And there is no translation in Russian...

    • 2 years ago
  16. klimenkov Group Title
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    Hope, you enjoyed this problem.

    • 2 years ago
  17. experimentX Group Title
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    hope that works for you!! i see things in russian when i look english in mirror!!

    • 2 years ago
  18. klimenkov Group Title
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    The translation is worse for me than the original text. My English wants to be better.

    • 2 years ago
  19. experimentX Group Title
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    Woops sorry!!

    • 2 years ago
  20. klimenkov Group Title
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    No problem. I think you just want to help me. Thank you.

    • 2 years ago
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