klimenkov
  • klimenkov
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.
Mathematics
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
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: \[1 - \left(1 - \frac{m}{n}\right)^r\]
klimenkov
  • klimenkov
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.
anonymous
  • anonymous
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?

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More answers

klimenkov
  • klimenkov
Holy ... . Let me think about your answers...
klimenkov
  • klimenkov
@experimentX you was so close...
anonymous
  • anonymous
Am I still missing something?
klimenkov
  • klimenkov
@wio I think about your answer. I think it must be right.
klimenkov
  • klimenkov
Yes. You are right.
experimentX
  • experimentX
why was it ?? \[ \huge {\sum\limits_{k=1}^r \binom{m}{k}\binom{n-m}{r-k}\over \binom{n}{r}} \]
klimenkov
  • klimenkov
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 \(A_n^k=k!C_n^k\).
klimenkov
  • klimenkov
Actually it equals \[\frac1{C_n^r}\cdot\sum_{i=0}^{r-1}C_{n-m}^iC_m^{r-i}\]
experimentX
  • experimentX
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.
anonymous
  • anonymous
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.
klimenkov
  • klimenkov
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!
klimenkov
  • klimenkov
It is a pity that this article is in English. And there is no translation in Russian...
klimenkov
  • klimenkov
Hope, you enjoyed this problem.
experimentX
  • experimentX
@klimenkov http://translate.google.com/translate?sl=en&tl=ru&js=n&prev=_t&hl=en&ie=UTF-8&layout=2&eotf=1&u=http%3A%2F%2Fen.wikipedia.org%2Fwiki%2FVandermonde%27s_identity&act=url
experimentX
  • experimentX
hope that works for you!! i see things in russian when i look english in mirror!!
klimenkov
  • klimenkov
The translation is worse for me than the original text. My English wants to be better.
experimentX
  • experimentX
Woops sorry!!
klimenkov
  • klimenkov
No problem. I think you just want to help me. Thank you.

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