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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.
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: \[1 - \left(1 - \frac{m}{n}\right)^r\]
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.
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?
Holy ... . Let me think about your answers...
@experimentX you was so close...
Am I still missing something?
@wio I think about your answer. I think it must be right.
why was it ?? \[ \huge {\sum\limits_{k=1}^r \binom{m}{k}\binom{n-m}{r-k}\over \binom{n}{r}} \]
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 \(A_n^k=k!C_n^k\).
Actually it equals \[\frac1{C_n^r}\cdot\sum_{i=0}^{r-1}C_{n-m}^iC_m^{r-i}\]
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.
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.
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!
It is a pity that this article is in English. And there is no translation in Russian...
Hope, you enjoyed this problem.
hope that works for you!! i see things in russian when i look english in mirror!!
The translation is worse for me than the original text. My English wants to be better.
No problem. I think you just want to help me. Thank you.