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klimenkov
 4 years ago
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.
klimenkov
 4 years ago
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.

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0The 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
 4 years ago
Best ResponseYou've already chosen the best response.0If I have bought more than \(nm\) tickets, so \(r>nm\) I will 100% win. But your probability is not equal to 1 even if \(r=n\). Think again.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Right, 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: \[1\frac{m}{n1}\] So we get \[1  \prod_{k=0}^{r1} \left(1  \frac{m}{nk}\right)\] Umm how about that?

klimenkov
 4 years ago
Best ResponseYou've already chosen the best response.0Holy ... . Let me think about your answers...

klimenkov
 4 years ago
Best ResponseYou've already chosen the best response.0@experimentX you was so close...

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Am I still missing something?

klimenkov
 4 years ago
Best ResponseYou've already chosen the best response.0@wio I think about your answer. I think it must be right.

experimentX
 4 years ago
Best ResponseYou've already chosen the best response.0why was it ?? \[ \huge {\sum\limits_{k=1}^r \binom{m}{k}\binom{nm}{rk}\over \binom{n}{r}} \]

klimenkov
 4 years ago
Best ResponseYou've already chosen the best response.0My answer was \[1\frac{C_{nm}^r}{C_n^r}=1\frac{A_{nm}^r}{A_n^r}\]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
 4 years ago
Best ResponseYou've already chosen the best response.0Actually it equals \[\frac1{C_n^r}\cdot\sum_{i=0}^{r1}C_{nm}^iC_m^{ri}\]

experimentX
 4 years ago
Best ResponseYou've already chosen the best response.0yeah ... 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
 4 years ago
Best ResponseYou've already chosen the best response.0There 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
 4 years ago
Best ResponseYou've already chosen the best response.0Yes you are both right! You can get the Vandermonde's identity solving this problem! Just put the "=" sign between 2 solutions! \[\frac1{C_n^r}\cdot\sum_{i=0}^{r1}C_{nm}^iC_m^{ri}=1\frac{C_{nm}^r}{C_n^r}\]If you do some simplifications you will get what Wiki says about  the Vandermonde's identity!

klimenkov
 4 years ago
Best ResponseYou've already chosen the best response.0It is a pity that this article is in English. And there is no translation in Russian...

klimenkov
 4 years ago
Best ResponseYou've already chosen the best response.0Hope, you enjoyed this problem.

experimentX
 4 years ago
Best ResponseYou've already chosen the best response.0hope that works for you!! i see things in russian when i look english in mirror!!

klimenkov
 4 years ago
Best ResponseYou've already chosen the best response.0The translation is worse for me than the original text. My English wants to be better.

klimenkov
 4 years ago
Best ResponseYou've already chosen the best response.0No problem. I think you just want to help me. Thank you.
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