## 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. one year ago one year ago

1. wio Group Title

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. klimenkov Group Title

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.

3. wio Group Title

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?

4. klimenkov Group Title

5. klimenkov Group Title

@experimentX you was so close...

6. wio Group Title

Am I still missing something?

7. klimenkov Group Title

8. klimenkov Group Title

Yes. You are right.

9. experimentX Group Title

why was it ?? $\huge {\sum\limits_{k=1}^r \binom{m}{k}\binom{n-m}{r-k}\over \binom{n}{r}}$

10. klimenkov Group Title

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 $$A_n^k=k!C_n^k$$.

11. klimenkov Group Title

Actually it equals $\frac1{C_n^r}\cdot\sum_{i=0}^{r-1}C_{n-m}^iC_m^{r-i}$

12. experimentX Group Title

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.

13. wio Group Title

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.

14. klimenkov Group Title

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!

15. klimenkov Group Title

16. klimenkov Group Title

Hope, you enjoyed this problem.

17. experimentX Group Title
18. experimentX Group Title

hope that works for you!! i see things in russian when i look english in mirror!!

19. klimenkov Group Title

The translation is worse for me than the original text. My English wants to be better.

20. experimentX Group Title

Woops sorry!!

21. klimenkov Group Title

No problem. I think you just want to help me. Thank you.