## klimenkov 2 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.

1. wio

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

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

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

5. klimenkov

@experimentX you was so close...

6. wio

Am I still missing something?

7. klimenkov

8. klimenkov

Yes. You are right.

9. experimentX

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

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

11. klimenkov

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

12. 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.

13. wio

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

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

16. klimenkov

Hope, you enjoyed this problem.

17. experimentX
18. experimentX

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

19. klimenkov

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

20. experimentX

Woops sorry!!

21. klimenkov

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