How to approach the next statistic problem? English is not my native language.
This problem occurred to me this weekend when I was in a tennis championship; There were 120 couples and they separated us into 6 groups, so that at the end, 40 couples took a medal (gold and silver)
After the prizes, a raffle of sports accessories was made among the 120 couples, and some of these accessories were won by couples who also had taken a medal. Obviously, this happened because of 120 couples, 40 couples are a third of the total, it was very likely that some of the winners were couples who had also won a medal.
The question is: Knowing that of all the accessories drawn, four of them fell into the hands of couples who also had a medal, it is possible to know how many accesories where drawn approximately, or al least what is the minimum number of accessories that were drawn? 
In the raffle,  the couple who won an award could no longer opt for the other prizes.
I really not sure how to do this... Thank you 
 A: Well, you know for sure there were at least 4 prizes (accessories). 
You might model the number $X$ of prizes won by medalists as 
$X \sim \mathsf{Binom}(n,1/3),$ where $n$ prizes are awarded. 
Intuitively, the estimated proportion of medalists winning prizes is 
$\hat p=X/n=4/n=1/3,$ so it seems that there may have been around $n = 12.$
More formally, the probability of seeing 4 medalists win prizes is
$$f(4; n, 1/3) = P(X = 4) = {n \choose 4}(1/3)^4(1-(1/3))^{n-4},$$ and that is maximized when
$n = 11$ or $12,$ as shown in the figure below (made using R statistical
software). You can do hand computations to verify a few values around $n = 12$
for yourself. [By looking at the ratio $f(4, n, 1/3)/f(4, n-1, 1/3),$
you could do a mathematical proof.]
n = 1:30;  like = dbinom(4, n, 1/3)
plot(n, like, pch=19, ylim=c(0,.25))
abline(v=12, col="red");  abline(h=seq(0,.25,by=.05), col="green")


However, for the information you provide, there is no way to know exactly how many prizes were given.
