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I can't seem to get my head around this: A random drawing will be held for which there are 900 tickets sold, for which there will be one winning ticket drawn. If you purchased 25 tickets, what is the probability that you will win.

I am pretty sure this follows a binomial distribution, but I don't know how to adopt the models to solve the problem.

Any help / directin would be appreciated. I am looking more to undertsand the problem domain than to get an answer.

Thank you in advance,

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    $\begingroup$ Related: stats.stackexchange.com/q/6945/2970 $\endgroup$
    – cardinal
    Sep 28, 2012 at 13:29
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    $\begingroup$ This is much simpler than a binomial distribution. If there are $n$ equally likely outcomes, the probability of each is $1/n$, and the probability of a set $S$ of them is $|S|/n$. This tells you the probability the winning ticket will be among those in the set $S$. $\endgroup$ Sep 28, 2012 at 14:28
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    $\begingroup$ It's not that this problem is much simpler than a binomial distribution, it's that the events corresponding to each individual ticket winning are mutually exclusive. If one ticket wins, the others can't. There is no binomial distribution because the state of each ticket is not independent of the state of the other tickets. $\endgroup$ Sep 28, 2012 at 16:07
  • $\begingroup$ But @Max, that makes it simpler. Surely figuring out 25/900 is simpler than the binomial expansion? $\endgroup$
    – Peter Flom
    Sep 29, 2012 at 19:04
  • $\begingroup$ @PeterFlom, My comment was absolutely unnecessary (I was slightly bored at work). I was trying to say that what's important isn't that we can use something much simpler than a binomial distribution, it's that the binomial distribution is not appropriate for this problem in the first place. I was being more explicit about why the binomial distribution is not appropriate. $\endgroup$ Sep 29, 2012 at 19:39

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To tidy this up and solve the problem (i don't like leaving items unanswered), as the users above have noted correctly the answer is 25/900.

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