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I have a question on the probabilities of winning a prize in a raffle. There is one round where 5 prizes are drawn (1st, 2nd, 3rd, 4th and 5th price). 1 ticket can only win 1 prize but a participant can buy several tickets and therefore win several prizes at once.

Suppose we have 1000 tickets in one round and I bought 10 tickets. What is the probability that I win the 1st prize, 2nd prize, 3rd prize, 4th prize or 5th prize.

My initial guess was: probabililty to win 1st prize = 10/1000, and probability to win 2nd prize = 10/1000 and so on. But this is not correct. Any suggestions?

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1 Answer 1

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Hint: Consider your first ticket on its own. What is the probability that this ticket does not win any prize? If this first ticket does not win a prize, what is the probability that your second ticket also does not win a prize? Now extend this reasoning --- What is the probability that none of your tickets wins a prize?

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  • $\begingroup$ Thanks for the response. The probability that ticket 1 is not winning = 999/1000, so the probability of not winning a price with 10 tickets then equals (999/1000)^10 = 0.96. Correct? In that case, the probability of winning a price is 0.04. So the probability for 1st price = 0.04/5, 2nd price = 0.04/5, 3rd price = 0.04/5, 4th price = 0.04/5 or 5th price = 0.04/5? $\endgroup$
    – Henk
    Jul 30, 2023 at 11:41
  • $\begingroup$ Go through the reasoning more slowly. Don't try to skip any steps. $\endgroup$
    – Ben
    Jul 30, 2023 at 13:44
  • $\begingroup$ Not sure how this is solving the problem for me but thanks for your help. I will continue my search on this forum. $\endgroup$
    – Henk
    Jul 30, 2023 at 15:59
  • $\begingroup$ @Henk consider the draw of the prizes as an urn full of 1000 marbles among which 5 are coloured (the prizes), and we are allowed to draw ten times a marble out of this urn. In the first draw, there are 1000 marbles inside and we have 5/1000 probability of winning and 995/1000 of not winning. In the second draw there are 999 marbles inside (1 less of course), and that changes the probabilities. $\endgroup$ Feb 5 at 16:42

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