In a certain lottery, 10,000 tickets are sold and 10 prizes are awarded. What is the probability of not getting a prize if you buy 2 tickets.

The answer to this question is simple enough: $$ \frac {9990 \choose 2}{10000 \choose 2} $$ I understand this answer, since there are 2 ways of picking from the 9990 tickets that don't give you a ticket. And 2 ways of picking some ticket from the 10,000 tickets. But, I wanted to verify this by calculating the probability of getting a ticket.

I used a similar method:

$$ \frac{10 \choose 2}{10000 \choose 2} $$

However, when I use a calculator to add them up, they don't sum to 1. I repeated this exercise for buying 10 tickets, in which case the answers for not getting a ticket and getting a ticket were respectively:

$$ \frac {9990 \choose 10}{10000 \choose 10} $$ $$ \frac{10 \choose 10}{10000 \choose 10} $$

In this case as well, the probability don't sum to 1. I understand that I'm actually decreasing the numerator in the second case. But I can't understand intuitively why this is wrong.

While calculating the probability of not getting a ticket, we divide the number of ways we can pick a ticket that doesn't give you a prize by the number of ways you can pick any ticket. So, while calculuating the probability of getting a ticket, shouldn't we divide the number of ways we can pick a ticket that gives you a prize divided the total number of ways you can pick a ticket?

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    $\begingroup$ These are good questions to ask. What if you buy $20$ tickets; what are the outcomes and their probabilities? $\endgroup$ – David K Feb 15 at 12:08

Note that $\frac{\binom{10}{2}}{\binom{10000}{2}}$ is the probability of getting two prizes! Hence, you are missing the case where you get just one price. If you add that, you do indeed get $1$:

$$ \frac{\binom{9990}{2}}{\binom{10000}{2}} +\frac{\binom{10}{2}}{\binom{10000}{2}} +\frac{\binom{9990}{1}\cdot\binom{10}{1}}{\binom{10000}{2}} = 1. $$

  • $\begingroup$ Shouldn't you also show the correct probability for winning at least one prize? $\endgroup$ – Wolfgang Kais Feb 15 at 13:13
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    $\begingroup$ Winning at least one prize is the same as winning exactly one or exactly two prices, so the last two summands. $\endgroup$ – Christoph Feb 15 at 13:51
  • $\begingroup$ Wow, that was really useful! I also used this logic to answer the question posed by @David in the comments, and they summed to 1 as well. Thank you. $\endgroup$ – WorldGov Feb 15 at 14:00

You can also solve by Binomial as the numbers are large. Hypergeometric distribution used by @Christoph gives the exact probabilities:

Probability that you will get a prize $= \frac{1}{1000}$

probability that you will not get a prize $= \frac{999}{1000}$

If you buy two tickets, the probability that you will not get a prize is

$\approx {2\choose0}(\frac{1}{1000})^0(\frac{999}{1000})^2$ you will get a sum of 1 in this too.

  • $\begingroup$ You are using the symbol "$=$" for an approximation. Please stress the fact that this answer does not give the exact probabilities! $\endgroup$ – Christoph Feb 15 at 12:08
  • $\begingroup$ @Christoph: Done!! $\endgroup$ – Satish Ramanathan Feb 15 at 12:10
  • $\begingroup$ Why a downvote, I am introducing another way to look at the problem!! $\endgroup$ – Satish Ramanathan Feb 15 at 12:27
  • $\begingroup$ I downvoted because this is not an answer to the question "Why do my probabilities do not sum up to $1$?" $\endgroup$ – Christoph Feb 15 at 12:35
  • $\begingroup$ Although it does not answer the question that the OP asked, it is still a valid answer to the problem posed by OP. I am going to leave it even if it receives further downvotes. $\endgroup$ – Satish Ramanathan Feb 15 at 12:38

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