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?
Best Answer
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 $1$:
$$ \frac{\binom{9990}{2}}{\binom{10000}{2}} +\frac{\binom{10}{2}}{\binom{10000}{2}} +\frac{\binom{9990}{1}\cdot\binom{10}{1}}{\binom{10000}{2}} = 1. $$