# Use of the classic definition of probability for lotteries without replacement

Consider a lottery where 5 numbers are extracted from 100 without replacement for each session. What is the probability that a number $n$ is extracted in a session?

If I use combinatorics, I can define it as:

$\frac{\binom {99} {4}}{ \binom {100} {5}}$ that it gets exactly $\frac{5}{100}$.

Could have I used directly the formula cases favorable / cases possible = 5/90, even if events are not equiprobable (ie, the probability of having $n$ as the first number of the five extracted is different from the probability of having $n$ as second, third, fourth or fifth)

• I do not understand why the order does not matter and the probability is always 1/90 irrespective of the number extraction order – floatingpurr Sep 25 '16 at 10:05
• For the combination, the order does not matter. If you are interested in the probability that $n$ occurs at the first,second,third,... time, the probabilities are somewhat surprisingly equal, each being $\frac{1}{100}$. For example, $n$ occurs at the second draw : $\frac{99}{100}\cdot\frac{1}{99}=\frac{1}{100}$. – Peter Sep 25 '16 at 22:55
• Sorry, that I intitially assumed the probabilities would not be equal. The only thing that we must assume is that in each draw every not drawn ball has the same probability to be drawn next. – Peter Sep 25 '16 at 22:57