Probability of an event if items are drawn simultaneously or not A box contains 10 balls, with each ball labeled from 1 to 5 (there are two balls labeled with a 1, two labeled with a 2, and so on). 3 are drawn without replacement. What is the probability of drawing one ball labeled with 1?
I tried two different approaches to this problem. The first one:
If I draw a ball labeled with not a one, then I draw a second ball labeled with not a one, and finally draw a ball labeled with a one, the probability of this event is:
$\frac{8}{10} \times \frac{7}{9} \times \frac{2}{8} = \frac{7}{45}$
If I don't care about the order in which the balls are drawn, then $\frac{7}{45}$ is the probability of drawing a ball labeled with a 1. If I do care about the order, then I must multiply my result times 3, because the ball with the one could be drawn at the first, second or third try. Therefore, the probability is $\frac{7}{15}$
Now, if I try to solve the problem thinking of it as a hypergeometric distribution, with $N = 10$, $n = 3$, $k = 2$ and $x = 1$, I get that the probability of drawing one ball labeled with a 1 is:
$\binom{2}{1} \binom{10 - 2}{3 - 1}/\binom{10}{3} = \frac{7}{15}$
So this leads me to ask: is the hypergeometric distribution only valid when the order of the items being drawn matters? In that case, we would be dealing with permutations rather than combinations. Am I right?
The other thing which I don't get is why does the probability change if the order matters or not. Because, if the order doesn't matter is because all items are being drawn at once. And it seems counter intuitive that just because the items are drawn all at once or one by one the probability changes. 
 A: I think the issue here is that when you say "I draw a ball...then I draw a second ball...finally I draw a ball" what you're doing is assigning an order to the draws. If you do not care about the order, then drawing (not 1)(not 1)(1),(not 1)(1)(not 1),(1)(not 1)(not 1) all amount to the same thing in your sample space. Therefore, to get the probability without order you actually need to do $3{7 \over 45}={7 \over 15}$ which coincides, as it should, with what the hypergeometric distribution gives you. Your first order the experiment to make things easier to grasp, and then multiply by $3$ to reflect the fact that the order does not matter. 
If the order were to matter then (not 1)(not 1)(1) and (1)(not 1)(not 1) would be to very different things, but the probability of the event "drawing a ball labelled 1" is unchanged, since by the same reasoning you used, it is: $$3 \left ( \frac{8}{10} \times \frac{7}{9} \times \frac{2}{8} \right )= 3\frac{7}{45}={7 \over 15}$$
Another way to think of the hypergeometric distribution, to enforce the fact that the order does not matter, is to think that you close your eyes, put your hand in the box and draw 3 balls at the same time. You do not know (and do not care), what order they came out in. 
