I've recently started studying statistics again, and I've just come across sampling without replacement. My book states that if we have a elements of type I and b elements of type II, then the probability of picking x elements of type I can be calculated according to the "classic definition of probability"
$p = {y \over z}$
It also states that since we're going to pick x elements of the a available of type I, then the remainding n-x elements picked can be picked in a variety of ways from the b elements of type II. This I agree with. The book then states that the amount of successfull picks, that is picks of type I, is calculated by multiplying the amount of ways to pick x elements of type I and the amount of ways to pick n-x elements of type II. In other words, if
$p = {y \over z}$
then
$y = {a \choose x}{b \choose n-x}$
We then divide with $z = {{a+b}\choose n}$ to get the probability.
Thing is that I don't really see how the amount of successfull picks can be ${a \choose x}{b \choose n-x}$. I get why we need to use the amounts of ways to pick x elements out of a, but why do we need the amount of ways to pick n-x elements out of b?
EDIT:
To clarify my question, I'd like to provide an example: Say that I want to play in the lottery. There are 50 lottery tickets, 5 of which are winning. I am going to buy 5 tickets, and I want to know the probability that 2 of them are winning. The amount of ways to pick the winning tickets are then
${5 \choose 2}$
and the amount of ways to pick a losing ticket is
${45 \choose 3}$
Are then both of these amounts considered to be "successful attempts", even if I only want to know the probability of me picking two winning tickets?
The probability of picking two winning tickets is $$\frac{\text{# of ways to pick 2 winning tickets}}{\text{# of ways of picking tickets}}.$$
If you don't take into account that the other three (non-winning) tickets may be chosen in a variety of ways, you're not counting some of the possibilities and the numerator won't be as large as it should!
To test this, think about the probability of picking zero winning tickets. If you don't take into account the unsuccessful attempts, you'd get zero here, and this probability is certainly not zero.