# Intuition behind probabilities for sampling without replacement

I am not understanding the explanation here in regards to expectations and probabilities for sampling without replacement. I am not seeing why we don't treat each draw as a unique random variable. For example, if there is only one ball left, then Pr(Xn = red) can either only be 1 or 0. We don't know which one it is until we get there, but clearly Pr(Xn = red) does not equal p. (Let's assume originally we have 3 reds and 7 blues.)

The below excerpt implies that Pr(Xi = red) = p for all i. And thus, it implies Pr(Xn = red) = 0.30, not 1 or 0. • You are mixing up different probabilities. At the last draw, the probabilities you give (zero or one) are necessarily conditional on all the preceding draws. You can analyze the situation this way, but it requires analyzing the entire tree of all possible sequences of draws and how the probabilities change after each draw--and that's a huge and complex endeavor. It pays, then, to study the clever methods people have found to achieve simple answers to such complex questions. – whuber Feb 25 at 16:10
• Ok thanks. I mean I understand the way he's describing it. I just don't understand why it is viewed that way. Like if I were to say, before we even drew any balls, that the Pr(X7 = red) = p, that I agree with. But why wouldn't we view Pr(X7 = red) given X1-6? I don't understand when to view things in which way and when to use the different probabilities. – confused Feb 25 at 16:32