Probability logic that I don't understand Here is the full question:
$5$ balls are drawn in succession without replacement from an urn containing $5$ red balls and $6$ blue balls. How many possible outcomes are there?
I've been pondering whether or not I should solve it this way. Let $n$ be the number of possible outcomes. Then:
$$n = 2\cdot 2\cdot 2\cdot 2\cdot 2\cdot 2$$
but isn't there one outcome that would use up all the red balls before drawing for the sixth sample space? Wouldn't it make it like this?
$$n = 2\cdot 2\cdot 2\cdot 2\cdot 2\cdot 1$$
(my reasoning: the sixth sample space would only be equal to one since all the red balls have been drawn and only one color is left)
I'm terribly confused on how I should tackle this. Will it affect the final answer? I would be grateful if someone could give me the correct answer and an explanation.
 A: There are multiple "reasonable" ways of counting the "different outcomes" in the sample space of the experiment, "Choose five balls from an urn containing 5 blue and 6 red balls":

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*"How many blue/red balls were there"? This sample space will have $6$ outcomes in it, of the form $(b$ blue, $r$ red$)$: $$\Omega = \{ (0, 5), (1, 4), (2, 3), (3, 2), (4, 1), (5, 0) \}.$$ This is the "coarsest" description of the sample space, and not all of these will have equal likelihood. For instance, it's clear that the outcome $(0$ blue, $5$ red$)$ is more likely than the outcome $(5$ blue, $0$ red$)$, since there are more red than blue balls in the urn.


*"What was the specific sequence of blue/red balls observed?" This sample space $\Omega$ will have $2^5 = 32$ (not $2^6 = 64$) different outcomes in it, corresponding to all possible five-letter "words" made with the letters $B$, $R$ (for instance, $BRRBR$) and seems to be the one OP is thinking of based on their calculation $2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2$ (which has one too many 2's). Again, the examples $BBBBB$ and $RRRRR$ show that not all of these 32 outcomes are equally likely. However, this sample space is useful for counting out the sequences with exactly $b$ blue balls: there will be $_5 C_b$ sequences with exactly $b$ $B$'s in them.


*If we actually want to calculate the probabilities directly from a sample space of equally likely outcomes, we should proceed as follows. Number the balls $1, 2, 3, ..., 10, 11$, so that balls $1, 2, 3, 4, 5$ are blue and balls $6, 7, 8, 9, 10, 11$ are red. Then our sample-space $\Omega$ consists of all ordered sequences of five-element subsets of $\{1, 2, 3, ..., 10, 11\}$ selected without replacement (like, for instance, $(11, 4, 6, 3, 1)$), of which there are $_{11} P_5 = 55440$. Every individual outcome in this sample space is equally likely, and so this sample space is best for calculating probabilities of specific events.
A: The number of red and blue balls left in the urn is uniquely determined by the number of red balls drawn. There may be anywhere from 0 to 5 red balls drawn. So there are 6 possible outcomes.
