# 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.

• Does the order of drawing the balls matter or just how many blue ones you get? – Ross Millikan Jun 13 at 21:06
• It is being presumed that all balls of the same color are indistinguishable from each other. This means (for example) that if all of the balls drawn are blue, no distinction is made as to which blue ball was not drawn. As such, assuming, as per the comment of @RossMillikan, that the order that the balls are drawn is irrelevant, then the answer of Jephph pertains. – user2661923 Jun 13 at 21:16
• Re my previous comment, assuming that you accept the answer of Jephph, it is irrelevant that some of the $(6)$ possibilities are more likely than others. Your stated question was merely how many possibilities are there, rather than what are the relative probabilities of each of these possibilities. – user2661923 Jun 13 at 21:19

1. "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.
2. "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.
3. 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.