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A bag contains red and black balls. We choose 3 balls at random. What is the sample space of this experiment?

We $3! = 6$ ways of choosing 3 balls.

But when listing them I get 8 ways:

$S = \{RRR, RRB, RBR, RBB, BBB, BRB, BRR, RRB\}$

Which answer is correct?

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  • $\begingroup$ It depends a bit how many balls of each colour are in the bag. Your question is consistent with having just two of each. (Pedantic point: And because you don't say 'only red and black' what is to prevent the bag having blue, green, yellow balls in it?) $\endgroup$ – Mark Bennet Mar 15 '14 at 22:26
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You have $3!$ ways of ordering three different balls you have already drawn (first ball has three places to choose from etc.): $$ \{ABC, ACB, CAB, CBA, BCA, BAC\} $$ You have $2^3$ different color orders you can make when you draw balls from a bag (first ball can be one of two different colours etc.): $$ \{RRR, RRB, RBR, RBB, BRR, BRB, BBR, BBB\} $$ Correctly identifying what method to use to count exactly what you want to count (and not something else) is difficult and more or less only comes with experience. Keep up the work, and hopefully it sticks sooner or later.

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Look at the symmetry cases- BBR is the same as RBB, as an example.

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