Given a box of 6 red and 6 blue balls, what is the probability that I will pick 3 red and 3 blue?

My guess is $(^6C_3 + ^6C_3)/^{12}C_3$

but do I need to account for ordering? Why or why not?

If I don't try the above method, could this be what I can do:

$$P = (\frac{6}{12} \times \frac{5}{12} \times \frac{4}{12})^2 \times 3!$$

Meaning there's a probability that I can pick red, red, red, and then blue, blue, blue, and should I multiply by $3!$ to account for the ordering, or should it be divide?

  • 2
    $\begingroup$ Are you choosing $6$ uniformly at random? If so, there are $\binom {12}6$ total ways to do it, and $\binom 63^2$ ways to respect the color restriction. It's up to you whether order matters or not...nothing in the problem suggests that it does, but it is your problem. $\endgroup$ – lulu Mar 26 '18 at 19:03
  • 1
    $\begingroup$ The probability $\dfrac{\binom{6}{3}+\binom{6}{3}}{\binom{12}{3}}$ is the probability that when you select three balls from your box that they are all red or they are all blue. Keep in mind the difference in when you want to multiply versus when you want to add. $\endgroup$ – JMoravitz Mar 26 '18 at 19:05
  • $\begingroup$ As for whether order matters or not, the problem of finding the probability that you pull a red followed by a red followed by a red followed by a blue followed by... is a different question than finding the probability that among the six pulls, three of them are red and three are blue (which could occur in the order rrrbbb, rrbrbb, rrbbrb, etc...). In the latter interpretation, whether or not you run the calculations as though order mattered will not make a difference in the final answer, both yield the same result. $\endgroup$ – JMoravitz Mar 26 '18 at 19:10
  • $\begingroup$ Thanks for clarifying. I always get caught up with the ordering. If I don't want to do it the counting way, is there another way I could do it? see my edit above $\endgroup$ – oldselflearner1959 Mar 26 '18 at 19:16
  • $\begingroup$ Closer... but no, not quite. Instead, it would be $\frac{6\times 5\times 4\times 6\times 5\times 4}{12\times 11\times 10\times 9\times 8\times 7}\times\binom{6}{3}$. The denominators should have gotten smaller, not all been at $12$. Also, the "ordering" here would have been deciding which order of rrrbbb, rrbrbb, rbbrrb, bbrrrb, etc... the balls appeared in. The point in the calculation being, the first ball drawn will be out of $12$, the second ball drawn will be out of $11$, and the third out of $10$ etc... while the first time pulling red would be out of $6$, second time out of 5 etc... $\endgroup$ – JMoravitz Mar 26 '18 at 19:37

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