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I don't know if I'm doing this right.

Suppose you have 12 marbles. 5 are red and 7 are blue. If you randomly choose 4 marbles at once, what is the probability you will get exactly 3 red marbles?

I did the equation: $(5/12)(4/11)(3/10)(7/9)$

Then I tried to do it in combination form: ${5 \choose 3}$${7 \choose 1}$ divided by ${12 \choose 4}$. But it doesn't yield the same result.

Did I do something wrong in the first equation? Or set up my combinations wrong?

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Your first answer is the probability that the first three marbles are red and the fourth one is blue assuming that the marbles are drawn in succession without replacing the previously drawn marbles. Can you figure out how to account for other placements of the blue marble? –  Dilip Sarwate Sep 17 '12 at 22:02
    
Thanks for pointing out the error everyone. I should be more careful next time. –  Anon Sep 17 '12 at 23:41
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2 Answers 2

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Your first expression is incorrect. It gives the probability that if you take out the marbles one at a time, you get red, red, red, blue in that order.

You could use the same reasoning to find the probability of the order red, red, blue, red; the probability of the order red, blue, red, red; and the probability of the order blue, red, red, red. These will each turn out to multiply out to the same answer as your first calculation. Now add up. You get $4$ times your first expression. That is correct, and agrees with your second calculation.

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Your first expression is the probability of drawing 3 red marbles then a blue one. You also have to take into account the probability of getting a blue marble then 3 reds, etc. The second expression does this because the choose function ignores order.

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