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A bag contains 3 red marbles and 5 green marbles. If 3 marbles are drawn at random, what is the probability that more green marbles were drawn than red?

Do I have to use factorials with this?

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closed as not a real question by Marvis, Ittay Weiss, Davide Giraudo, Neal, Fabian Jan 7 '13 at 21:31

It's difficult to tell what is being asked here. This question is ambiguous, vague, incomplete, overly broad, or rhetorical and cannot be reasonably answered in its current form. For help clarifying this question so that it can be reopened, visit the help center.If this question can be reworded to fit the rules in the help center, please edit the question.

Supa, because you vandalized your post (by replacing it with "blah..." and "delete...") I am going to lock it for one day. This will prevent you or anyone else from editing it, though it has the additional effect of preventing any new answers for that day as well. Do not vandalize your posts again, or you will be suspended. –  Zev Chonoles Jan 7 '13 at 23:05
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1 Answer 1

Hint: To get a majority green marbles, you must have either $2$ or $3$. Calculate the probability of each of these, and they are mutually exclusive, so you can add them.

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so 5/8 + 4/7? That would give me 2 green marbles which would be more than red? –  Supa Jan 7 '13 at 20:41
@Supa: that would make the probability greater than 1, which cannot be true. The chance to get $3$ green is $\frac 58 \cdot \frac 47 \cdot \frac 36=\frac {60}{336}$. Each fraction is the chance to get a green marble with the bag holding what is left at that point. –  Ross Millikan Jan 7 '13 at 21:03
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