Given a bag containing 10 red balls and 10 green balls. When you draw 6 balls, without replacement, what is the probability that you will have at least 1 red and 1 green ball.
I attempted this initially by figuring about the probability of drawing at least 1 red, which is 1 - P(no red). (http://mathforum.org/library/drmath/view/69151.html) Then I just figured that P(>=1 red) * P(>=1 green) = P(>= 1 red & >= 1 green) but since they aren't independent. (i.e. You can't draw 6 red & 6 green when drawing 6 total) The only way I could concieve a solution would be to subtract the overlapping events.
This leaves me with
P(>= 1 red & >= 1 green) = P(>= 1 red)*P(>= 1 green) - P(=6 red)*P(=6 green) - P(=6 red)*P(=5 green) - ... - P(=1 red)*P(=6 green)
Which I think is "Correct" but I feel as though there should be a more intuitive way to calculate this.