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.
Best Answer
There certainly is a more intuitive way at looking at this:
Consider the inverse of this statement, where we try and calculate the probability that not at least one green and red ball are chosen. Therefore, all the balls chosen will be either green or red.
So, for all green:
$$P =\frac{10}{20}*\frac{9}{19}*\frac{8}{18}*\frac{7}{17}*\frac{6}{16}*\frac{5}{15} = \frac{7}{1292}$$
Since you have the same number of red balls, it will be the same probability.
Therefore, the total probability is: $$\frac{7}{1292}*2 = \frac{14}{1292}$$
Now we simply subtract this from the total probability, $1$:
$$1 - \frac{14}{1292} = \frac{639}{646}$$
Comment if you have any questions.