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Ok, considering I managed to accidentally press enter and submit this thread before even writing the title, this has to do with probability.
Mod note: I added a title... Not very informative, but it's a title.
I have 3 buckets of tennis balls, and each bucket has red, green and blue balls . If I pick out 4 balls from random buckets, what is the probability that I'll have at least 2 red balls and 1 green ball?
I'm so terrible at probability that it's not even funny... So this is my crude guess:
The probability of picking a red, then red, then green is 1/3^3, but we can pick these colours in any order, and there are 3C2 ways of picking them, but we also pick out a 4th ball, so there are 4C3 ways of arranging the 3 balls we need to pick out. So my guess is that we have a probability of
[tex]P = \frac{3C2\cdot4C3}{3^3} = \frac{12}{27}[/tex]
I'd be amazed if this were correct though, so if any of you may, please let me know where I've gone wrong.
Mod note: I added a title... Not very informative, but it's a title.
Homework Statement
I have 3 buckets of tennis balls, and each bucket has red, green and blue balls . If I pick out 4 balls from random buckets, what is the probability that I'll have at least 2 red balls and 1 green ball?
The Attempt at a Solution
I'm so terrible at probability that it's not even funny... So this is my crude guess:
The probability of picking a red, then red, then green is 1/3^3, but we can pick these colours in any order, and there are 3C2 ways of picking them, but we also pick out a 4th ball, so there are 4C3 ways of arranging the 3 balls we need to pick out. So my guess is that we have a probability of
[tex]P = \frac{3C2\cdot4C3}{3^3} = \frac{12}{27}[/tex]
I'd be amazed if this were correct though, so if any of you may, please let me know where I've gone wrong.
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