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December 23rd, 2015, 08:11 AM   #1
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Probability With Marbles

There are 11 green marbles, 4 red marbles, and 5 blue marbles. You select 4 marbles without replacement. What are these probabilities?

1. 2 green marbles and 2 red marbles
2. 1 green marble and 3 red marbles

I'm also interested in the probability of 4 red marbles and the probability of at least 1 blue marble, but I can do those myself.
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December 23rd, 2015, 08:41 AM   #2
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1. $\displaystyle \frac{\binom{11}{2}\cdot\binom{4}{2}}{\binom{11+4+ 5}{4}}$
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December 23rd, 2015, 06:28 PM   #3
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Hi skaa,

I am unfamiliar with this notation. Could you please give a brief explanation? If I wanted to solve this problem, I would've had to go through the arduous process of adding the products of each possible sequence that qualifies.
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December 24th, 2015, 05:51 AM   #4
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$\displaystyle \binom{N}{M}$ is called a combination. It means the number of ways you can take M marbles from a bag of N without replacing them. It evaluates to N!/M!/(N-M!).

So, for example, if you take two marbles from a bag containing four labeled {A,B,C,D}; you can get {A,B}, {A,C}, {A,D}, {B,C}, {B,D} or {C,D}. That's six combinations, or 4!/2!/(4-2)! = (4*3*2*1)/(2*1)/(2*1)=6.

If you take 4 marbles from 11+4+5=20, there are $\displaystyle \binom{20}{4}$=4845 ways to do it.

To get 2 of the 11 green, and 2 of the 4 red, there are $\displaystyle \binom{11}{2}*\binom{4}{2}$=330 ways.

The chances of getting that combination are thus 330/4845~=.0681.

The second one can be done the same way.
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December 24th, 2015, 06:13 AM   #5
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I see!
I had just been using the equivalent formula: N! / ((N - M)!(M!))
Glad there is a simpler way to write it! Haha. Makes the process understandable (:

Thank you
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