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There are $12$ balls in a bag. $3$ of them are red, $4$ of them are green, and $5$ of them are blue. We randomly take out $3$ balls from the bag at the same time. What is the probability that all three balls are of the same colour?

My answer: $(3/12)^3 + (4/12)^3 + (5/12)^3$. Is this correct?

EDIT: My explanation is that since there are the odds of getting a red ball are 3/12, I simply multiplied it 3 times since I consider multiplication to be the "and" operator. Similarly, I consider addition to be the "or" operator. So, the way I think of it is:

(1 red and 1 red and 1 red) or (1 green and 1 green and 1 green) or (1 blue and 1 blue and 1 blue).

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You might want to explain your answer. – Did Mar 19 '13 at 6:28
I just did. Thanks! – chromozonex Mar 19 '13 at 6:35
The problem with your answer is that the probability of picking $3$ red balls isn't $\left( \frac{3}{12} \right)^3$ but rather it's $\frac{3}{12} \cdot \frac{2}{11} \cdot \frac{1}{10}$. Picking the balls out at the same time doesn't change this but it's there to suggest you're probably more interested in the combinations than you are the permutations. – muzzlator Mar 19 '13 at 6:41
By that logic, I get $17/264$ and not $3/44$ as you've described below. Or am I still doing something wrong? – chromozonex Mar 19 '13 at 6:47
Check your arithmetic. – muzzlator Mar 19 '13 at 6:49
up vote 4 down vote accepted

There are ${5\choose{3}} + {4 \choose 3} + {3 \choose 3}$ total ways of having a success. There are a total of $12 \choose 3$ different ways of choosing 3 balls.

So the answer is $$\dfrac{{5\choose{3}} + {4 \choose 3} + {3 \choose 3}}{12 \choose 3} = \dfrac{3}{44} $$

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" the same time." doesn't magically improve a chance of ball to be picked somehow.

You still pick one by one, but you don't know the order, but it doesn't matter with current question (It would be important if you were asked all balls different color for example).

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+1, Important point, the language is usually just there to be suggestive of placing more importance in the combination you're left with rather than in what order you got them. On the other hand it wouldn't be important for the question are all balls different colour, that would simply be $\frac{5 \cdot 4 \cdot 3}{12 \choose 3}$ by considering combinations again. – muzzlator Mar 19 '13 at 6:55

For $3$ balls of red color $$\mathrm{Prob}=\frac{3}{12}\cdot\frac{2}{11}\cdot\frac{1}{10}$$

For $3$ balls of green color $$\mathrm{Prob}=\frac{4}{12}\cdot\frac{3}{11}\cdot\frac{2}{10}$$ For $3$ balls of blue color $$\mathrm{Prob}=\frac{5}{12}\cdot\frac{4}{11}\cdot\frac{3}{10}$$

Total probability of all ball same color is $$ \frac{3}{12}\cdot\frac{2}{11}\cdot\frac{1}{10}+\frac{4}{12}\cdot\frac{3}{11}\cdot\frac{2}{10}+\frac{5}{12}\cdot\frac{4}{11}\cdot\frac{3}{10} $$

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This is a question of picking up the balls all together so probability of picking up any 3 balls( b = 12C3 ) and success cases ( a = 5C3 + 4C3 +3C3 )

answer= a/b

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Combination ($\binom n r$) can only be used when we can differentiate the balls even when they have the same colour, but if they are all of a particular colour they cannot be distinguished and hence the number of ways to pick up $r$ balls from a set of $n$ balls of a particular colour is only $1$ and not $binom n r$, and the probability is $\dfrac r n$.

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