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).
Best Answer
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 $$\dfrac{{5\choose{3}} + {4 \choose 3} + {3 \choose 3}}{12 \choose 3} = \dfrac{3}{44} $$