We have $5$ red, $3$ blue and $2$ green balls. We take $3$ of them at random. What is the probability that they are of different colors?

Question

We have $5$ red, $3$ blue and $2$ green balls (balls of the same color are indistinguishable). We take $3$ of them at random. What is the probability that they are all of a different colors?

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I have two ideas how to calculate and I think 2nd is right?

First: The number of all outcomes is $n = {10 \choose 3} = 120$ and the number of good outcomes is $m = 5\cdot 3\cdot 2 $ so $$P= {m\over n} = 0,25$$

Second: Probability of choosing $1$ red/blue/green ball is ${5\over 10}$ /${3\over 10}$/${2\over 10}$. So choosing each color is ${3\over 100}$, but we have $6$ different arrangements of good choices that give us $$P = 6\cdot {3\over 100} = 0,18$$

Now which one is correct? Or it is no one? If you offer a new solution then please write down a sample space and probability function in your solution.

Answer 1

Your first method is correct.

Let's correct your second method. We are selecting balls without replacement. Thus, the probability of selecting one red, one blue, and one blue in that order is $$\frac{5}{10} \cdot \frac{3}{9} \cdot \frac{2}{8} = \frac{1}{24}$$ As you said, there are $3!$ orders in which we could select the balls, which gives a probability of $\frac{1}{4}$, which agrees with the answer obtained using your first method.

Answer 2

I believe the first is correct and the second is flawed.

It looks like you don't consider fact that once you picked a red ball the chance to pick another red ball decreases.

For instance the chance to pick red then blue then green is in fact $\frac{5}{10}\cdot\frac{3}{9}\cdot\frac{2}{8}$