$1.$ With replacement: There are three ways this can happen, red, red, red; blue, blue, blue; and green, green, green.
The probability of red, red, red is $\left(\dfrac{5}{19}\right)^3$. Find similar expressions for the other two colurs, and add up.
$2.$ Without Replacement: The probability of red, red, red is $\left(\dfrac{5}{19}\right)\left(\dfrac{4}{18}\right)\left(\dfrac{3}{17}\right)$. Find similar expressions for the other two colours, and add up.
Remark: We could also do the problem by a counting argument. Let's look at the first problem. Put labels on the balls to make them distinct. Then for with replacement, there are $19^3$ strings of length $3$ of our objects. (The three objects in the string are not necessarily distinct.) All of these $19^3$ strings are equally likely.
There are $5^3$ all red strings, $11^3$ all blue, and $8^3$ all green. This gives probability $\dfrac{5^3+6^3+8^3}{19^3}$.
One can do a similar calculation for the without replacement case.
For without replacement, there is a third approach. We can choose $3$ objects from $19$ in $\dbinom{19}{3}$ ways. And we can choose $3$ of the same colour in $\dbinom{5}{3}+\dbinom{6}{3}+\dbinom{8}{3}$ ways.
We have $r$ red, $w$ white, and $b$ blue. The first question you have is about the probability that all the red are drawn before any white are drawn.
Forget about the blues, they are irrelevant. Imagine that we
only record whether a drawn ball is red and white. A blue is a non-event. All the red are drawn before any white if and only if the first $r$ recorded draws are red. There are $\binom{r+w}{r}$ equally likely records. Only $1$ of them is all red.
So our probability is $\dfrac{1}{\binom{r+w}{r}}$.
For the "before $2$ white are drawn" the argument is similar. The event we are concerned with can happen in two ways: (i) all the red are drawn first or (ii) there is exactly one white in the first $r$ draws, and then a red (again, we can forget about the blues).
We have already found the probability of (i).
For the probability of (ii), the number of ways to draw $r-1$ red and $1$ white in the first $r$ recorded draws is $\binom{r}{r-1}\binom{w}{1}$. So the probability this has happened is $\frac{rw}{\binom{r+w}{r}}$. Given this happened, there are $w$ balls of significance left, of which one is red. Thus the probability of (ii) is $\frac{rw}{\binom{r+w}{r}}\frac{1}{w}=\frac{r}{\binom{r+w}{r}}$.
Add the probabilities of (i) and (ii).
Best Answer
Let each pair of balls of a particular colour be in a box of that colour.
Since the balls of a given colour are identical, choosing $4$ balls of different colours is equivalent to choosing four out of the given six boxes. This can be done in $6\choose4$ ways.
Among the unfavourable outcomes, it is possible that exactly two balls are of the same colour. Then, only three boxes would be selected and the pair of balls would be taken out from one of them. This can be done in $6\choose3$ $\times$ $3\choose1$ ways.
Finally, it is possible to get two pairs of balls of the same colour. This can be done by selecting any two boxes and picking out the pairs from both of them. This can be done in $6\choose2$ ways.
Thus, the probability of getting four balls of distinct colours is $$\text{Probability}=\frac{\binom{6}{4}}{\binom{6}{4}+\binom{6}{3}\times\binom{3}{1}+\binom{6}{2}}=\frac{15}{15+20\times3+15}=\frac{15}{90}=\frac16$$
Edit:
It turns out that the balls of a given colour are not supposed to be identical.
Then, the number of ways of choosing any four balls from the twelve distinct balls is $12\choose4$.
For the balls to have distinct colours, the number of ways to choose four boxes as above is $6\choose4$. From each box, we have a choice of two balls. Thus, the total number of ways of picking four balls with different colours is $\binom{6}{4}\times2^4$.
Then, $$\text{Probability}=\frac{\binom{6}{4}\times2^4}{\binom{12}{4}}=\frac{48}{99}=0.4848...$$