First of all, for clarity, I will use the following variables: $r$ for the number of red balls, $y$ for yellow, $g$ for green, and $b$ for blue. I will denote the total number of balls by $n$.
Out of the total number of permutations of 9 balls selected from the $n$ total, we only want to consider certain permutations. We want at least one of the first seven selections to be yellow, one of the remaining seven selections (one is taken by the yellow ball) to be green, three of the remaining seven selections to be red, and one of the 4 still remaining selections to be blue. The remaining four unspecified selections could be anything.
There are ${{y}\choose{1}}\cdot{{7}\choose{1}}$ ways of a yellow ball taking one of the first seven positions.
There are ${{g}\choose{1}}\cdot{{7}\choose{1}}$ ways of a green ball taking one of the first eight positions, after one has already been specified as yellow.
EDIT: There are ${{r}\choose{3}}\cdot{{b}\choose{1}}\cdot{{7}\choose{4}}\cdot{{4}\choose{1}}$ ways of three red balls and one blue taking four of the remaining seven positions. (And we choose one of the four positions to be taken by the blue ball.)
There are still three positions left, and we have already placed six balls, leaving us ${_{n-6}}P_3$ possible ways to fill the remaining positions.
Thus the total number of possible permutations that satisfy the requirements is ${{y}\choose{1}}\cdot{{7}\choose{1}}\cdot{{g}\choose{1}}\cdot{{7}\choose{1}}\cdot{{r}\choose{3}}\cdot{{b}\choose{1}}\cdot{{7}\choose{4}}\cdot{{4}\choose{1}}\cdot{_{n-6}}P_3$
out of a total of ${_n}P_9$ permutations.
Probability of getting a red or white ball on first draw:$\frac{1}{2}$
probability of getting the remaining ball that is either red or white $\frac{1}{3}$
probability both of these happen: $\frac{1}{2}\cdot \frac{1}{3}\approx0.167$
Best Answer
By symmetry, a particular order is just as likely as its reversal. And in the reversal, we pick at least one red and one green ball before we pick a blue one. But the probability of this is easy to calculate:
The probability that red is the first colour to be picked, followed by green, is
$$\frac{10}{60}\times \frac{20}{50}=\frac{1}{15}$$
And the probability that green is the first colour to be picked, followed by red, is
$$\frac{20}{60}\times \frac{10}{40}=\frac{1}{12}$$
So the answer is $\frac{1}{15}+\frac{1}{12}=\frac{3}{20}$.