Your method counts every favorable selection twice, once for each way you could designate one of the balls from the color that is selected twice as the ball of that color.
Let's compare your approach with the correct solution.
Solution: The number of ways we may select four of the $4 + 5 + 6 = 15$ balls is
$$\binom{15}{4}$$
If at least one ball of each color is selected, then since four balls are being selected from three colors, exactly one color must be selected twice with the other two colors each selected once.
Two white balls, one red ball, and one black ball are selected: We must select two of the four white balls, one of the five red balls, and one of the six black balls, which can be done in
$$\binom{4}{2}\binom{5}{1}\binom{6}{1}$$
ways.
One white ball, two red balls, and one black ball are selected: We must select one of the four white balls, two of the five red balls, and one of the six black balls, which can be done in
$$\binom{4}{1}\binom{5}{2}\binom{6}{1}$$
ways.
One white ball, one red ball, and two black balls are selected: We must select one of the four white balls, one of the five red balls, and two of the six black balls, which can be done in
$$\binom{4}{1}\binom{5}{1}\binom{6}{2}$$
ways.
Number of favorable cases: Since the three cases above are mutually exclusive and exhaustive, the number of favorable cases is found by adding the number of selections for each case.
$$\binom{4}{2}\binom{5}{1}\binom{6}{1} + \binom{4}{1}\binom{5}{2}\binom{6}{1} + \binom{4}{1}\binom{5}{1}\binom{6}{2}$$
Probability: Assuming each ball is selected randomly, the probability of selecting at least one ball of each color is
$$\frac{\dbinom{4}{2}\dbinom{5}{1}\dbinom{6}{1} + \dbinom{4}{1}\dbinom{5}{2}\dbinom{6}{1} + \dbinom{4}{1}\dbinom{5}{1}\dbinom{6}{2}}{\dbinom{15}{4}}$$
What error did you make?
By designating a ball of each color as the ball reserved for that color, you count each favorable selection twice, once for each way you could have designated one of those balls as the ball of that color. To borrow @lulu's example, you count the selection $W_1, W_2, R_1, B_1$ twice:
\begin{array}{c c}
\text{designated balls} & \text{additional ball}\\ \hline
W_1, R_1, B_1 & W_2\\
W_2, R_1, B_1 & W_1
\end{array}
Notice that
$$\color{red}{\binom{2}{1}}\binom{4}{2}\binom{5}{1}\binom{6}{1} + \binom{4}{1}\color{red}{\binom{2}{1}}\binom{5}{2}\binom{6}{1} + \binom{4}{1}\binom{5}{1}\color{red}{\binom{2}{1}}\binom{6}{2} = \binom{4}{1}\binom{5}{1}\binom{6}{1} \cdot 12$$
where the factor $\color{red}{\binom{2}{1}}$ represents the number of ways you could have designated one of the two balls from the color that is selected twice as the representative ball of that color.
Best Answer
Personally, I find this method much simpler.
Red-white-black need to be drawn in 2-1-1 ,1-2-1 or 1-1-2 pattern, so simply apply the hypergeometric formula:
$$455\times \dfrac{{6\choose 2}{4\choose 1}{5\choose1}+{6\choose 1}{4\choose2}{5\choose 1} +{6\choose1}{4\choose 1}{5\choose 2}}{15\choose 4} = 240$$
NOTE
Make a note that whenever you try to "fractionate" the combinations, by fulfilling the conditions in instalments the way you did, you will always overcount.