To find the probability distribution of $X_4$ means to find the probability that $X_4=0$, the probability that $X_4=1$, the probability that $X_4=2$, and so on.
1) Since there are only $7$ balls, $P(X_4\ge 2)=0$.
2) $P(X_4=1)$ is the sum of the probabilities of three scenarios you mentioned. You are on the right track to calculate it.
3) $P(X_4=0) = 1- P(X_4=1)$, which you get from (2).
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
tl;dr
Do you think order matters? No. Why? Because if we select $RBB$ or $BRB$ or $BBR$ any of them occuring is favorable, and we are done, you might as well consider order but that would cancel eventually. So total possibilities in three turns: $$\binom{10}3$$ Ways to do favorable event: $$\binom62\binom41$$ Probability: $$P=\frac{\binom62\binom41}{\binom{10}3}=\frac12=50\%$$
Now let me consider order:
Total ways: $$10*9*8$$ Favourable ways: $$RBB:\binom41\binom61\binom51\\ BRB:\binom61\binom41\binom51\\ BBR:\binom61\binom51\binom41$$ Or simply no. of ways of arranging $B,R,B$ times "you know": $$\frac{3!}{2!}6*5*4=6*5*4*3$$ Probability is: $$\frac{6*5*4*3}{10*9*8}=\frac12=50\%$$