Suppose you have $B$ balls in $C$ colours, with $b_j$ balls of colour $j$, and draw $E$ balls. For any subset $S$ of $\{1,2,\ldots, C\}$, let $b_S = \sum_{j \in S} b_j$. The
probability $P(S)$ that you get all the balls of all the colours in $S$ is $0$ if $E < b_S$ and ${{B-b_S} \choose {E-b_S}}/{B \choose E}$ if $E \ge b_S$. By the inclusion-exclusion principle, the probability that you get all the balls of at least one colour is
$ \sum_S (-1)^{|S|-1} P(S)$, where the sum is over all nonempty subsets $S$ of $\{1,2,\ldots, C\}$ with $b_S \le E$.
Just use generating functions , lets say that one of the color has $f$ balls in a color , so the generating function of this color is $$\binom{f}{1}x$$
Lets assume that we have $K$ distinct color of balls and they have $f_1 ,f_2,f_K$ balls.Moreover , we want to choose $I$ balls with distinct colors , so find the coefficient of $x^I$ in the expansion of $$\bigg[1+\binom{f_1}{1}x\bigg]\bigg[1+\binom{f_2}{1}x\bigg]..\bigg[1+\binom{f_K}{1}x\bigg]$$.
Moreover, we know that the denominator is $\binom{N}{I}$ ,because all balls are seem distinct when computing probability , so the answer is $$\frac{[x^I]\bigg[\bigg(1+\binom{f_1}{1}x\bigg)\bigg(1+\binom{f_2}{1}x\bigg)..\bigg(1+\binom{f_K}{1}x\bigg)\bigg]}{\binom{N}{I}}$$
$\mathbf{EDITION:}$ For $I=2$
Generating function for $10$ blue ball : $1+\binom{10}{1}x$
Generating function for $139$ white ball : $1+\binom{139}{1}x$
Generating function for $44$ green ball : $1+\binom{44}{1}x$
Generating function for $1$ red ball : $1+\binom{1}{1}x$
where $1$ represent $x^0$ ,i.e, we did not make selection , so by expansion
In link , we see that the coefficient of $x^2$ is $8139$ , so the answer is $$\frac{8139}{C(194,2) =18,721}= 0,434752417$$
Best Answer
We find a simple formula for the expected number of different colours. For this we use the method of Indicator Random Variables.
For $i=1$ to $K$, let $I_i=1$ if colour $i$ is drawn at least once, and let $I_i=0$ otherwise. Then the number $Y$ of colours drawn is $I_1+I_2+\cdots+I_K$, and by the linearity of expectation we have $$E(Y)=E(I_1)+E(I_2)+\cdots +E(I_K).$$
$E(I_i)$ is defined as: $$E(I_i) = 1 \cdot Pr(I_i=1) + 0 \cdot Pr(I_i=0)$$ $$E(I_i) = 1 \cdot Pr(I_i=1) $$
Hence, we need to find $\Pr(I_i=1)$ to solve for $E(I_i)$. However, it turns out to be simpler to find $\Pr(I_i=0)$ and then use the fact that $\Pr(I_i=1)=1-\Pr(I_i=0)$.
In general, there are $b=\binom{N}{n}$ equally likely ways to choose $n$ balls. More specifically, there are $N-N/K$ balls not of colour $i$, so there are $a=\binom{N-N/K}{n}$ ways to choose $n$ balls, none of colour $i$.
It follows that $\Pr(I_i=1)=1-\frac{a}{b}$ and therefore $$E(Y)=K\left(1-\frac{a}{b}\right).$$