The total number of combinations is:
$$\dbinom{5+6+7}{5}=\dfrac{18!}{5!\cdot13!}=8568$$
The number of combinations with no red balls is:
$$\dbinom{6+7}{5}=\dfrac{13!}{5!\cdot8!}=1287$$
The number of combinations with no white balls is:
$$\dbinom{5+7}{5}=\dfrac{12!}{5!\cdot7!}=792$$
The number of combinations with no blue balls is:
$$\dbinom{5+6}{5}=\dfrac{11!}{5!\cdot6!}=462$$
The number of combinations with no red balls and no white balls is:
$$\dbinom{7}{5}=\dfrac{7!}{5!\cdot2!}=21$$
The number of combinations with no red balls and no blue balls is:
$$\dbinom{6}{5}=\dfrac{6!}{5!\cdot1!}=6$$
The number of combinations with no white balls and no blue balls is:
$$\dbinom{5}{5}=\dfrac{5!}{5!\cdot0!}=1$$
So the probability of a combination without at least one ball of each color is:
$$\dfrac{1287+792+462-21-6-1}{8568}=\dfrac{2513}{8568}$$
And the probability of a combination with at least one ball of each color is:
$$1-\dfrac{2513}{8568}=\dfrac{6055}{8568}\approx0.7067$$
The events are:
- three balls of one colour, and one each of the other two.
- two balls of two colours, and one of the remainder.
We count the ways to select which colours belong to the group, then select balls for each.
- $\binom 3 1$ ways to select a colour, $\binom 73$ ways to select three balls of that colour, then $\binom 7 1\binom 71$ ways to select one ball of each of the other colours.
- $\binom 3 2$ ways to select two colours, $\binom 72\binom 72$ ways to select two balls of each of those colours, then $\binom 7 1$ ways to select one ball of each of the remaining colour.
$$\dfrac{\binom 3 1\binom 73\binom 7 1\binom 71+\binom 3 2\binom 72\binom 72\binom 7 1}{\binom{21}{5}}$$
Best Answer
Let $A,B,C$ denote the events that we draw "at least 1 red", "at least 1 black", "at least 1 white" ball in 7 draws.
We're looking for $A\cap B\cap C$.
Using inclusion-exclusion we get: $$ A\cup B\cup C=A+B+C - A\cap B - A \cap C - B\cap C + A\cap B\cap C $$
We further have that $A\cup B\cup C = \Omega = \binom{20}7$.
Since all two element cuts still are difficult to calculate, we apply inclusion-exclusion to them as well: $$A\cup B =A+B - A\cap B $$ We then use that $A\cup B = \binom{20}7-A^C\cap B^C $ and this event is $\binom{20}7 -120$ for $A,B$ and $\binom{20}7$ else.
Finally, we can also calculate $A,B,C$ using their complement.
Substituting in we now have:
$$ A\cup B\cup C=A+B+C - A\cap B - A \cap C - B\cap C + A\cap B\cap C \\ \Leftrightarrow\\ A\cup B\cup C=A+B+C -A-B+A\cup B + A \cup C-A-C +B\cup C - B - C + A\cap B\cap C \\ \Leftrightarrow\\ A\cap B\cap C = -A \cup C + A - A\cup B + A\cup B\cup C + B - B\cup C + C \\ \Leftrightarrow\\ A\cap B\cap C = -\binom{20}7 + A -(\binom{20}7 -120) + \binom{20}7 + B - \binom{20}7 + C \\ \Leftrightarrow\\ A\cap B\cap C = -\binom{20}7 + \binom{20}7-\binom{15}7 -(\binom{20}7 -120) + \binom{20}7 + \binom{20}7-\binom{15}7 - \binom{20}7 + \binom{20}7-\binom{10}7 \\ \Leftrightarrow\\ A\cap B\cap C = 64650 $$
And so we have that $\mathbb{P}(A\cap B\cap C) = 64650/77520$.
If we denote with $\mathcal{H}_{\mathcal{n},(\mathcal{N}_1,...,\mathcal{N}_m)}\left(\{\left(a_1,\ldots,a_m\right)\}\right)$ the hypergeometric distribution on $m$ categories with category $i$ having $\mathcal{N}_i$ elements when we draw exactly $a_i$ elements of category $i$, then we can also brute-force calculate it: $$\begin{align} \mathbb{P}(A\cap B\cap C) &=\sum_{i+j+k = 7\\ i,j,k\ge 1} \mathcal{H}_{\mathcal{7},(5,5,10)}\left(\{\left(i,j,k\right)\}\right) \\ &=\sum_{i=1}^{7-2} \sum_{j=1}^{7-i-1} \mathcal{H}_{\mathcal{7},(5,5,10)}\left(\{\left(i,j,7-i-j\right)\}\right) \end{align}$$